The work done in the spring is calculated to be 2.8 J
What is Hooke's law?Hooke's law states that, the extension of a given material is directly proportional to the applied force as long as the elastic limit is not exceeded . First, we must bear in mind that the material must remain within the elastic limit for us to apply the Hooke's law in solving the problem.
Now;
From Hooke's law;
F = Ke
F = force applied
K = force constant
e = extension
F = W = mg = 2.50 - kg * 9.8 m/s^2 = 24.5 N
K = 24.5 N/ 2.76 * 10^-2
K = 888 N/m
e = F/K
F = W = 1.25 - kg * 9.8 m/s^2 = 12.25 N
e = 12.25 N/ 888 N/m = 0.014 m or 1.4 cm
Work done by an external agent = 1/2 Kx^2
= 0.5 * 888 * (8 * 10^-2)^2
= 2.8 J
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a quantity of gas is contained in a sealed container of fixed volume the temperature of the gas is increased
The average kinetic energy of the gas's molecules directly relates to its temperature. Faster moving particles will more frequently and violently hit the container walls. Because of this, the force acting on the container's walls increases, increasing the pressure.
What happens when the temperature of a gas is increased?We know that temperature is proportional to the average kinetic energy of a sample of gas. The proportionality constant is (2/3)R and R is the gas constant with a value of 0.08206 L atm K-1 mol-1 or 8.3145 J K-1 mol-1.The average kinetic energy and the velocity of the gas particles striking the container walls rise as the temperature rises. As the temperature rises, the pressure must as well since pressure is the force the particles per unit of area exert on the container.The molecules in a gas gain more energy and can move more quickly as it is heated. The pressure will rise, and there will be greater impacts on the container walls. On the other hand, cooling the molecules will cause them to slow down and lower the pressure.To learn more about gas volume and temperature, refer to:
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When the volume of gas increased the KE of gases particles increases and hit the walls of container which increases the pressure on the walls .
What if gas will increase?
The gas expands in a closed container . The molecules strike and hit each other therefore the pressure at the walls increase.
According to ideal gas law , when volume is cnstant the Pressure and temperature both will increase.
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a. A light wave moves through glass (n = 1.5) at an angle of 15°. What angle will it have when it moves from the glass into water (n = 1.33)? (4 points)
b. Draw a ray diagram to locate the image of the arrow, as refracted through the lens shown. Write 2 - 3 sentences describing the type of image and its size relative to the object. What type of mirror could be used to form an image of the same type and size? (8 points)
c. An object is located 65 cm from a concave mirror with a focal length of 45 cm. What is the image distance? Is the image real or virtual? (6 points)
Answer in detail with the correct units and steps to solve. Will mark brainliest.
A wave is a phenomenon that does not cause a permanent displacement in the particles of the medium through which it passes. And it transfers energy from one end of the medium to the other. Examples of waves include light waves, sound waves, water waves, x-rays, radiowaves, etc. Thus the required answers for each part of the question are:
a. The angle that the light wave would have is [tex]7.5^{o}[/tex].
b. The type of mirror that can be used is a plane mirror.
c. The image distance is 146.3 cm.
ii. The image formed by the mirror is a real image.
a. When a ray of light passes from one medium to another, then refraction occurs. The refraction depends on the refractive index of the medium considered.
Thus from Snell's law, we have:
refractive index, n, = [tex]\frac{Sin i}{Sin r}[/tex]
where: i is the angle of incidence, and r is the refracted angle.
Now given that n = 1.5, and i = 15.
