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slava [35]
2 years ago
6

A solar-powered car is traveling at constant speed around a circular track. what happens to the centripetal acceleration of the

car if the speed is tripled?
Physics
1 answer:
dimulka [17.4K]2 years ago
4 0
The centripetal acceleration of the car is related to its speed by the relationship
a_c =  \frac{v^2}{r}
where ac is the centripetal acceleration, v is the car's speed and r is the radius of the circular orbit. 

Looking at the formula, we can see that if we triple the speed (v'=3v), the centripetal acceleration becomes
a_c'= \frac{v'^2}{r}= \frac{(3v)^2}{r}= 9\frac{v^2}{r}=9 a_c
so, the centripetal acceleration increases by a factor 9.
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When a car accelerates from a standing start, the crash test dummy appears to be pressed backward into the seat cushion. Which o
Setler [38]

<u>Answer:</u>

Option: D. Gravity is pulling the crash test dummy in the direction the car is moving.

<u>Explanation: </u>

When a car accelerates from a standing start, the crash test dummy appears to be pressed backward into the seat cushion because the gravity is pulling the crash test dummy in the direction the car is moving.  

Basically when the car is starting, the person inside is in static position and the car is going to move. So it is putting a force on the person to move on the same speed. But as the person is sitting static hence gravity is pulling him behind from moving. Hence, The dummy appears to be pressed backward.

7 0
2 years ago
A 250 GeV beam of protons is fired over a distance of 1 km. If the initial size of the wave packet is 1 mm, find its final size
Margarita [4]

Answer:

The final size is approximately equal to the initial size due to a very small relative increase of 1.055\times 10^{- 7} in its size

Solution:

As per the question:

The energy of the proton beam, E = 250 GeV =250\times 10^{9}\times 1.6\times 10^{- 19} = 4\times 10^{- 8} J

Distance covered by photon, d = 1 km = 1000 m

Mass of proton, m_{p} = 1.67\times 10^{- 27} kg

The initial size of the wave packet, \Delta t_{o} = 1 mm = 1\times 10^{- 3} m

Now,

This is relativistic in nature

The rest mass energy associated with the proton is given by:

E = m_{p}c^{2}

E = 1.67\times 10^{- 27}\times (3\times 10^{8})^{2} = 1.503\times 10^{- 10} J

This energy of proton is \simeq 250 GeV

Thus the speed of the proton, v\simeq c

Now, the time taken to cover 1 km = 1000 m of the distance:

T = \frac{1000}{v}

T = \frac{1000}{c} = \frac{1000}{3\times 10^{8}} = 3.34\times 10^{- 6} s

Now, in accordance to the dispersion factor;

\frac{\delta t_{o}}{\Delta t_{o}} = \frac{ht_{o}}{2\pi m_{p}\Delta t_{o}^{2}}

\frac{\delta t_{o}}{\Delta t_{o}} = \frac{6.626\times 10^{- 34}\times 3.34\times 10^{- 6}}{2\pi 1.67\times 10^{- 27}\times (10^{- 3})^{2} = 1.055\times 10^{- 7}

Thus the increase in wave packet's width is relatively quite small.

Hence, we can say that:

\Delta t_{o} = \Delta t

where

\Delta t = final width

3 0
2 years ago
If the cold temperature reservoir of a Carnot engine is held at a constant 306 K, what temperature should the hot reservoir be k
Paraphin [41]
Efficiency η of a Carnot engine is defined to be: 
<span>η = 1 - Tc / Th = (Th - Tc) / Th </span>
<span>where </span>
<span>Tc is the absolute temperature of the cold reservoir, and </span>
<span>Th is the absolute temperature of the hot reservoir. </span>

<span>In this case, given is η=22% and Th - Tc = 75K </span>
<span>Notice that although temperature difference is given in °C it has same numerical value in Kelvins because magnitude of the degree Celsius is exactly equal to that of the Kelvin (the difference between two scales is only in their starting points). </span>

<span>Th = (Th - Tc) / η </span>
<span>Th = 75 / 0.22 = 341 K (rounded to closest number) </span>
<span>Tc = Th - 75 = 266 K </span>

<span>Lower temperature is Tc = 266 K </span>
<span>Higher temperature is Th = 341 K</span>
6 0
2 years ago
Block b rests upon a smooth surface. if the coefficients of static and kinetic friction between a and b are μs = 0.4 and μk = 0.
aliina [53]

Given

Weight of the block A, Wa = 20 lb, weight of block B Wb = 50 lb. Applied force to block A, P = 6lb, coefficient of static friction µs = 0.4, coefficient of kinetic friction µk = 0.3. If a force P is applied to the body, no relative motion will take place until the applied force is equal to the force of friction Ff, which is acting opposite to the direction of motion. Magnitude of static force of friction between block A and block B, Fs = µsN, where N is reaction force acting on block A. Now, resolve the forces Fx = max. P = (mA + mB)a,

 

6 = (20 / 32.2 + 50 / 32.2)a

 

2.173a = 6

 

A = 2.76 ft/s^2

 

To check slipping occurs between block A and block B, consider block A:

P – Ff = mAaA

6 – Ff = 1.71

Ff = 4.29 lb

 

And also,

N = wA. We know static friction,

Fs = µsN

Fs = 0.4 x 20

Fs = 8lb

Frictional force is less than static friction. Ff < Fs

<span>Therefors, acceleration of block A, aA = 2.76 ft/s^2, acceleration of block B aB = 2.76 ft/s^2</span>

6 0
2 years ago
A teacher is demonstrating energy transformation with a basketball. He drops the ball from a fixed height and challenges his stu
Luda [366]

The ball will bounce at a height lower than the height it was dropped.

Answer: Option B.

<u>Explanation:</u>

When a basket ball is thrown from a particular height, it bounces back. But the height it bounces back at is not exactly the same height from where it was thrown.

With further bounces, the energy of the basket ball goes on decreasing and the bounces go on getting smaller. This shows that there is a change in the energy of the basket ball with every bounce that the ball makes. Some energy lost from the ball gets absorbed by the court and some of the energy is changed into thermal energy.

5 0
2 years ago
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