In Milikan’s experiment, a drop of radius of 1.64 μm and density 0.851 g/cm3 is suspended in the lower chamber when a downward-pointing electric field of 1.92 * 105 N/C is applied.

What is the weight of the drop?

Find the charge on the drop, in terms of e.

How many excess or deficit electrons does it have?

Answers

Answer 1

(a) The weight of the drop is 1.54 x 10⁻²⁵ N,

(b) The the charge on the drop, in terms of e is 5 x 10⁻¹²e and

(c) The excess electrons is 5 x 10⁻¹² electron.

Weight of the drop

The weight of the drop is calculated as follows;

Volume of the drop; V = ⁴/₃πr³

V = ⁴/₃π(1.64 x 10⁻⁶)³ = 1.845 x 10⁻¹⁷ m³

mass = density x volume

mass =  0.851 g/cm³ x  1.845 x 10⁻²³ cm³ = 1.572 x 10⁻²³ g =  1.57 x 10⁻²⁶ kg.

Weight = 1.57 x 10⁻²⁶ kg x 9.8 m/s² = 1.54 x 10⁻²⁵ N.

Charge on the drop

q = F/E

q = (1.54 x 10⁻²⁵ N)/(1.92 x 10⁵)

q = 8.01 x 10⁻³¹ C

1.6 x 10⁻¹⁹ C = 1e

8.01 x 10⁻³¹ C = ?

= 5 x 10⁻¹²e

Excess electron on the drop

1.6 x 10⁻¹⁹ C ------- 1 electron

8.01 x 10⁻³¹ C ------- ?

= 5 x 10⁻¹² electron

Thus, the weight of the drop is 1.54 x 10⁻²⁵ N, the the charge on the drop, in terms of e is 5 x 10⁻¹²e and the excess electrons is 5 x 10⁻¹² electron.

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Related Questions

Technetium -99 (half-life = 6.01) is used in the medical imaging. How many half-lives would go by in 44.0h?

Answers

The number of half-lives that would go by in 44 h is 7 half-lives

What is half life?

This is the time taken for half a substance to decay

How to determine the number of half life

The number of half-lives that will elaspe after 44 h can be obtained as illustrated below:

Half-life (t½) = 6.0 hTime (t) = 44 hNumber of half-lives (n) =?

n = t / t½

n = 44 / 6.01

n = 7

Thus, 7 half-lives will elaspe after 44 h

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Consider the following isobaric process for air, modelled as a Calorically Perfect Ideal Gas
(CPIG), from state 1 to state 2. P1 = 200 kPa, T1 = 500 K, T2 = 600 K. Show that the
condition satisfies the second law of thermodynamics. (Show all relevant steps involved).

Answers

These given conditions satisfy the second law of thermodynamics.

As the process is isobaric

So there will be a straight line of P= 200kPa in P-v and P-T planes

P1 = P2 = 100kPa

For perfect ideal gas, v-T plane:

[tex]v = (\frac{R}{P}) T[/tex]

[tex]v_{1} = (\frac{R}{P_{1} }) T_{1}[/tex] = 287 × 500/200000 = 0.717 m³/kg

[tex]v_{2} = (\frac{R}{P_{2} }) T_{2}[/tex] = 287 × 600/200000 = 0.861m³/kg

As it is the calorically perfect gas

de = [tex]c_{v}[/tex]dT

Integration on both sides

e2 - e1 = [tex]c_{v}[/tex](T2 - T1)

           = ( 716.5J/kg/K) (600-500)

           = 71650 J/kg

also,

Tds = de + Pdv

Tds = [tex]c_{v}[/tex]dT +Pdv

For ideal gas

V = RT/P        

dv = Rdt/P - RTdp/P²

Tds = [tex]c_{v}[/tex]dT + Rdt - RTdp/P

ds = ([tex]c_{v}[/tex] + R)dT/T - RdP/P

ds = ([tex]c_{v} + c_{p} -c_{v}[/tex])dT/T - RdP/P

ds = [tex]c_{p}[/tex]dT/T - RdP/P

Integration on both sides

s2 - s1 = [tex]c_{p}[/tex]ln (T2/T1) - R ln (P2/P1)

Since P is constant

s₂ - s₁ = [tex]c_{p}[/tex] ln (T2/T1)

           = 1003.5 ln (600/500)

           = 1003.5 × 0.182

           = 182.95 J/kg/K

w = Pdv

[tex]w_{12}[/tex] = P(v₂ - v₁)

     = 2,00,000 ( 0.861 - 0.717)

     = 28,800 J/kg

de = δq -δw

δq = de + δw

q₁₂ = (e₂ - e₁) +  w₁₂

    =  71,650 + 28,800 = 1,00,450 J/kg

Now in this process, the gas is heated from 500 K to 600 K. We would expect at a minimum that the surroundings were at 600 K.