Then;
n = [tex]\frac{Sin i}{Sin r}[/tex]
1.5 = [tex]\frac{Sin 15}{Sin r}[/tex]
Sin r = [tex]\frac{0.2588}{1.5}[/tex]
= 0.17253
r = [tex]Sin^{-1}[/tex] 0.17253
= 9.936
r ≅ [tex]10^{o}[/tex]
Since the light wave now moves from the glass into water, the determined refracted angle now becomes its angle of incidence in water. So that;
n = [tex]\frac{Sin i}{Sin r}[/tex]
1.33 = [tex]\frac{Sin 10}{Sin r}[/tex]
Sin r = [tex]\frac{0.17365}{1.33}[/tex]
= 0.1306
r = [tex]Sin^{-1}[/tex] 0.1306
= 7.504
r = [tex]7.5^{o}[/tex]
Therefore, the angle that the light wave would have is [tex]7.5^{o}[/tex].
b. The image formed would be the same size as that of the object. And also the same distance as that of the object to the pole of the lens.
The type of mirror that can be used is a plane mirror.
The ray diagram is attached to this answer.
c. From the mirror formula;
[tex]\frac{1}{f}[/tex] = [tex]\frac{1}{u}[/tex] + [tex]\frac{1}{v}[/tex]
where; f is the focal length of the mirror, u is the object's distance to the mirror, and v is the image's distance to the mirror.
Given; u = 65 cm, and f = 45 cm, then:
[tex]\frac{1}{45}[/tex] = [tex]\frac{1}{65}[/tex] + [tex]\frac{1}{v}[/tex]
[tex]\frac{1}{v}[/tex] = [tex]\frac{1}{45}[/tex] - [tex]\frac{1}{65}[/tex]
[tex]\frac{1}{v}[/tex] = [tex]\frac{20}{2925}[/tex]
v = [tex]\frac{2925}{20}[/tex]
v = 146.25 cm
The image distance is 146.3 cm.
ii. The image formed by the mirror is a real image.
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please help me with this physics question ASAP
Answer:
See below
Explanation:
With switches open, the circuit is a simple series circuit ....the ammeters will have the same readings
V = IR
I = V/R = 5 / (10+5+5) = .25 A
b) With S1 closed 5 ohm and 10 ohm in parallel become = 5 *10 / (5+10) = 3.33 ohm
then the series circuit current becomes
5 v / ( 10 + 3.33 + 5 ) = ammeter 1 = .273 amps
ammeter 2 will get a portion of this ...the smaller resistor will get 2/3 ...the 10 ohm resistor will get 1/3 .273 * 10 / 15 =.182 amps
Which type of wave interaction is shown in the photo?
The wave interaction that is shown in the photo is refraction as light moves from air to water.
What is refraction?Refraction refers to the change in the frequency of a wave and the direction of the wave as it moves from one medium to another. We know that waves makes a body under water to look slightly different than when it is in air.
Thus, the wave interaction that is shown in the photo is refraction as light moves from air to water.
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Two vectors A and B have precisely equal magnitudes. In order for the magnitude of A + B to be larger than the magnitude of A - B by the factor n, what must be the angle between them?
Answer:
[tex]\alpha=arccos[\frac{(a^2+b^2)(n-1)}{2ab(n+1)}].[/tex]
Explanation:
1) for A+B: a²+b²-2abcos(π-α);
2) for A-B: a²+b²-2abcos(α);
3) according to the condition (A+B):(A-B)=n, then
[tex]n=\frac{a^2+b^2-2abcos(\pi-a)}{a^2+b^2-2abcos(a)}; \ = > \ n=\frac{a^2+b^2+2abcos(a)}{a^2+b^2-2abcos(a)}; \ = > \ cos(\alpha)=\frac{(a^2+b^2)(n-1)}{2ab(n-1)}.[/tex]
Define a dipole. Hence, write the expression for calculating the electric moment of a dipole
Answer:
Explanation:
An electric dipole is formed by two point charges +q and −q connected by a vector a. The electric dipole moment is defined as p = qa
Two planets X and Y travel counterclockwise in circular orbits about a star, as seen in the figure.
The radii of their orbits are in the ratio 4:3. At some time, they are aligned, as seen in (a), making a straight line with the star. Five years later, planet X has rotated through 88.0°, as seen in (b). By what angle has planet Y rotated through during this time?
