Which of the following accurately describes the behavior of water when subjected to temperature change? Question 9 options: A) The volume of water will decrease if heated from 6°C to 7°C. B) The volume of water will increase if cooled from 3°C to 2°C. C) A mass of water will contract if cooled from 1°C to 0°C. D) A mass of water will expand if heated from 0°C to 2°C.

Answers

Answer 1

The behavior of water when subjected to temperature change is that the volume of water will increase if cooled from 3° to 2°C.

The chemical compound water, which can exist in the gaseous, liquid, and solid phases, is made up of the elements hydrogen and oxygen in the ratio 2: 1 i.e. 2 atoms of hydrogen and 1 atom of oxygen.

In general

Volume of water depends on the temperature and is directly proportional to it.

Thus, as the temperature rises, the molecules of water gain energy and move more quickly, which causes the molecules to spread apart and increase the volume of the liquid.

When water cools, it initially contracts (decreases in volume) until a temperature of about four degrees Celsius (4°C).

But at temperatures below 4.0° C, water undergoes an abnormal expansion that causes its volume to start to rise.

This ability is related to the formation of hexagonal structures, which take up a lot of room and increase the volume of the water, as a result of strong hydrogen bonding between water molecules at a lower temperature.

Hence, the volume of water will increase if cooled from 3° to 2°C

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Related Questions

A battery has an emf of ε = 15.0 V. THe terminal voltage of the battery is Vt = 11.6 V when it is delivering P = 20.0 W of power to an external load resistor R. (a) What is the value of R? (b) What is the internal resistance r of the battery?

Answers

AnswerHM???

Explanation:

I dONT KNOW

what is the mathematical formula associated with newton's 2nd law of motion?​

Answers

Answer:

F= m x a

Explanation:

Force (f) = mass (m) x acceleration (a)

A high-voltage direct-current generating station delivers 10 MW of power at 250 kV to a city, as depicted in Fig. P2.12. The city is represented by resistance RL and each of the two wires of the transmission line between the generating station and the city is represented by resistance RTL. The distance between the two locations is 2000 km and the transmission lines are made of 10 cm diameter copper wire. Determine (a) how much power is consumed by the transmission line and (b) 12 V I0 _

Answers

Answer:

The answer is below

Explanation:

The resistivity of copper is ρ = 1.72 * 10⁻⁸ Ωm, diameter d = 10 cm = 0.1 m

The resistance (R) of transmission line is given as:

Rtl = ρL / A; where ρ = resistivity of copper = 1.72 * 10⁻⁸ Ωm, L = length of transmission line = 2000 km = 2000000 m, A is the area of the wire = πd²/4 = π(0.1)²/4

[tex]R_{tl}=\frac{\rho L}{A}=\frac{1.72*10^{-8}*2000000}{\pi*0.1^2/4}=4.4 \ ohm[/tex]

Power = [tex]\frac{V_L^2}{R_L}[/tex]

Power = 10 MW = 10 * 10⁶ W

[tex]10*10^6=\frac{(250*10^3)^2}{R_L} \\\\R_L=\frac{(250*10^3)^2}{10*10^6} \\\\R_L=6250\ ohm[/tex]

[tex]I_L=\frac{V_L}{R_L} \\\\I_L=\frac{250*10^3}{6250} =40\ A[/tex]

a) Since there are two tranmission lines, the power consumed by the lines is:

[tex]P_{TL}=2*I_L^2*R_{TL}=2*40^2*4.4=14080\ W[/tex]

b) The energy generated by the source = 10 * 10⁶ W + 14080 W = 10014080 W

Fraction used = 10 * 10⁶ / 10014080 * 100% = 99.86%

A crate rests on a flatbed truck which is initially traveling at 13.6 m/s on a level road. The driver applies the brakes and the truck is brought to a halt in a distance of 38.1 m. If the deceleration of the truck is constant, what is the minimum coefficient of friction between the crate and the truck that is required to keep the crate from sliding

Answers

Answer:

0.248

Explanation:

Initial speed u = 13.6

Final speed v = 0

Distance s = 38.1

We have umg = ma

We make u subject of the formula

u = a/g

V² = u² + 2as

a = v²-u²/2s

We substitute the values into the above

a = 0-(13.6)²/2*38.1

a = 184.96/76.2

a = 2.427m/sec

Remember that

u = a/g

u = 2.427/9.8

= 0.2476

This is approximately 0.248

This is the minimum coefficient of friction required to keep the crate from sliding.

a car acceleration from rest to 90km/h in 10 seconds. what is its acceleration in meter per second square?​

Answers

Answer:

2.5 m/s^2

Explanation:

First, convert 90 km/hr into m/s:

90/3.6 = 25 m/s

vf = final velocity = 25 m/s

vi = initial velocity = 0 m/s

t = time = 10 seconds

a = acceleration, unknown

Then, find a using the following equation:

(vf - vi)/t = a

(25 m/s)/10 s = 2.5 m/s^2

a = 2.5 m/s^2

Hope this helps!! :)

Blythe and Geoff compete in a 1.00-km race. Blythe's strategy is to run the first 600 m of the race at a constant speed of 4.10 m/s, and then accelerate to her maximum speed of 7.30 m/s, which takes her 1.00 min, and then finish the race at that speed. Geoff decides to accelerate to his maximum speed of 8.30 m/s at the start of the race and to maintain that speed throughout the rest of the race. It takes Geoff 3.00 min to reach his maximum speed. Assume all accelerations are constant.

Required:
a. Calculate the time of Blythe's run.
b. Calculate the time of Geoff s run.

Answers

Answer:

Explanation:

1.00 km = 1000 m .

Blythe's run : -------

Time taken to run 600 m speed

= distance / speed

T₁= 600 / 4.10 = 146.34 s

Next time T₂ = 1 min = 60 s

acceleration of Blythe

a = (7.30 - 4.10) / 60 = .053 m /s²

displacement during acceleration

= ut + 1/2 at²

= 4.10 x 60 + .5 x .053 x 60²

= 246 + 95.4

= 341.4 m

Rest of the distance to be covered = 1000 - ( 600 + 341.4 )

= 58.9 m

Time taken to cover this distance

T₃= 58.9 / 7.3 = 8.06 s

Total time = T₁ + T₂ + T₃ = 214.4 s

Geoff s run : ---------

initial acceleration during first 3 min

= (8.3 - 0 ) / (3 x 60 )

= .046 m /s²

displacement

s = ut + 1/2 a t²

= 0 + .5 x .046 x ( 3 x 60 )²

= 745.2 m

Rest of the distance of race

= 1000 - 745.2 = 254.8 m

This distance is covered at speed of 8.3 m/s

time taken to cover this distance

T₂ = 254.8 / 8.3

= 30.7 s

Total time taken to complete the race

= 180 + 30.7

= 210.7 s .

PROBLEM 5 (Problem 4-145 in 7th edition) Consider a well-insulated horizontal rigid cylinder that is divided into two compartments by a piston that is free to move but does not allow either gas to leak into the other side. Initially, one side of the piston contains 1 m3 of N2 gas at 500 kPa and 120oC while the other side contains 1 m3 of He gas at 500 kPa and 40oC. Assume the piston is made of 8 kg of copper initially at the average temperature of the two gases on both sides. Now thermal equilibrium is established in the cylinder as a result of heat transfer through the piston. Using constant specific heats at room temperature, determine the final equilibrium temperature in the cylinder. What would your answer be if the piston were not free to move

Answers

Answer:

The answer is "[tex]\bold{83.8^{\circ} \ C}[/tex]".