Let’s check for second law satisfaction.

s₂ - s₁ ≥ q₁₂ / Tₓ

182.95 ≥ 1,00,450 / 600 K

182.95 J/kg/K ≥ 167.41 J/kg/K

Hence this condition satisfies the second law of thermodynamics

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A 29.0 kg beam is attached to a wall with a hi.nge while its far end is supported by a cable such that the beam is horizontal.
If the angle between the beam and the cable is θ = 57.0° what is the vertical component of the force exerted by the hi.nge on the beam?

Answers

The vertical component of the force exerted by the hi.nge on the beam is 114.77 N.

Tension in the cable

Apply the principle of moment and calculate the tension in the cable;

Clockwise torque = TL sinθ

Anticlockwise torque = ¹/₂WL

TL sinθ  =  ¹/₂WL

T sinθ  =  ¹/₂W

T = (W)/(2 sinθ)

T = (29 x 9.8)/(2 x sin57)

T = 169.43 N

Vertical component of the force

T + F = W

F = W - T

F = (9.8 x 29) - 169.43

F = 114.77 N

Thus, the vertical component of the force exerted by the hi.nge on the beam is 114.77 N.

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A baseball player holds a 1.42 N baseball in his hand, a distance of 34.0 cm from the elbow joint as shown in the figure below. The biceps attached at a distance of 2.75 cm from the elbow exerts an upward force of 12.6 N on the forearm. Consider the forearm and hand to be a uniform rod with a mass of 1.20 kg. Calculate the net torque acting on the forearm and hand.

Answers

Answer:

90.3N

Explanation:

⊥mg = (0.170 m)(1.20 kg) 9.81 m/s

τ ball = r⊥Wball = (0.340 m)(1.42 N) = − 0.483 N ⋅m

F − 2.001− 0.483 N ⋅m = 0

F = 2.484 N ⋅m

0.0275 m = 90.3 N

The net torque acting on the forearm and hand is 90.3N

What is torque?

Torque is a measure of the pressure that can motivate an object to rotate about an axis. simply as pressure is what causes an object to accelerate in linear kinematics, torque is what reasons an item to collect angular acceleration. Torque is a vector amount.

⇒mg = (0.170 m)(1.20 kg) 9.81 m/s

⇒torque = rweight of the ball

⇒ (0.340 m)(1.42 N) = − 0.483 N ⋅m

⇒F = − 2.001− 0.483 N ⋅m = 0

⇒F = 2.484 N ⋅m

⇒0.0275 m = 90.3 N

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Glucose solution is administered to a patient in a hospital. The density of the solution is 1.308 kg/l. If the blood pressure in the vein is 35.7 mmHg, then what is the minimum necessary height of the IV bag above the position of the needle?

Answers

The minimum  necessary height of the IV bag above the position of the needle is 0.37 m.

we know that,

P = ρgh

where,

P = 35.7mmHg

  = 4759.609 Pa

g = 9.8[tex]\frac{m}{s^{2} }[/tex]

ρ = 1.308 kg / m^3

now, substituting all the values, we get,

4759.609 = 1.308 × 9.8 × h

h = 0.37 m

The minimum  necessary height of the IV bag above the position of the needle = 0.37 m.

what is an IV bag ?

A reagent, also called as an analytical reagent, is a substance or compound that is added to a system in chemistry to bring about a chemical reaction or examine to see if one happens. Even though the terms "reagent" and "reactant" are frequently used synonymously, "reactant" refers to a substance that is consumed during a chemical reaction.

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Object A has the same mass as Object B but Object A is traveling faster. What can we say about the momentum of Object A compared to that of Object B?

Answers

Answer:

Object A's momentum is larger

Explanation:

as the formula for momentum goes:

P = M * V

where P is momentum, M is mass, V is velocity.

so where Va (Object A's velocity) is larger than Vb (Object B's velocity) we get:

( i ) Va > Vb

as the masses of both objects are equal, we mark:

( ii) Ma = Mb = M

we multiply both sides of ( i ) by M to get:

( iii ) Va × M > Vb × M

and we finally get:

( iv ) Pa > Pb

Answer: Object A's momentum of larger than compared to that of object B.

Explanation:

Planet-X has a mass of 4.74×1024 kg and a radius of 5870 km.
1. What is the Second Cosmic Speed i.e. the minimum speed required for a satellite in order to break free permanently from the planet?
2. If the period of rotation of the planet is 16.6 hours, then what is the radius of the synchronous orbit of a satellite?