Answer: Two planets X and Y travel counterclockwise in circular orbits about a star, as seen in the figure. The radii of their orbits are in the ratio 4:3. At some time, they are aligned, as seen in (a), making a straight line with the star. Five years later, planet X has rotated through 88.0°, as seen in (b). Then, at an angle 135.48°, the planet Y rotated through during this time.
Explanation: To find the answer, we need to know about the Kepler's third law of planetary motion.
What is Kepler's third law of planetary motion?Kepler's third law of planetary motion states that, the square of the period of revolution is proportional to the cube of the orbital radius of the elliptical path.It can be expressed as,T² ∝ r³
How to solve the problem?We have given with the ratio of the radii of their orbits as,4:3.planet X rotated through an angle of 88°.thus,[tex]\frac{r_1}{r_2}=\frac{4}{5} \\\frac{T_1}{T_2} =(\frac{r_1}{r_2})^{3/2}\\[/tex]
As we know that,[tex]T=\frac{2\pi }{w}[/tex] where, w is the angular velocity.
Angular displacement is the angle swept by the position vector of a particle in a given interval of time.[tex]\alpha[/tex] =wt.
We can rewrite our equation as,[tex]\frac{T_x}{T_y}=\frac{w_y}{w_x}\\thus,\\\frac{w_y}{w_x}=(\frac{r_1}{r_2})^{3/2}[/tex]
We have to find the angle rotated by planet Y during 5 yrs. So, we can rewrite the above equation in terms of angular displacement.[tex]\frac{\alpha _y}{\alpha _x} = (\frac{r_1}{r_2})^{3/2}\\where,\\\alpha _x=\frac{88^0}{5 yrs} because,\\here, angle \beta_x =88^0.\\[/tex]
Thus, the angle rotated by planet Y during 5 yrs will be [tex]\beta _y[/tex] =[tex]\alpha _y=\frac{88}{5yrs} *(\frac{4}{3} )^{3/2}=\frac{135.48^0}{5yrs} .\\thus,\\\beta _y=135.48^0.[/tex]
Thus, we can conclude that the angle rotated by planet Y during 5 yrs will be 135.48 degrees.
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The planet Y then rotated through at this time at an angle of 135.48°.
In order to understand the solution, we must be familiar with Kepler's third law of planetary motion.
What does the third law of planetary motion by Kepler say?According to Kepler's third law of planetary motion, the elliptical path's orbital radius is proportional to the cube of the square of the revolution's period.It can be stated as follows:T² ∝ r³
How can the issue be resolved?The ratio of their orbital radii that we have provided is 4:3.Planet X rotated at an 88° angle. thus,[tex]\frac{R_1}{R_2}=\frac{4}{5} \\\frac{T_1}{T_2}=(\frac{4}{5} ) ^{\frac{3}{2} }[/tex]
As we are aware,[tex]T=\frac{2\pi }{w}[/tex]
where w is the angle of rotation per time.
The angle that a particle's position vector sweeps over in a specific amount of time is known as the angular displacement.[tex]\alpha[/tex]=wt.
Our equation can be rewritten as,[tex]\frac{w_y}{w_x} =(\frac{4}{5} ) ^{\frac{3}{2} }[/tex]
We have to find the angle that planet Y rotated at over the course of five years. Consequently, we can express the equation above in terms of angular displacement.[tex]\frac{\alpha _y}{\alpha _x}=(\frac{4}{3} ) ^{\frac{3}{2} } , where\\\alpha _x=\frac{88}{5yrs} \\[/tex]
So, during a period of five years, planet Y will rotate at an angle,[tex]\alpha _y=\frac{88}{5yrs} *(\frac{4}{3} ) ^{\frac{3}{2} }=\frac{135.48}{5yrs}[/tex]
Thus, we may infer that planet Y will revolve at an angle of 135.48 degrees during the course of five years.