Explanation:

Formula for calculating the mass in He:

[tex]\to m = \frac{PV}{RT}\\[/tex]

        [tex]= \frac{500 \times 1}{ 2.0769 \times (40 + 273)}\\\\ = \frac{500 }{ 2.0769 \times 313}\\\\ = \frac{500 }{ 650.0697}\\\\= 0.76914 \ Kg[/tex]

Formula for calculating the mass in [tex]N_2[/tex]:

[tex]\to m = \frac{PV}{RT}\\[/tex]

        [tex]= \frac{500 \times 1}{ 0.2968 \times (120+ 273)}\\\\ = \frac{500 }{ 0.2968 \times 393}\\\\ = \frac{500 }{ 116.6424}\\\\= 4.2866\ Kg[/tex]

by using the temperature balancing the equation:

[tex]T' = \frac{mcT (He) + mcT ( N_2 )}{ mc (He) + mc ( N_2)}[/tex]

    [tex]= \frac{0.76914 \times 3.1156 \times 313 + 4.2866 \times 0.743 \times393}{ 0.76914 \times 3.1156 + 4.2866 \times 0.743} \\\\ = 357 \ \ K \approx 83.8^{\circ} \ C[/tex]

A flat screen tv uses 120 watts. How much energy is used up if it is left on for 15 min?
A.) 4j
B.) 15j
C.) 0.67j
D.) 108,000j

Answers

Answer:

d

Explanation:

Four-wheel drive trucks do not stop better on icy
roads than a car. Is what law of motion (Newton's laws)

Answers

It would be Newton’s second law of motion

How do you think that changing the mass of the pendulum bob will affect the period of the pendulum swing?​

Answers

(Mass does not affect the pendulum's swing. The longer the length of string, the farther the pendulum falls; and therefore, the longer the period, or back and forth swing of the pendulum. The greater the amplitude, or angle, the farther the pendulum falls; and therefore, the longer the period.)

Which of the following hydrocarbons are SATURATED hydrocarbons?

I. alkanes II. alkenes III. alkynes IV. cycloalkanes

A. I and IV

B. II and III

C. I and III

D. II and IV

Answers

Answer:

i think c

Explanation:

A mole of a monatomic ideal gas at point 1 (101 kPa, 5 L) is expanded adiabatically until the volume is doubled at point 2. Then it is cooled isochorically until the pressure is 20 kPa at point 3. The gas is now compressed isothermally until its volume is back to 5 L (point 4). Finally, the gas is heated isochorically to return to point 1.
a. Draw the four processes and label the points in the pV plane.
b. Calculate the work done going from 1 to 2.
c. Calculate the pressure and temperature at point 2.
d. Calculate the temperature at point 3.
e. Calculate the temperature and pressure and point 4.
f. Calculate the work done going from from 3 to 4.
g. Calculate the heat flow into the gas going from 3 to 4. g

Answers

Answer:

(a). Check attachment.

(b). 280.305 J.

(c). 31.81 kpa; 38.26K.

(d). 24.05K.

(e). 24.05k; 40kpa.

(f). -138.6J.

Explanation:

(a). Kindly check the attached picture for the diagram showing the four process.

1 - 2 = adiabatic expansion process.

2 - 3 = Isochoric process.

3 - 4 = isothermal process.

4 - 1 = isochoric process.

(b). Recall that the process from 1 to is an adiabatic expansion process.

NB: b = 5/3 for a monoatomic gas.

Then, the workdone = (1/ 1 - 1.66) [ (p1 × v1^b)/ v2^b × v2 - (p1 × v1)].

= ( 1/ 1 - 5/3) [ (101 × 5^5/3) × 10^1 -5/3] - 101 × 5.

Thus, the workdone = 280.305 J.

(c). P2 = P1 × V1^b/ V2^b = 101 × 5^5/3/ 10^5/3 = 31.81 kpa.

T2 = P2 × V2/ R × 1 = 31.81 × 10/ 8.324 = 38.36k.

(d). The process 2 - 3 is an Isochoric process, then;

T3 = T2/P2 × P3 = 38.26/ 31.82 × 20 = 24.05K.

(e). The process 3 - 4 Is an isothermal process. Then, the temperature at 4 will be the same temperature at 3. Tus, we have the temperature; point 3 = point 4 = 24.05k.

The pressure can be determine as below;

P4 = P3 × V3/ V4 = 20 × 10/ 5 = 200/ 5 = 40 kpa.