Answers

The minimum speed required for a satellite in order to break free permanently from the planet and the radius of the synchronous orbit of a satellite are 7.3 Km/s and 3.1 × 10⁴ km respectively.

To find the answer, we need to know about the escape velocity and time period of revolving satellite.

What's the expression of escape velocity of a satellite?Mathematically, escape velocity= √(2GM/R)G = gravitational constant, M = mass of planet, R= radius of the planetHere, M = 4.74×10²⁴kg, R = 5870 kmEscape velocity=  √(6.67×10^(-11)×4.74×10²⁴/5.870×10⁶)

          = 7.3 Km/s

What's the expression of time period of a circularly orbiting satellite?T= {2π×r^(3/2)}/√(GM)r= (T/2π)⅔× (GM)^(1/3)r is the radius of the orbitWhat's the radius of the circular orbit, if the time period of the satellite is 16.6 hours?T = 16.6 hours = 16.6×3600 second = 59760sr = (59760/2π)^⅔× (6.67×10^(-11)×4.74×10²⁴)^(1/3)

        = 3.1 × 10⁴ km

Thus, we can conclude that the escape velocity and the radius of the synchronous orbit of a satellite are 7.3 Km/s and 3.1 × 10⁴ km respectively.

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Tuning an Instrument. A musician tunes the C-string of her instrumeut to a fundamental frequency of 65.4 Hz. The vibrating portion of the string is 0.600 m long and has a mass of 14.4 g. (a) With what tension must the musician stretch it? (b) What percent increase in tension is needed to increase the frequency from 65.4 Hz to 73.4 Hz, corresponding to a rise in pitch from C to D?

Answers

(a) The tension the musician must stretch it is 147.82 N.

(b) The percent increase in tension is needed to increase the frequency is 26%.

Tension in the string

v = √T/μ

where;

v is speed of the waveT is tensionμ is mass per unit length = 0.0144 kg / 0.6 m = 0.024 kg/m

v = Fλ

in fundamental mode, v = F(2L)

v = 2FL

v = 2 x 65.4 x 0.6 = 78.48 m/s

v = √T/μ

v² = T/μ

T = μv²

T = 0.024 x (78.48)²

T = 147.82 N

When the frequency is 73.4 Hz;

v = 2FL = 2 x 73.4 x 0.6 = 88.08 m/s

T = μv²

T = (0.02)(88.08)²

T = 186.19 N

Increase in the tension

= (186.19 - 147.82)/(147.82)

= 0.26

= 0.26 x 100%

= 26 %

Thus, the tension the musician must stretch it is 147.82 N.

The percent increase in tension is needed to increase the frequency is 26%.

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The decibel level of the sound from a certain hair dryer is measured at 60 dB. Find the intensity of the sound.

Answers

Based on the calculations, the sound intensity level is equal to 1.0 × 10⁻⁵ W/m².

How to determine intensity of the sound?

Mathematically, sound intensity level can be calculated by using this formula:

[tex]\beta = 10 log(\frac{I}{I_{ref}} )[/tex]

Where:

I is the intensity of the sound.

Note: The reference value of sound intensity is equal to 1.0 × 10⁻¹² W/m².

Rewriting the formula, we have:

β/10 = logI - logIo

Substituting the parameters into the formula, we have;

60/10 = logI - log(1.0 × 10⁻¹²)

6 = logI + 12

logI = 6 - 12

logI = -6

I = 1.0 × 10⁻⁵ W/m².

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All moving objects have ____________.
A. Force
B. Distance
C. Momentum
D. Time

Answers

Answer:

I think it's force

Explanation:

Mark as brainliest if it is right

Answer: A

Explanation:

Force can cause a stationary object to start moving or a moving object to change its speed or direction or both

If the velocity of an object is 9 m/s and its momentum is 72 kgm/s, what is its mass

Answers

An object with a velocity (v) of 9 m/s and a linear momentum (p) of 72 kg.m/s, has a mass (m) of 8 kg.

What is momentum?

In Newtonian mechanics, linear momentum, or simply momentum, is the product of the mass and velocity of an object.

It is a vector quantity, possessing a magnitude and a direction.

The mathematical expression for momentum is:

p = m . v

where,

p is the linear momentum of the object.m is the mass of the object.v is the velocity of the object.

An object has a velocity (v) of 9 m/s and its linear momentum (p) is 72 kg.m/s. We will use the definition of linear momentum to calculate the mass of the object.

p = m . v

m = p / v

m = (72 kg.m/s) / (9 m/s) = 8 kg

An object with a velocity (v) of 9 m/s and a linear momentum (p) of 72 kg.m/s, has a mass (m) of 8 kg.