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Derive the following equations of motion
1. v = at
2. s = ut + at²
3. v² = u² + 2as
according to definition of acceleration
a=v-u/t
t=v-u/a(equation 1)
according to the formula of average velocity
v+u/2*s/t
s=v+u/2*t(equation 2)
now putting the value of t in equation 2
s=v+u/2*v-u/a
s=v^2-u^2/2a
v^2=u^2+2as
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Three strings, attached to the sides of a rectangular frame, are tied together by a knot as shown in the figure. The magnitude of the tension in the string labeled C is 56.3 N.
Calculate the magnitude of the tension in the string marked A. (You'll need to get the various positions from the graph. The ends of the strings are exactly on one of the tic marks.)
The magnitude of the tension in the string marked A is 52.5N
Generally, the equation for is mathematically given as
Let's take θ be an angle at A
So, tanθ = 3/8
Let's take α be an angle at B (Below X)
tanα = 5/4
Let's take β be an angle at C (Below x)
tanβ = 1/6
First we take the Horizontal Components
74.9cos(9.46°) = Acos(20.6°) + Bcos(51.3°)
By solving the equation, we get
A = 78.9 - 0.668B … (1)
Now, we take the vertical components
74.9sin(9.46°) + Asin(20.6°) = Bsin(51.3°)
By solving the equation, we get
40.07 = 1.015B
B = 39.5N
By substituting the value of B in equation (1)
A = 78.9 - 0.6668× 39.5
A = 52.5N
Hence, the magnitude of the tension in the string marked A is 52.5N
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A boat moves through the water with two forces acting on it. One is a 1,575-N forward push by the water on the propeller, and the other is a 1,200-N resistive force due to the water around the bow. (Review attachment)
(a) What is the acceleration of the 1,100-kg boat?
______ m/s2
(b) If it starts from rest, how far will the boat move in 20.0 s?
______ m
(c) If it starts from rest, how far will the boat move in 20.0 s?
______ m/s
(a) The acceleration of the 1,100-kg boat is 0.341 m/s².
(b) The distance covered by the boat is 68.2 m.
(c) The speed of the boat is 6.82 m/s.
Acceleration of the boat
Net force on the boat = 1,575 N - 1,200 N = 375 N
F(net) = ma
a = F(net)/m
a = 375/1100
a = 0.341 m/s²
Distance moved in 20 ss = ut + ¹/₂at²
s = 0 + ¹/₂(0.341)(20)²
s = 68.2 m
Speed of the boat in 20 sv = u + at
v = 0 + 0.341(20)
v = 6.82 m/s
Thus, the acceleration of the 1,100-kg boat is 0.341 m/s², the distance covered by the boat is 68.2 m and the speed of the boat is 6.82 m/s.
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When this surgical procedure is used to reduce the risk of stroke it will correct stenosis in the artery the most common cause of this condition is the buildup of plaque that forms in the artery name this procedure
Endarterectomy is used to reduce the risk of stroke and correct the stenosis in the artery.
What is carotid artery stenosis?The primary blood vessels that supply the brain with blood and oxygen are the carotid arteries.
The narrowing of these arteries is referred to as carotid artery disease. Carotid artery stenosis is another name for it. The main factor causing constriction is atherosclerosis.
This fat deposit reduces the blood flow to the brain which cause a stroke.
Carotid endarterectomy is a surgical treatment to remove plaque, an accumulation of fatty deposits that causes a carotid artery to become narrowed.
Hence, an Endarterectomy is used to reduce the risk of stroke, it will correct stenosis in the artery.
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Calculate the speed of a satellite moving in a stable circular orbit about the Earth at a height of 4930 km .
The speed of the satellite moving in a stable circular orbit about the Earth is 5,916.36 m/s.
Speed of the satellite
v = √GM/r
where;
M is mass of EarthG is universal gravitation constantr is distance from center of Earth = Radius of earth + 4930 kmv = √[(6.626 x 10⁻¹¹ x 5.97 x 10²⁴) / ((6371 + 4930) x 10³)]
v = 5,916.36 m/s
Thus, the speed of the satellite moving in a stable circular orbit about the Earth is 5,916.36 m/s.