(f) workdone = xRT ln( v4/v3) = 1 × 8.314 × 24.05 × ln (5/10) = - 138.6 J

If a satellite is orbiting the Earth in elliptical motion, then it will move _______________ (slowest, fastest) when its closest to the Earth. While moving towards the Earth (along the path from D to A) there is a component of force in the __________________ (same, opposite) direction as the motion; this causes the satellite to ___________________ (slow down, speed up). While moving away from the Earth (along the path from A to D) there is a component of force in the _________________ (same, opposite) direction as the motion; this causes the satellite to ___________________ (slow

Answers

Answer:fastest,same,slow down,opposite,slow

Explanation:

A satellite move fastest when its closest to the Earth. The other correct options are same direction, speed up, opposite direction and slow.

Velocity of a satellite around the planet.

If a satellite is orbiting the Earth in elliptical motion, then it will move fastest when its closest to the Earth (based on Kepler's, law).

While moving towards the Earth (along the path from D to A) there is a component of force in the  same direction as the motion; this causes the satellite to speed up.

While moving away from the Earth (along the path from A to D) there is a component of force in the opposite direction as the motion; this causes the satellite to slow.

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Which statement is true of a glass lens that diverges light in air?
A.
It is thick near the center and thin at the edges.
B.
It is thin near the center and thick at the edges.
C.
It is uniformly thick.
D. It is uniformly thin.

Answers

C it is uniformly thick

Answer: it is thin near the center and thick at the edges

Explanation: took the test on Plato :)

A 4.0-g bead carries a charge of 20 μC. The bead is accelerated from rest through a potential difference V, and afterward the bead is moving at 3.0 m/s. What is the magnitude of the potential difference V? *
1 point
900 V
400 V
200 V
400 kV

Answers

Answer:

The magnitude of the potential difference is 900 V.

Explanation:

Given;

mass of the bead, m = 4.0 g = 0.004 kg

charge of the bead, Q = 20 μC = 20 x 10⁻⁶ C

final velocity of the bead, v = 3 m/s

What is the magnitude of the potential difference V?

Apply the principle of conservation of energy;

The electric potential energy at the beginning is equal to kinetic energy of the bead  at the end of the journey.

qV = ¹/₂mv²

[tex]V = \frac{mv^2}{2q} \\\\V = \frac{0.004 \ \times \ (3)^2}{2(20 \times 10^{-6})}\\\\V = 900 \ V[/tex]

Therefore, the magnitude of the potential difference is 900 V.

The magnitude of the potential difference (V) is equal to 900 Volts.

Given the following data:

Mass of bead = 4.0 g to kg = 0.004 kgCharge of bead = 20 μC = [tex]20 \times 10^{-6} \;C[/tex]Final velocity of bead = 3 m/s

To determine the magnitude of the potential difference (V):

How to calculate the potential difference (V).

We would apply the law of conservation of energy, which states that the electric potential energy possessed by the bead at the beginning is equal to the kinetic energy possessed by the bead at the end of the journey:

[tex]qV = \frac{1}{2} mv^2\\\\V = \frac{\frac{1}{2} mv^2}{q} \\\\V = \frac{\frac{1}{2} \times 0.004 \times 3.0^2}{20 \times 10^{-6}} \\\\V = \frac{ 0.002 \times 9}{20 \times 10^{-6}}[/tex]

V = 900 V.

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Two strings with linear densities of 5 g/m are stretched over pulleys, adjusted to have vibrating lengths of 0.50 m, and attached to hanging blocks. The block attached to string 1 has a mass of 20 kg and the block attached to string 2 has a mass of M. Listeners hear a beat frequency of 2 Hz when string 1 is excited at its fundamental frequency and string 2 is excited at its third harmonic. What is one possible value for mass M

Answers

Answer:

2.18 kg

Explanation:

The frequency of a wave in a stretched string f = n/2L√(T/μ) where n = harmonic number, L = length of string, T = tension = mg where m = mass of object on string and g = acceleration due to gravity = 9.8 m/s² and μ = linear density of string.