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A 69-kg base runner begins his slide into second base when he is moving at a speed of 3.2 m/s. The coefficient of friction between his clothes and Earth is 0.70. He slides so that his speed is zero just as he reaches the base.
(a) How much mechanical energy is lost due to friction acting on the runner?
J
(b) How far does he slide?
m

Answers

The mechanical energy lost is 353.28 Joules . The distance he slid off is 0.16m.

we know:-

Mass = 69 kg

Speed = 3.2 m/s

Coefficient of friction ( ratio of friction force and normal force ) = 0.70

Acceleration due to gravity, g = 9.8 m/s^2

(a) To determine the amount of mechanical energy that is lost because of friction acting on the runner, we would calculate the change in kinetic energy:

[tex]KE = \frac{1}{2} m (v-u)^{2}[/tex]

      [tex]= \frac{1}{2} 69 ( 3.2-0 )^{2}[/tex]

      [tex]= 353.28[/tex] Joules

Mechanical energy = 353.28 Joules

(b) To determine how far (distance) the runner slide:

acceleration = ug

                     [tex]= 3.2[/tex] × [tex]9.8[/tex]

                     [tex]= 31.36[/tex] [tex]m/s^{2}[/tex]

distance ,

[tex]V^{2} = U^{2} + 2aS[/tex]

[tex]( 3.2)^{2} = 0 + 2 ( 31.36) S[/tex]

[tex]10.24 = 62.72 S[/tex]

[tex]S = {\frac{10.24}{62.72} }[/tex]

Distance, S = 0.16 m  

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What adaptation of a cactus protects it from predators? a round cactus with many spines Broad leaves Sharp spines Thick stems Yellow flowers

Answers

A round cactus with many spines is the adaptation of a cactus that protects it from predators.

A cactus, unlike other plants, has unique adaptations in its roots, leaves, and stems that allow it to flourish in hot and dry surroundings.

The one adaptation that protects the cactus from predators is spines.

A cactus does not have any parts that resemble leaves if you could look at one closely.

Instead, the leaves are transformed into spines, which protrude from the plant's tiny bumps known as areoles.

Herbivores that live in the desert may be enticed to eat the cactus. The spines prevent these predators by modifying leaves into spines.

Other than protection, Spines perform many functions like

1) Since evaporation is a problem in a desert since water is scarce, the spines prevent excessive evaporation.

2)  The spines also impede airflow and prevent evaporation by trapping air.

3) Collecting dew from the early-morning fog is another crucial job that the spines do.

The gathered dew turned into liquid water and ran down to the earth below. The plant then absorbs this water.

Hence, the adaptation of a cactus that protects it from predators is round cactus with many spines.

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An Airbus A380-800 passanger airplane is cruising at constant altitude on a straight line with a constant speed. The total surface area of the two wings is 395 m2. The average speed of the air just below the wings is 259 m/s, and it is 288 m/s just above the surface of the wings.
What is the mass of the airplane? The average density of the air around the airplane is ρ^air = 1.21 kg/m^3.

Answers

The mass of the airplane of area of two wings 395m², and the average speed of lower and upper surface of the wings are, 259 and 288m/s is 386.8×10^3 kg.

To find the answer, we need to know about the Bernoulli's principle.

How to find the mass of the airplane?The Bernoulli's principle states that, the sum of pressure energy, kinetic energy and potential energy of an incompressible, non-viscous, fluid in streamlined flow is a constant.It can be expressed as,

                   P+ [tex]\frac{1}{2}[/tex] ρv²+ρgh=a constant.

Instead of ρ we take d as density.

We have given that,

                    [tex]A= 395 m^2\\v_l=295 m/s\\v_u=288m/s\\density=1.21kg/m^3[/tex]

We equate the principle for lower and upper surfaces of the wing like,

                     [tex]P_1+\frac{1}{2}v_1^2d+dgh_1=P_2+ \frac{1}{2}v_2^2d+dgh_2\\here\\h_1=h_2\\thus\\P_1-P_2=\frac{1}{2}d(v_2^2-v_1^2)\\P_1-P_2=\frac{1}{2}*1.21(288^2-259^2)=9597.12 atm\\[/tex]

Thus, the mass of the airplane from the above equation will be,

                       [tex]\frac{F}{A}=9597.12 atm\\\\ F=9597.12*395m^2=37.9*10^5 N\\\\mg=37.9*10^5 N\\\\m=\frac{37.9*10^5 N}{9.8}\\\\ m=386.8*10^3kg[/tex]

Thus, we can conclude that, the mass of the airplane is 386.8×10^3 kg.

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The mass of an airplane with two wings that are 395m2 in size and average wing surface speeds of 259 and 288 m/s is 387x 10^3 kg.

We need to be aware of the Bernoulli principle in order to determine the solution.