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The motor of a weed trimmer spins
at 9,000 rpm. The amount of time
required for the motor to reach this
speed would NOT be affected by...
A. the distribution of
trimmer line inside the
spool
B. the mass of the spool
containing the trimmer
line
C. the direction in which
the torque is applied
The speed of the trimmer is not affected by the the distribution of trimmer line inside the spool. Option A
What is a weed trimmer?The weed trimmer is a device that is used to trim the grasses on a lawn or a field. This device has a rotating shaft that does the actual trimming of the grasses.
The speed of the trimmer is not affected by the the distribution of trimmer line inside the spool thus it does not affect the amount of time required to reach 9,000 rpm speed.
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9.80 1. The density of mercury is 13600 kg/m³. What is this value in g/cm³? 2. Find the mass of water which will fit in a large tank measuring 2 m x 1 m x 20 cm. Density of water is 1000 kg/m³ or 1.0 g/cm³. 3. Find the volume of a lump of softwood whose mass is 120 g. Density of softwood is 0.6 gcm-3 or 600 kgm-³. -3
Answer:
1. 13..6 grams per centimeters cubed.
2. Mass = 400kg
3. Volume = 200cm = 2m
Explanation:
1. The conversion for kg/m^3 to g/cm^3 is divide by 1000.
2. [tex]density=\frac{mass}{volume}[/tex]
[tex]1000=\frac{mass}{2 * 1 * 0.2}[/tex]
[tex]1000*0.4=mass[/tex]
[tex]400kg = mass[/tex]
3. [tex]density=\frac{mass}{volume}[/tex]
[tex]0.6=\frac{120}{volume}[/tex]
[tex]volume=\frac{120}{0.6}[/tex]
[tex]volume= 200cm[/tex]
The light beam shown in the figure below makes an angle of = 24.8° with the normal line NN' in the linseed oil. Determine the angles and '. (The refractive index for linseed oil is 1.48.)
The angle of refraction of the light beam is determined as 16.46 ⁰.
Angle of refraction of the light beamn = sin i / sin r
where;'
n is refractive indexi is angle of incidencer is angle of refractionAngle between the ray and the normal = incident angle = 24.8⁰
1.48 = sin (24.8) / (sin r)
sin r = sin (24.8) / (1.48)
sin r = 0.283
r = sin ⁻¹(0.283)
r = 16.46 ⁰
Thus, the angle of refraction of the light beam is determined as 16.46 ⁰.
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What is the fastest possible speed called in our universe and what is the equation for it?
Answer:
The speed of light traveling through a vacuum is exactly 299,792,458 meters (983,571,056 feet) per second. That's about 186,282 miles per second — a universal constant known in equations as "c," or light speed.
[tex]s\frac{d}{t}[/tex]
Explanation:
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Forces always act in ________. A. Solitude B. Pairs C. Unpredictable ways D. Interesting ways
By how much does the gravitational potential energy of a 55- kg pole vaulter change if her center of mass rises about 4.0 m during the jump?
Express your answer to two significant figures and include the appropriate units.
The change in the gravitational potential energy of the pole vaulter is 2,156 J.
Change in the gravitational potential energy
The change in the gravitational potential energy of the pole vaulter is calculated as follows;
ΔP.E = mg(Δh)
where;
m is massΔh is change in heightΔP.E = (55)(9.8)(4)
ΔP.E = 2,156 J
Thus, the change in the gravitational potential energy of the pole vaulter is 2,156 J.
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Given that the acceleration of gravity at the surface of Mars is 0.38 of what it is on Earth, and that Mars' radius is 3400 km , determine the mass of Mars.
