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TiliK225 [7]
3 years ago
11

a train engineer started the train from a standstill and sped up to 5 meters per second, she then rounded a corner at a constant

speed. She then sped the train up to 10 meters per second on the next straightaway. Next, the train chugged around another bend at 10 m/sec, and then its speed slowed to 5 meters per second after coming out of the bend. Finally, the train reached the station, and the engineer applied the brakes and brought the engine to a stop. How many times did the train accelerate? Once a carousel is in motion, the constant movement of the carousel horse around the center of the circle can best be described as
Physics
1 answer:
olasank [31]3 years ago
3 0

for acceleration we can define that rate of change in velocity is know as acceleration

So whenever velocity of train is changing with time we can say train is accelerating

Now here if initially train is standstill then after some time its speed is 5 m/s

so here the train is accelerated first time

Then on straight path its speed changed from 5 m/s to 10 m/s so here train gets accelerated second time

After this train chugged around a curve with same speed 10 m/s

SO here since train is moving in curve so here its direction of velocity is continuously changing and this type of acceleration is known as centripetal acceleration

SO this is accelerated Third time

Then its speed decreases and it comes to speed of 5 m/s from 10 m/s

So here it is acceleration of train for Fourth time

Then finally train comes to stop so again its speed changed from 5 m/s to 0

so this is acceleration of train Fifth time

So total train will accelerate 5 times in whole path

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Answer:

B. 17m/s

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In this case, after 3 seconds, the speed of the car will be 17 m/s. This inference was got by tracing the position of 3s in the x-axis to the value on the y-axis. Doing this, the best inference for the speed of the car after 3 seconds is 17m/s.

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2 years ago
Sedimentary rock turns into magnum through which process
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3 years ago
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3 0
3 years ago
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The work done by an external force to move a -6.70 μc charge from point a to point b is 1.20×10−3 j .
ASHA 777 [7]

Answer:

108.7 V

Explanation:

Two forces are acting on the particle:

- The external force, whose work is W=1.20 \cdot 10^{-3}J

- The force of the electric field, whose work is equal to the change in electric potential energy of the charge: W_e=q\Delta V

where

q is the charge

\Delta V is the potential difference

The variation of kinetic energy of the charge is equal to the sum of the work done by the two forces:

K_f - K_i = W + W_e = W+q\Delta V

and since the charge starts from rest, K_i = 0, so the formula becomes

K_f = W+q\Delta V

In this problem, we have

W=1.20 \cdot 10^{-3}J is the work done by the external force

q=-6.70 \mu C=-6.7\cdot 10^{-6}C is the charge

K_f = 4.72\cdot 10^{-4}J is the final kinetic energy

Solving the formula for \Delta V, we find

\Delta V=\frac{K_f-W}{q}=\frac{4.72\cdot 10^{-4}J-1.2\cdot 10^{-3} J}{-6.7\cdot 10^{-6}C}=108.7 V

4 0
3 years ago
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The parking brake on a 1200kg automobile has broken, and the vehicle has reached a momentum of 7800kg.M/s. What is the velocity
AysviL [449]
Ok so the equation for momentum is:
v=p/m

So you would do:
7800/1200=6.5

So the answer is:
6.5 m/s

Hope this helps :)
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