How can I determine an airplane's mass?According to the Bernoulli's principle, the total amount of pressure energy, kinetic energy, and potential energy in a streamlined flow of an incompressible, non-viscous fluid is constant.It can be stated as follows:

                             P+ [tex]\frac{1}{2}[/tex]dv²+dgh = constant.

We substitute d for to represent density.

We've done that,

                           [tex]v_1=259m/s\\v_2=288m/s\\A=395m^2\\d=1.21kg/m^3[/tex]

We compare the governing idea for the wing's bottom and upper surfaces to:

                            [tex]P_1+\frac{1}{2}dv_1^2+dgh= P_2+\frac{1}{2}dv_2^2+dgh\\\\P_1+\frac{1}{2}dv_1^2=P_2+\frac{1}{2}dv_2^2\\\\P_1-P_2=\frac{1}{2}d(v_2^2-v_1^2)\\\\P_1-P_2=9597 atm[/tex]

Consequently, using the aforementioned equation, the airplane's mass will be,

                        [tex]F/A= 9597 atm\\mg=9597*395 =38*10^5N\\m=\frac{38*10^5}{9.8} = 387*10^3kg[/tex]

Consequently, we can say that the mass of an airplane with two wings that are 395m2 in size and average wing surface speeds of 259 and 288 m/s is 387 x 10^3 kg.

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An action force is 50 N to the left. The reaction force must be:
A. 50 N right
B. 50 N down
C. 50 N left
D. 50 N up

Answers

According to Newton's Third Law:"For every action,there exists an equal and opposite reaction"

So an action force of 50 N to the left, results in a 50 N reaction force to the right.

Option A

What is the symbol for a variable resistor?
A. Horizontal wire
B. 2 parallel wires
C. Line with a zig zag pattern with a diagonal line drawn across it
D. Circle with an X through it

Answers

The symbol for a variable resistor is a line with a zig zag pattern with a diagonal line drawn across it.

What is a variable resistor ?

A resistor whose electric resistance value may be altered is known to as a variable resistor. A variable resistor, which works usually by sliding a contact (wiper) over a variable resistor, is basically an electro-mechanical transducer.

The flow of electrical current is restricted by a resistor. The resistivity of a fixed resistor is steady. By adjusting a slider's position, the resistance of this resistor can be altered. Some volume controls and dimmer switches employ variable resistors.

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A contestant in a winter games event pulls a 36.0 kg block of ice across a frozen lake with a rope over his shoulder as shown in Figure 4.29(b). The coefficient of static friction is 0.1 and the coefficient of kinetic friction is 0.03.

Figure 4.29
(a) Calculate the minimum force F he must exert to get the block moving.
40.873

(b) What is its acceleration once it starts to move, if that force is maintained?


m/s2

Answers

(a) The minimum force F he must exert to get the block moving is 38.9 N.

(b) The acceleration of the block is 0.79 m/s².

Minimum force to be applied

The minimum force F he must exert to get the block moving is calculated as follows;

Fcosθ = μ(s)Fₙ

Fcosθ = μ(s)mg

where;

μ(s) is coefficient of static frictionm is mass of the blockg is acceleration due to gravity

F = [0.1(36)(9.8)] / [(cos(25)]

F = 38.9 N

Acceleration of the block

F(net) = 38.9 - (0.03 x 36 x 9.8) = 28.32

a = F(net)/m

a = 28.32/36

a = 0.79 m/s²

Thus, the minimum force F he must exert to get the block moving is 38.9 N.

The acceleration of the block is 0.79 m/s².

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Glucose solution is administered to a patient in a hospital. The density of the solution is 1.308 kg/l. If the blood pressure in the vein is 35.7 mmHg, then what is the minimum necessary height of the IV bag above the position of the needle?

Answers

The minimum necessary height of the IV bag above the position of the needle, if the density of the solution is 1.308kg/L, and the pressure in the vein is 35.7 mmHg is, 2.78m.

To find the answer, we need to know more about the pressure exerted by a liquid column.

How to find the height of IV bag above the position of needle?Consider a liquid of density ρ contained in a vessel of height h, the pressure exerted by the liquid column at the bottom of the vessel is given by ,

                                                [tex]P=[/tex] ρgh

In our question, it is given that,

                                           [tex]density=1.308 kg/L\\\\P=35.7 mmHg.\\[/tex]

Thus, the height of the bag h will be,

                                       [tex]h=\frac{P}{density*g} =\frac{35.7}{1.308*9.8}\\\\ h=2.78 m[/tex]

Thus, we can conclude that, the height of the Iv bag is 2.78 m above the position of needle.

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If the solution has a density of 1.308 kg/L and the vein pressure is 35.7 mmHg, the minimum height of the IV bag above the needle position is 2.78 m.