The mass of the planet Mar, given the data from the question is 6.45×10²³ Kg
Data obtained from the questionAcceleration due to gravity of Earth = 9.8 m/s²Acceleration due to gravity of Mar (g) = 0.38 × 9.8 = 3.724 m/s²Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²Radius (r) = 3400 Km = 3400 × 1000 = 3400000 mMass (M) =? How to determine the mass of Marg = GM / r²
Cross multiply
GM = gr²
Divide both sides by G
M = gr² / G
M = (3.724 × 3400000²) / 6.67×10¯¹¹
M = 6.45×10²³ Kg
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It's a snowy day and you're pulling a friend along a
level road on a sled. You've both been taking physics,
so she asks what you think the coefficient of friction
between the sled and the snow is. You've been
walking at a steady 1.5 m/s, and the rope pulls up
on the sled at a 35 ° angle. You estimate that the
mass of the sled, with your friend on it, is 57 kg and
that you're pulling with a force of 75 N
The coefficient of friction between the sled and the snow is 0.119.
To find the answer, we need to know about the friction.
How to find the coefficient of friction between the sled and the snow?Whenever a body moves over the surface of another body, a force come into play, which acts parallel to the surface of contact and oppose the relative motion. This opposing force is called friction.To solve the problem, we have to draw the free body diagram of the given system.We have given with the following values,[tex]a=0\\\alpha =35^0\\T=75N\\m=57kg[/tex]
Here, acceleration will be equal to zero, because the velocity is given as constant.
Thus, from the diagram, we can write the balancing equations as follows,[tex]ma=Tcos\alpha -f\\\where\\f=kN\\\N+Tsin\alpha=mg\\Thus,\\N=mg-Tsin\alpha[/tex]
Substituting N in f and f in the equation of ma, then we get,[tex]ma= Tcos\alpha -k(mg-Tsin\alpha )[/tex]
Substituting values, we get the coefficient of friction as,[tex]0=(75*cos35)-k((57*9.8)-(75sin35))\\\\k((57*9.8)-(75sin35))=(75*cos35)\\\\515.6k=61.44\\\\k=\frac{61.44}{515.6}=0.119[/tex]
Thus, we can conclude that, the coefficient of friction between the sled and the snow is 0.119.
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The sled's coefficient of friction with the snow is 0.119.
We must understand the friction in order to choose the solution.
How can I determine the sled and snow's coefficient of friction?A force that works parallel to the surface of contact and opposes the relative motion is present whenever one body moves over the surface of another body. Friction is the name for this opposing force.We must create the given system's free body diagram in order to solve the issue.The values that we have provided are[tex]\alpha =35\\T=75N\\m=57kg\\a=0[/tex]
Because the velocity is specified as constant in this case, the acceleration will be equal to zero.
Consequently, we can express the balancing equations as follows using the diagram:[tex]ma=Tcos\alpha -f\\ where,f=kN\\N+Tsin\alpha =mg\\ thus,\\N=mg-Tsin\alpha[/tex]
When we substitute N for f and f in the equation for ma, we obtain,[tex]ma=Tcos\alpha -k(mg-Tsin\alpha )[/tex]
By substituting values, we obtain the friction coefficient as.[tex]k=0.119[/tex]
As a result, we may say that there is 0.119 coefficient of friction between the sled and the snow.
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Please show work if possible! Thank you!!
A 2.0 x 103 kg roller coaster travels around a vertical 24-m radius loop. If the coaster has a tangential speed of 18 m/s at the lowest point of the loop, what is the normal force that is exerted on the coaster by the track at this point?
a. 5.3 x 10^4 N
b. 4.7 x 10^4 N
c. 3.0 x 10^4 N
d. 2.7 x 10^4 N
B. The normal force that is exerted on the coaster by the track at the lowest point is 4.7 x 10⁴ N.
Normal force exerted on the coaster at the lowest point
Fₙ = mg + mv²/r
where;
m is mass of the coasterv is speed of the coasterr is radius of the pathFₙ = (2,000 x 9.8) + (2,000 x 18²)/24
Fₙ = 46,600 N
Fₙ = 4.7 x 10⁴ N
Thus, the normal force that is exerted on the coaster by the track at the lowest point is 4.7 x 10⁴ N.
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Which of the following is an example of an offensive action?