For string 1, its fundamental frequency f  is when n = 1. So,

f = 1/2L√(T/μ) =  1/2L√(mg/μ)

Now for string 1, L = 0.50 m, m = 20 kg and μ = 5 g/m = 0.005 kg/m

substituting the values of the variables into f, we have

f = 1/2L√(mg/μ)

f = 1/2 × 0.50 m√(20 kg × 9.8 m/s²/0.005 kg/m)

f = 1/1 m√(196 kgm/s²/0.005 kg/m)

f = 1/1 m√(39200 m²/s²)

f = 1/1 m × 197.99 m/s

f = 197.99 /s

f = 197.99 Hz

f ≅ 198 Hz

For string 2, at its third harmonic frequency f'  is when n = 3. So,

f' = 3/2L√(T/μ) =  3/2L√(mg/μ)

Now for string 2, L = 0.50 m, m = M kg and μ = 5 g/m = 0.005 kg/m

substituting the values of the variables into f, we have

f' = 3/2L√(Mg/μ)

f' = 3/2 × 0.50 m√(M × 9.8 m/s²/0.005 kg/m)

f' = 3/1 m√(M1960 m²/s²kg)

f' = 3/1 m√M√(1960 m²/s²kg)

f' = 3/1 m √M × 44.27 m/s√kg

f' = 132.81√M/s√kg

f' = 132.81√M Hz/√kg

Since the frequency of the beat heard is 2 Hz,

f - f' = 2 Hz

So, 198 Hz - 132.81√M Hz/√kg = 2 Hz

132.81√M Hz/√kg = 198 Hz - 2 Hz

132.81√M Hz/√kg = 196 Hz

√M Hz/√kg = 196 Hz/138.81 Hz

√M/√kg = 1.476

squaring both sides,

[√M/√kg] = (1.476)²

M/kg = 2.178

M = 2.178 kg

M ≅ 2.18 kg

how does the strength of the forces that hold the basic particles of a substance together relate to the temperature at which the substance changes state​

Answers

The kinetic energy keeps the molecules apart and moving around, and is a function of the temperature of the substance. ... Increasing the pressure on a substance forces the molecules closer together, which increases the strength of intermolecular forces.

is a step in the scientific method. The step that follows this step involves forming

Answers

Answer: read this hope this helped

Explanation: A hypothesis is a possible explanation for a set of observations or an answer to a scientific question. ... The next step in the scientific method is to test the hypothesis by designing an experiment. This includes creating a list of materials and a procedure— a step-by-step explanation of how to conduct the experiment.

Any change in the cross section of the vocal tract shifts the individual formant frequencies, the direction of the shift depending on just where the change in area falls along the standing wave. Constriction of the vocal tract at a place where the standing wave of a formant exhibits minimum-amplitude pressure oscillations generally causes the formant to drop in frequency; expansion of the tract at those same places raises the frequency. Three other major tools for changing the shape of the tract in such a way that the frequency of a particular formant is shifted in a particular direction are the jaw, the body of the tongue and the tip of the tongue. Moving the various articulatory organs in different ways changes the frequencies of the two lowest formants over a considerable range [18].

One way to increase formant frequency is to ________ the vocal tract at a place where the standing wave of a formant frequency exhibits minimum-amplitude pressure oscillations.

a. Stretch
b. Vibrate
c. Contract
d. Expand

Answers

Answer:

The correct answer is option D.

Explanation:

It is stated in the question that constriction of the vocal tract at a place where the standing wave of a formant exhibits minimum-amplitude pressure oscillations generally causes the formant to drop in frequency so to increase formant frequency, the vocal should expand where the standing wave of a formant exhibits minimum-amplitude pressure oscillations. The answer is D.

I hope this helps.

A block of mass m1 = 19.5 kg slides along a horizontal surface (with friction, μk = 0.35) a distance d = 2.6 m before striking a second block of mass m2 = 8.25 kg. The first block has an initial velocity of v = 6.5 m/s.
(a) Assuming that block one stops after it collides with block two, what is block two's velocity after impact in m/s?
(b) How far does block two travel, d2 in meters, before coming to rest after the collision?