We need to learn more about the pressure that a liquid column exerts in order to determine the solution.

How can I determine the height of IV bag above where the needle is?The pressure exerted by the liquid column at the bottom of the vessel is given by for a liquid with density enclosed in a vessel of height h.

                            [tex]P=[/tex] ρgh

In our inquiry, it is assumed that,

                        [tex]Density=1.308kg/L\\P=35.7mmHg[/tex]

As a result, the bag's height will be,

                           [tex]h=\frac{P}{density*g}\\\\ h=2.78m[/tex]

As a result, we may say that the I.V. bag is 2.78 m above the ground.

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Two bullets are fired at the same time with the same kinetic energy.
If one bullet has twice the mass of the other, what is the ratio of the speed of the lighter bullet to the speed of the heavier?
Which can do the most work?

Answers

The ratio of the speed of the lighter bullet to the speed of the heavier will be four times the heavier bullet.

Both can do same amount of work.

What is kinetic energy?

Kinetic energy of a body is the energy due to the motion of the body.

Kinetic energy = mv²/2

where m is mass and v is velocity of the object.

Since both objects have the same kinetic energy but one bullet has twice the mass of the other, the ratio of the speed of the lighter bullet to the speed of the heavier will be four times the heavier bullet.

The kinetic energy of both bullets is the same. Hence, they can do equal amount of work.

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A moonshiner makes the error of filling a glass jar to the brim and capping it tightly. The moonshine expands more than the glass when it warms up, in such a way that the volume increases by 0.6% (that is, ΔV/V0 = 6 ✕ 10-3) relative to the space available. Calculate the force exerted by the moonshine per square centimeter if the bulk modulus is 2.1 ✕ 109 N/m2, assuming the jar does not break.

Answers

The force exerted per square centimeter is 126 N/cm².

What is pressure?

Pressure is the force acting per unit area.

Pressure = force/area

Based on the data given:

volume increase, ΔV/V0 = 6 * 10⁻³

Bulk Modulus, B = 2.1 * 10⁹ N/m²

Bulk modulus B of a material is ratio of change in pressure and change in volume as given below:

B = ΔP/ [(ΔV/V)]

Solving for ΔP;

ΔP = B * [(ΔV/V)]

ΔP = (2.1 * 10⁹ N/m²) * (6 * 10⁻³)

ΔP = 1.26 * 10⁶ N/m²

Converting to per square centimeter

ΔP = (1.26 * 10⁶ N/m²)/10⁴

ΔP = 126 N/cm²

In conclusion, the force exerted per square centimeter is a measure of the pressure.

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A gas is confined to a vertical cylinder by a piston of mass 2 kg and radius 1 cm. When 5J of heat are added, the piston rises by 2.4 cm. Find: (a) the work done by the gas; (b) the change in its internal energy. Atmospheric pressure is 105Pa

Answers

The work done by gas is 0.753 J and change in internal energy is 4.247J

So we are given that mass is 2kg , radius 1 cm and the amount of heat is 5 cm

The piston raised by 2.4cm

As we know that Work done is PΔV

Where ΔV is change in volume

Therefore ΔV =  πr^2 h = π x (.01)^2 x .024 =7.53×10^(-6)m^3

Here pressure is 10^5 pa

So W = [tex]10^5\times7.53\times10^-(6)[/tex]

Therefore W = 0.753 J

Now coming to change in internal energy

Change in Internal Energy = Heat Added - Energy lost in work

5J - 0.753 J = 4.247J

Hence the change in internal energy is 4.247 J

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A treasure chest full of silver and gold coins is being lifted from a pirate ship to the shore using two ropes as shown in the figure. The mass of the treasure chest is 75.6 kg.

Tension in rope A:

7.42×102 N

Tension in rope B:

7.52×102 N

What is the tension in rope C?

Answers

Tension in the rope C is 1.24× 10² N.

To find the answer, we need to know about the horizontal component of tension in the rope B.

What's the angle made by the rope B by horizontal?From the figure of the answer, in the triangle PQR, tan(θ)= PQ/QR = 6/1= 6 θ= tan inverse of 6 = 80.5°

What's the horizontal component of the tension in rope B?

Horizontal component= tension in rope B × cos80.5°

= 7.52×10² N × cos80.5°

= 1.24×10² N

What's the tension in the rope C?From the figure, we have found that the tension in rope C = horizontal component of the tension in rope BSo, tension in rope C= 1.24×10² N

Thus, we can conclude that the tension in the rope C is 1.24× 10² N.

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State the newton's law of motion and give
application of each law.