An example of an offensive action is B. kicking a ball into the goal
Something that is offensive upsets or embarrasses people because it is rude or insulting.
What is a example of an offensive action?Offensive is an arranged campaign or plan of action, normally created by the military or planned to achieve some specific political aim or goal. A military plan to strike is an example of a military offensive. A plan to fight the war on medications is an instance of an offensive.
What are the types of offensive operations?The four types of offensive operations are move to contact, attack, exploitation, and pursuit.
Leaders direct these offensive operations sequentially and in combination to generate highest combat power and destroy the enemy.
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if f=30 m=4 calculate value of m in equation
m=f/a
Answer:
a = 7.5
Explanation:
m = f/a
f = 30
m = 4
Thus;
a = f/m
a = 30/4
a = 7.5
Three strings, attached to the sides of a rectangular frame, are tied together by a knot as shown in the figure. The magnitude of the tension in the string labeled C is 56.3 N.
Calculate the magnitude of the tension in the string marked A.
The magnitude of the tension in the string marked A and B is mathematically given as
A = 52.5 N
What is the magnitude of the tension in the string marked A?Generally, the equation for is mathematically given as
tan=3/8
negative x
[tex]B= tan\phi \\tan \phi=5/4[/tex]
negative x
C= tan=1/6
Hence, considering the Horizontal components
74.9cos(9.46) = A*cos(20.6) + B*cos(51.3)
A = 78.9 - 0.668B
Vertical components
74.9*sin(9.46) + Asin(20.6) = Bsin(51.3)
40.07 = 1.015B
B = 39.5 N
In conclusion, Sub the value of B is the equation of A
A = 78.9 - 0.668B
Sub
A = 78.9 - 0.668( 39.5 N)
A = 52.5 N
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What is the resistance of a carbon rod at 25.8 ∘C if its resistance is 0.0200 Ω at 0.0 ∘C ?
0.02020 ohm is the resistance of a carbon rod at 25.8 ∘C if its resistance is 0.0200 Ω at 0.0 ∘C.
What is a resistor?A resistor is an electrical component that controls or restricts how much electrical current can pass across a circuit in an electronic device. A specified voltage can be supplied via resistors to an active device like a transistor.
The temperature of the resistor varies based on the variation in the temperature. The equation that describes the relationship between the two of them is:
R = R0[1+ alpha(T-T0)] where:
R is the new resistance we are looking for
alpha is the temperature coefficient of resistance. For carbon rod, alpha = ₋ 4.8 x [tex]10^{-4}[/tex](1/°c)
T0 is the standard temperature =25.8°C
R0 is the resistance at T0 = 0.0200 ohms
T is the temperature at which we want to get R = 0
Substitute in the equation to get R as follows:
R = 0.0200 [1+( ₋ 4.8 x [tex]10^{-4}[/tex]) (0-25.8)] = 0.02020 ohm
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A 1500 kg aircraft going 35 m/s collides with a 2000 kg aircraft that is parked and they stick together after the collision and are going 15 m/s after the collision. If they skid for 112.1 m before stopping, how long (in seconds) did they skid? Hint: Are the aircraft moving at a constant velocity after the collision or do they experience an acceleration?
The two aircrafts are skidded for 14.9 s.
To find the answer, we need to know about the Newton's equation of motion.
What is the Newton's equation that relates velocity, distance, acceleration and time?As per Newton's equation of motion
V²-U²= 2aSV= U+atV= final velocity, U = initial velocity, S = distance, a= acceleration, t= timeWhat's the acceleration of the aircrafts that skidded from 35 m/s after the collision for 112.1 m before achieving 15 m/s?Here, U = 35 m/s, V = 15m/s, S= 112.1 mSo, 15²- 35²= 2a×112.1= 224.2a=> a= -1000/224.2= -4.5 m/s²
What's the time taken to stop if these aircrafts are de-accelerated by 4.5 m/s² from 35m/s?Here, U = 35 m/s, V=15 m/s a= -4.5 m/s²35= 15+(-4.5)t=> t= 20/4.5 = 4.4 s
Thus, we can conclude that the two aircrafts are skidded for 4.4 s.