Answers

Answer:

19.5 m/s

87.8 m

Explanation:

The acceleration of block one is:

∑F = ma

-m₁gμ = m₁a

a = -gμ

a = -(9.8 m/s²) (0.22)

a = -2.16 m/s²

The velocity of block one just before the collision is:

v² = v₀² + 2aΔx

v² = (8.25 m/s)² + 2 (-2.16 m/s²) (2.3 m)

v = 7.63 m/s

Momentum is conserved, so the velocity of block two just after the collision is:

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

m₁u₁ = m₂v₂

(18.5 kg) (7.63 m/s) = (7.25 kg) v

v = 19.5 m/s

The acceleration of block two is also -2.16 m/s², so the distance is:

v² = v₀² + 2aΔx

(0 m/s)² = (19.5 m/s)² + 2 (-2.16 m/s²) Δx

Δx = 87.8 m

Explanation:

By using conservation of linear momentum and also by equating work done to kinetic energy,  [tex]V_{2}[/tex]  = 15.36 m/s and [tex]d_{2}[/tex] = 4.32 meters

Parameters given are :

[tex]m_{1}[/tex] = 19.5 kg

friction, μk = 0.35

distance d = 2.6 m

mass [tex]m_{2}[/tex] = 8.25 kg.

initial velocity of [tex]U_{1}[/tex] = 6.5 m/s.

a.) Since we assumed that the block one stops after it collides with block two, the final velocity for block one will be zero. That is, [tex]V_{1}[/tex] = 0 so its final momentum = 0

Let us also assume that block two was initially at rest. Therefore, it initial velocity and its momentum will be equal to zero.

The formula to use will be :

[tex]m_{1}U_{1} = m_{2}V_{2}[/tex]

Substitute all the parameters into the formula above

19.5 x 6.5 = 8.25[tex]V_{2}[/tex]

Make [tex]V_{2}[/tex] the subject of formula

[tex]V_{2}[/tex] = 126.75/8.25

[tex]V_{2}[/tex] = 15.36 m/s

b.) Let us first calculate the work done in by block one.

The K.E = [tex]1/2mU^{2}[/tex]

substitute its mass and velocity into the formula

K.E = 1/2 x 19.5 x [tex]6.5^{2}[/tex]

K.E = 411.94 Joule

The work done = Kinetic energy

But the resultant Force F = force f - friction

where Frictional force = 0.35 x 19.5 x 9.8

Frictional force = 66.89N

Work done will be the product of resultant Force F and the distance travelled

(F - 66.89) x 2.6 = 411.94

F - 66.89 = 411.94/2.6

F - 66.89 = 158.44

F = 225.3 N

The second block will experience the same force which is equal to 225.3N

Find the kinetic energy of the second block.

K.E =  [tex]1/2mV^{2}[/tex]

K.E = 0.5 x 8.25 x 15.36^2

K.E = 973.2

Using The work done = Kinetic energy

225.3[tex]d_{2}[/tex] = 973.2

[tex]d_{2}[/tex] = 973.2/225.3

[tex]d_{2}[/tex] = 4.32 meters

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How do you classify flammable liquid to gas and solid?​

Answers

Answer:

A flammable liquid enjoys the attention of at least three different federal agencies: the DOT in matters of its transportation, OSHA as it might affect workplace safety, and the EPA concerning its cradle-to-grave management.A flammable liquid enjoys the attention of at least three different federal agencies: the DOT in matters of its transportation, OSHA as it might affect workplace safety, and the EPA concerning its cradle-to-grave management.Add to that a gaggle of state, regional, and local authorities; and we feel compelled to begin this blog entry with our favorite caveat: get expert advice before deciding what to do with that rusting drum of stale gasoline out back.

Explanation:

I hope it help you;)


what is a vector quantity?

Answers

Answer:

A quantity that has magnitude and direction. It's usually represented by an arrow whose direction is the same direction is the same as that of the quantity and whose length is proportional to the quantity's magnitude

Vector Quantity

A physical Quantity, which has magnitude, direction and units But must follow the traingle law of vector addition

Example:- Force, velocity acceleration displacement etc.

For a given substance, the molecules
move fastest when the substance is

Answers

Heat makes the molecules move faster

Answer:GAS

Explanation:

Plzz answer this question correctly

Answers

Answer:

by reducing friction.....

Plzz help me with this

Answers

Answer: d

Explanation:

Why are we seeing extremely old light from Canopus instead of light in real-time?