Answers

Answer:

In the first law, an object will not change its motion unless a force acts on it. In the second law, the force on an object is equal to its mass times its acceleration. In the third law, when two objects interact, they apply forces to each other of equal magnitude and opposite direction.

Explanation:

What force causes a bike to move forward?
A. Air resistance
B. Thrust
C. Friction
D. Gravity

Answers

The Answer is Option C. Friction.

Explanation:

The friction force acts in the forward direction on the rear wheel and it acts in the backward direction on the front wheel. The magnitude of friction force on the rear wheel can be more, equal or less than that on the front wheel.

Calculate the magnitude of the electric field at one corner of a square 1.10 m on a side if the other three corners are occupied by 4.05×10−6 C charges. What is the direction of the electric field?

Answers

The direction of magnetic field is south-east and the magnitude is

[tex]23.66[/tex] × [tex]10^{3} N/C[/tex].

Here, the magnitude of CD and BC will be cancelled, as they both are in the opposite direction and equal to each other.
the magnitude, towards the diagonal AC will result in CP, which is the direction of the electric field.


magnitude of electric field can be defined as :- The force per charge on the test charge is a straight forward way to define the size of the electric field.

To find the magnitude of the electric field use the formula

[tex]E = kq/ r^{2}[/tex]

inserting the values,

[tex]E = 9. 10^{9}[/tex] × [tex]4.05[/tex] × [tex]10^{-6} / 1.1 \sqrt{2}[/tex]

[tex]E= 36.45[/tex] × [tex]10^{3}[/tex] / [tex]1.54[/tex]

[tex]E = 23.66[/tex] × [tex]10^{3}[/tex] N/C

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Why should you use an iron hammer instead of a wooden
hammer?

Answers

Answer:

more durability and easier to pound a hammer with the heavy weight.

Explanation:

One should use an iron hammer instead of the wooden hammer because the force developed with the help of the iron hammer is far greater than the wooden hammer

What is Newton's second law?

Newton's Second Law states that The resultant force acting on an object is proportional to the rate of change of momentum.

The mathematical expression for Newton's second law is as follows

F = m*a

where F represents the force applied

m is the mass of the object

a is the acceleration of the object

As we know that force is a product of mass and acceleration, let us assume that both iron and wooden hammers have the same dimension , considering the acceleration applied is the same while hammering ,

As for the same dimension the mass of iron hammer would be greater as result force applied through the iron hammer would be grater

Since the iron hammer's force is so much stronger than the wooden hammer's force, one should use an iron hammer instead of a wooden hammer.

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An air-filled 39.1 μH solenoid has a length 4.0 cm and a cross-sectional area 0.60 cm2. How many turns are in this solenoid?
a.21000
b.144
c.12
d.120
e.1200

Answers

No. of turns in the solenoid is an option (b) 144.

The self-inductance of a long solenoid depends only on its physical properties (such as the number of turns of wire per unit length and the volume), and not on the magnetic field or the current.

Self-inductance of solenoid = 39.1 μH

                                           = 39.1 × [tex]10^-^6[/tex] H

Length of the solenoid = 4.0 cm

Cross-sectional area = 0.60 cm²

Expression for the self-inductance of a coil ;

L = µ₀N²A / [tex]l[/tex]

where,

L = Self- Inductance

N = No. of turns.

A = Cross-sectional area

[tex]l=[/tex] Length of the solenoid

L =( 4π × [tex]10^-^7[/tex] × N² × 0.60 ) / 4.0

39.1× 4.0 / 4π × [tex]10^-^7[/tex] × 0.60 = N²

N² = 2.07 × [tex]10^6[/tex]

N = 144

Therefore, the no. of turns of the solenoid is 144.

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Drag each label to the correct location on the image. Jessica is visiting a park with her mother. Jessica sits on a swing. Her mother pulls the swing to a height of 3 meters above the ground and lets it go. The image shows Jessica at three positions on the swing. Jessica‘s mass is 44 kilograms and the maximum velocity of the swing is 5 meters/second. What’s the energy she has at each position shown? Ignore friction and air resistance. Use g = 9.8 m/s2, PE = m × g × h, and KE = 0 joules KE = 550 joules PE = 862.4 joules

Answers

The  energy she has at each position shown are:

Position at maximum height -- 1,294 JPosition at minimum height --0 JPosition at maximum velocity - 550 JPosition at minimum velocity - 0 JWhat does velocity implies?

Velocity is known to be a term that connote the direction of any kind of a moving body or an object.

Note that the Speed is known to be a a scalar quantity and as such,  Velocity is said to be a vector quantity.