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describe one similarity and one difference between the velocity on the reference circle and the velocity on the pendulum
SHM can be acquired by perpendicular projection of uniform circular motion of a particle on its diameter such a particle is called reference particle and its circular path is called reference circle.
A pendulum reaches its maximum velocity when the block is at its lowest point (the pendulum is vertical and pointing straight down). We can then use the term for conservation of energy to determine the maximum height of the block.
What is the velocity at the bottom of a pendulum?
As the pendulum swings downward, gravity converts this potential energy into kinetic energy, so that at the bottom of the swing, the pendulum bob has zero potential energy, and its kinetic power, (1/2)mv2, equals the initial potential energy (mgh). (So the velocity, v, equals √(2gh).)
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Checking the what will reduce the possibility of having to rebuild or replaced the engine?
Checking the oil will reduce the possibility of having to rebuild or replaced the engine.
what is an engine ?
A device created to transform one or more sources of energy into mechanical energy is known as an engine or motor. Potential energy, heat energy, chemical energy, electric potential, and nuclear energy are all forms of energy that are readily available.
The lubricating function of engine oil is crucial. It shields and stops all the moving parts from rubbing against one another. Metal-on-metal wear would quickly kill your engine without lubrication.
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A certain kind of glass has an index of refraction of 1.640 for blue light of wavelength 440 nm and an index of 1.600 for red light of wavelength 690 nm. If a beam containing these two colors is incident at an angle of 30.0° on a piece of this glass, what is the angle between the two beams inside the glass?
The angle between the two beams inside the glass is mathematically given as
<(BR)= 0.563°
What is the angle between the two beams inside the glass?Generally, the equation for an angle for blue is mathematically given as
[tex]< B= arcsin( sin\theta / i )[/tex]
Therefore
<B= arcsin( sin30 / 1.660 )
<B= 17.53°
For angle for red
<R= arcsin(sin30 / 1.610)
<R= 18.09°
For the angle in between
<(BR)= 18.093 - 17.530
<(BR)= 0.563°
In conclusion, the angle between the two beams inside the glass
<(BR)= 0.563°
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A man pushing a crate of mass
m = 92.0 kg
at a speed of
v = 0.880 m/s
encounters a rough horizontal surface of length
ℓ = 0.65 m
as in the figure below. If the coefficient of kinetic friction between the crate and rough surface is 0.359 and he exerts a constant horizontal force of 299 N on the crate.
(a) Find the magnitude and direction of the net force on the crate while it is on the rough surface.
magnitude _______N
Direction?:
1. Same as the motion of the crate
2. opposite as the motion of the crate
(b) Find the net work done on the crate while it is on the rough surface.
___________J
(c) Find the speed of the crate when it reaches the end of the rough surface.
_________m/s
(a) The magnitude and direction of the net force on the crate while it is on the rough surface is 24.67 N, opposite as the motion of the crate.
(b) The net work done on the crate while it is on the rough surface is -16.04 J.
(c) The speed of the crate when it reaches the end of the rough surface is 0.65 m/s.
Magnitude of net force on the crateF(net) = F - μFf
F(net) = 299 - 0.351(92 x 9.8)
F(net) = -24.67 N
Net work done on the crateW = F(net) x L
W = -24.67 x 0.65
W = - 16.04 J
Acceleration of the cratea = F(net)/m
a = -24.67/92
a = - 0.268 m/s²
Speed of the cratev² = u² + 2as
v² = 0.88² + 2(-0.268)(0.65)
v² = 0.426
v = √0.426
v = 0.65 m/s
Thus, the magnitude and direction of the net force on the crate while it is on the rough surface is 24.67 N, opposite as the motion of the crate.
The net work done on the crate while it is on the rough surface is -16.04 J.
The speed of the crate when it reaches the end of the rough surface is 0.65 m/s.
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