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3 years ago

The diagram below represents a bicyclist at the top of a hill, with four points labeled W, X, Y, and Z. Assume that

Physics

1 answer:

KiRa [710]3 years ago

8 0

Answer:

Explanation:

Potential energy is energy that is built up and saved. When you are moving and letting out that energy, that is kinetic energy. You would be using the least kinetic energy when you aren't moving, which is at the top of the hill because you move faster the more you go down.

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If a body travels half its total path in the last 1.50 s of its fall from rest, find the total time of its fall (in seconds).

svetoff [14.1K]

Answer:

time to fall is 3.914 seconds

Explanation:

given data

half distance time = 1.50 s

to find out

find the total time of its fall

solution

we consider here s is total distance

so equation of motion for distance

s = ut + 0.5 × at² .........1

here s is distance and u is initial speed that is 0 and a is acceleration due to gravity = 9.8 and t is time

so for last 1.5 sec distance is 0.5 of its distance so equation will be

0.5 s = 0 + 0.5 × (9.8) × ( t - 1.5)² ........................1

and

velocity will be

v = u + at

so velocity v = 0+ 9.8(t-1.5) ..................2

so first we find time

0.5 × (9.8) × ( t - 1.5)² = 9.8(t-1.5) + 0.5 ( 9.8)

solve and we get t

t = 3.37 s

so time to fall is 3.914 seconds

6 0

3 years ago

Since when was the light we see now emanating from the quasar? Note that the distance between the Earth and the quasar is 598 Mp

lakkis [162]

Hi hi hi hi hi hi hi hi hi hi hi

8 0

3 years ago

The quartz crystal used in an electric watch vibrates with a frequency of 32,768 Hz. What is the period of the crystal's motion?

Elan Coil [88]

Answer:

Time period, $T=3.05\times 10^{-5}\ s$

Explanation:

Given that,

The quartz crystal used in an electric watch vibrates with a frequency of 32,768 Hz, f = 32768 Hz

We need to find the period of the crystal's motion. The relationship between the frequency and the time period is given by :

$T=\dfrac{1}{f}$

T is the time period of the crystal's motion.

Time period is given by :

$T=\dfrac{1}{32768}$

$T=3.05\times 10^{-5}\ s$

So, the time period of the crystal's motion is $3.05\times 10^{-5}\ s$ . Hence, this is the required solution.

8 0

3 years ago

Three observers watch a train pull away from a station toward the right of the platform. Observer A is in one of the train’s car

juin [17]

Observer A is moving inside the train

so here observer A will not be able to see the change in position of train as he is standing in the same reference frame

So here as per observer A the train will remain at rest and its not moving at all

Observer B is standing on the platform so here it is a stationary reference frame which is outside the moving body

So here observer B will see the actual motion of train which is moving in forward direction away from the platform

Observer C is inside other train which is moving in opposite direction on parallel track. So as per observer C the train is coming nearer to him at faster speed then the actual speed because they are moving in opposite direction

So the distance between them will decrease at faster rate

Now as per Newton's II law

F = ma

Now if train apply the brakes the net force on it will be opposite to its motion

So we can say

- F = ma

$a = \frac{-F}{m}$

so here acceleration negative will show that train will get slower and its distance with respect to us is now increasing with less rate

It is not affected by the gravity because the gravity will cause the weight of train and this weight is always counterbalanced by normal force on the train

So there is no effect on train motion

5 0

3 years ago

Calculate the speed of the car at each checkpoint by

lana66690 [7]

Answer: The speed at the first quarter checkpoint is 0.74 m/s. The speed at the second quarter checkpoint is 1.40 m/s. The speed at the third quarter checkpoint is 1.61 m/s. The speed at the finish line is 1.89 m/s.

Explanation: I did the assignment and got it correct :)

3 0

3 years ago