Answers

Answer:

Canopus is more than 300 light years away from earth. This means it takes the light we see more than 300 years to reach us.

An object initially traveling in a straight line with
a speed of 5.0 meters per second is accelerated
at 2.0 meters per second squared for 4.0 seconds.
The total distance traveled by the object in the
4.0 seconds is

Answers

Answer:

We conclude that the total distance traveled by the object in the 4 seconds is 36 m.

Explanation:

Given

Initial velocity u = 5.0 m/sAcceleration a = 2.0 m/s²Time t = 4 s

To determine

The total distance traveled by the object in the  4.0 seconds is

Important Tip:

We can determine the total distance traveled by the object in the  4.0 seconds by using the equation of motion such as

[tex]s=ut+\frac{1}{2}at^2[/tex]

where

s = distanceu = initial velocitya = accelerationt = time

substituting u = 5.0, a = 2, and t = 4 in the formula

[tex]s=ut+\frac{1}{2}at^2[/tex]

[tex]s=\left(5\right)\left(4\right)+\frac{1}{2}\left(2\right)\left(4\right)^2[/tex]

[tex]s=20+16[/tex]

[tex]s=36[/tex] m

Therefore, we conclude that the total distance traveled by the object in the 4 seconds is 36 m.

The total distance traveled by the object is 36 meters.

Given the following data:

Initial velocity = 5.0 m/s Acceleration = 2.0 [tex]m/s^2[/tex] Time = 4.0 seconds.

To find the total distance traveled by the object, we we  would use the second equation of motion.

Mathematically, the second equation of motion is given by the formula;

[tex]S = ut + \frac{1}{2}at^2[/tex]

Where:

S is the total distance traveled.

u is the initial velocity.

a is the acceleration.

t is the time measured in seconds.

Substituting the values into the formula, we have;

[tex]S = 5(4) + \frac{1}{2}(2)(4^2)\\\\S = 20 + 1(16)[/tex]

Total distance, S = 36 meters.

Therefore, the total distance traveled by the object is 36 meters.

Read more: https://brainly.com/question/8898885

please answer correctly

Answers

Answer:

616000 J.

Explanation:

Bi. Determination of the work done.

Force (F) = 220 N

Distance (s) = 2800 m

Workdone (Wd) =?

Workdone is simply defined as the product of force and the distance moved in the direction of the force. Mathematically, is can be expressed as:

Workdone = Force × distance

Wd = F × s

With the above formula, we can obtain the Workdone as follow:

Force (F) = 220 N

Distance (s) = 2800 m

Workdone (Wd) =?

Wd = F × s

Wd = 220 × 2800

Wd = 616000 J.

At what tempreture will the of and oC
be the same​

Answers

Answer:

-40 degrees

To find the temperature when both are equal, we use an old algebra trick and just set �F = �C and solve one of the equations. So the temperature when both the Celsius and Fahrenheit scales are the same is -40 degrees.

Explanation:

Hope it is helpful.....

Answer:

To find the temperature when both are equal, we use an old algebra trick and just set �F = �C and solve one of the equations. So the temperature when both the Celsius and Fahrenheit scales are the same is -40 degrees.

Even though Alice visits the wishing well frequently and always tosses in a coin for good luck, none of her wishes have come true. As a result, she decides to change her strategy and make a more emphatic statement by throwing the coin downward into the well. If the water is 5.43 m below the point of release and she hears the splash 0.85 seconds later, determine the initial speed at which she threw the coin. (Take the speed of sound to be 343 m/s.)

Answers

Answer:

Explanation:

Total time taken = 0.85 s .

Time taken by sound to travel 5.43 m + time taken by coin to fall by 5.43 m = .85

5.43 / 343 + time taken by coin to fall by 5.43 m = .85

time taken by coin to fall by 5.43 m = .85 - 5.43/343 = .834 s

Let the initial velocity of throw of coin = u

displacement of coin s  = 5.43 m

time take to fall t =  .834 s

s = ut + 1/2 gt²

5.43 = u x .834 + .5 x 9.8 x .834²

5.43 = u x .834 + 3.41

u x .834 = 2.02

u = 2.42 m /s .

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