Note also that from the question given,  Jessica's height of the swing 3 meters above the ground, therefore:

Jessica Position at maximum height :

PE = mgh

PE = 44kg x  9.8m/s² x 3

PE = 1,294 J

Jessica  Position at minimum height:

PE = 0 J

Jessica  Position at maximum velocity:

KE = 1/2 x mv²

KE = 1/2 x 44kg x (5m/s²)²

KE = 550 J

Jessica  Position at minimum velocity:

KE = 0 J

Therefore, The  energy she has at each position shown are:

Position at maximum height -- 1,294 JPosition at minimum height --0 JPosition at maximum velocity - 550 JPosition at minimum velocity - 0 J

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Light is incident normally on the short face of a 30∘−60∘−90∘ prism (Figure 1). A drop of liquid is placed on the hypotenuse of the prism.
a) If the index of the prism is 1.50, find the maximum index that the liquid may have for the light to be totally reflected.
Express your answer using three significant figures.

Answers

1.06 is the maximum refractive index that the liquid may have for the light to be totally reflected.

Only when a light source passes from a denser to a rarer medium can it completely reflect.

When the angle of incidence surpasses a specific critical value, specular reflection occurs in the more highly refractive of the two mediums at their interface, and this reflection is known as total reflection.

sin [tex]i_{c}[/tex] = μ[tex]_{r}[/tex] / μ[tex]_{d[/tex]

From the diagram

Angle of incidence = 60°

sin60°  ≥ sin[tex]i_{c}[/tex] = μ[tex]_{r}[/tex]/μ[tex]_{d}[/tex]

μ[tex]_{r}[/tex] ≤ μ[tex]_{d}[/tex] sin60°

μ[tex]_{r}[/tex] ≤ √1.5 × √3/2

   = 1.06

Hence, the maximum index that the liquid may have for the light to be totally reflected is 1.06

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A 40.0 kg beam is attached to a wall with a hi.nge and its far end is supported by a cable. The angle between the beam and the cable is 90°. If the beam is inclined at an angle of θ = 31.0° with respect to horizontal.
The horizontal component of the force exerted by the hi.nge on the beam = 8.662×101 N
What is the magnitude of the force that the beam exerts on the hi.nge?

Answers

The magnitude of the force that the beam exerts on the hi.nge will be,261.12N.

To find the answer, we need to know about the tension.

How to find the magnitude of the force that the beam exerts on the hi.nge?Let's draw the free body diagram of the system using the given data.From the diagram, we have to find the magnitude of the force that the beam exerts on the hi.nge.For that, it is given that the horizontal component of force is equal to the 86.62N, which is same as that of the horizontal component of normal reaction that exerts by the beam on the hi.nge.

                           [tex]N_x=86.62N[/tex]

We have to find the vertical component of normal reaction that exerts by the beam on the hi.nge. For this, we have to equate the total force in the vertical direction.

                           [tex]N_y=F_V=mg-Tsin59\\[/tex]

To find Ny, we need to find the tension T.For this, we can equate the net horizontal force.

                           [tex]F_H=N_x=Tcos59\\\\T=\frac{F_H}{cos59} =\frac{86.62}{0.51}= 169.84N[/tex]

Thus, the vertical component of normal reaction that exerts by the beam on the hi.nge become,

                    [tex]N_y= (40*9.8)-(169.8*sin59)=246.4N[/tex]

Thus, the magnitude of the force that the beam exerts on the hi.nge will be,

                 [tex]N=\sqrt{N_x^2+N_y^2} =\sqrt{(86.62)^2+(246.4)^2}=261.12N[/tex]

Thus, we can conclude that, the magnitude of the force that the beam exerts on the hi.nge is 261.12N.

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The hi.nge will be subjected to a force of 261.12N from the beam.

We must understand the tension in order to choose the solution.

How can the amount of force the beam applies on the height be determined?Let's use the provided information to create the system's free body diagram.We need to calculate the force the beam is exerting on the height using the diagram.For this, it is assumed that the horizontal component of force is 86.62N, the same as the horizontal component of the normal reaction that the beam exerts on the height.We need to identify the vertical component of the normal reaction the beam exerts on the height. We must equalize the total force acting in the vertical direction to achieve this.

                       [tex]N_y=F_v=mg-Tsin59[/tex]

Finding the tension T is necessary to determine Ny. Thus, we can use the net horizontal force to equate this.

                         [tex]F_H=N_x=Tcos59\\T=\frac{F_H}{cos59} =169.84N[/tex]

As a result, the normal reaction that the beam has on the height becomes, with a vertical component,

                  [tex]N_y=(40*9.8)-(169.84*sin59)=246.4N[/tex]

As a result, the force the beam applies on the height will be of the order of,

                        [tex]N=\sqrt{N_x^2+N_y^2} =261.12N[/tex]

Thus, we can infer that the force the beam applies to the height is 261.12N in size.

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