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Rasek [7]
3 years ago
12

A train has a length of 81.1 m and starts from rest with a constant acceleration at time t = 0 s. At this instant, a car just re

aches the end of the train. The car is moving with a constant velocity. At a time t = 11.6 s, the car just reaches the front of the train. Ultimately, however, the train pulls ahead of the car, and at time t = 36.3 s, the car is again at the rear of the train. Find the magnitudes of (a) the car's velocity and (b) the train's acceleration. g
Physics
1 answer:
arlik [135]3 years ago
4 0

Answer: a) vcar= 7 m/s ; b) a train= 0.65 m/s^2

Explanation: By using the kinematic equation for the car and the train we can determine the above values of the car velocity and the acceletarion of the train, respectively.

We have for the car

distance = v car* t, considering the length of train (81.1 m) travel by the car during the first 11.6 s

the v car =  distance/time= 81.1 m/11.6s= 7 m/s

In order to calculate the acceleration we have to use the kinematic equation for the train from the rest

distance train = (a* t^2)/2

distance train : distance travel by the car at constant speed

so distance train= (vcar*36.35)m=421 m

the a traiin= (2* 421 m)/(36s)^2=0.65 m/s^2

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Which of the following statements concerning moral development is not true?
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A proton (mass m = 1.67 × 10-27 kg) is being accelerated along a straight line at 2.50 × 1012 m/s2 in a machine. If the proton h
Veronika [31]

Answer:

(A) Speed will be 44.18\times 10^4m/sec

(b) Change in kinetic energy = 1560\times 10^{-19}      

Explanation:

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Acceleration of the proton a=2.50\times 10^{12}m/sec^2

Initial velocity u = 1.60\times 10^4 m/sec

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(a) From third equation of motion we know that

v^2=u^2+2as

v^2=(1.60\times 10^4)^2+2\times 2.5\times 10^{12}\times 0.039

v=44.18\times 10^4m/sec

(b) Initial kinetic energy KE_I=\frac{1}{2}mv^2=\frac{1}{2}\times 1.67\times 10^{-27}\times (1.6\times 10^4)^2

Final kinetic energy KE_F=\frac{1}{2}mv^2=\frac{1}{2}\times 1.67\times 10^{-27}\times (44.18\times 10^4)^2

So change in kinetic energy \Delta KE=KE_F-KE_I=\frac{1}{2}\times 1.6\times 10^{-27}\times 10^8\times (44.18^2-1.6^2)=1560\times 10^{-19}J

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A cat walks across the kitchen with a constant speed of 0.5 m/s. How long does it take the cat to walk 8 m?
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The temperature at a point (x, y) on a flat metal plate is given by T(x, y) = 54/(7 + x2 + y2), where T is measured in °C and x,
ANTONII [103]

Answer:

dt/dx = -0.373702

dt/dy =  -1.121107

Explanation:

Given data

T(x, y) = 54/(7 + x² + y²)

to find out

rate of change of temperature with respect to distance

solution

we know function

T(x, y) = 54 /( 7 + x² + y²)

so derivative it x and y direction i.e

dt/dx = -54× 2x / (7 +x² + y²)²    .........................1

dt/dy = -54× 2y / (7 + x² + y²)²      .........................2

now put the value point (1,3) as x = 1 and y = 3 in equation 1 and 2

dt/dx = -54× 2(1) / (7 +(1)² + (3)²)²  

dt/dx = -0.373702

and

dt/dy =  -54× 2(3) / (7 + (1)² + (3)²)²

dt/dy =  -1.121107

7 0
2 years ago
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