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3 years ago

10

Find the potential energy associated with a 69-kg hiker atop new hampshire's mount washington, 1900 m above sea level. take the

zero of potential energy at sea level.

1 answer:

Art [367]3 years ago

4 0

Eg =mgh
Eg =(69)(9.8)(1900)
Eg = 1284780J

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Problem: The frequency of an FM radio station is 89.3 MHz. Calculate its period. Part B: From the Library, select the general eq

vekshin1

Answer:

Time period, $T=1.11\times 10^{-8}\ s$

Explanation:

We have,

The frequency of an FM radio station is 89.3 MHz.

It is required to find the period of the wave.

The reciprocal of frequency is called time period of a wave. It can be given by :

$T=\dfrac{1}{f}\\\\T=\dfrac{1}{89.3\times 10^{6}\ Hz}\\\\T=1.11\times 10^{-8}\ s$

So, the period of the wave is $1.11\times 10^{-8}\ s$ .

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3 years ago

Just wondering if I did this right

Kisachek [45]

Yeah

All they are all correct

5 0

3 years ago

Whose geocentric model was accepted for 1400 years

zysi [14]

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8 0

4 years ago

In a game of pool, a cue ball rolls without slipping toward the stationary eight ball with a momentum of 0.23 kg. After the two

IrinaVladis [17]

This question involves the concepts of the law of conservation of momentum.

The magnitude of the final momentum of the eight ball is "0.22 N.s".

According to the law of conservation of momentum:

$P_{i1}+P_{i2}=P_{f1}+P_{f2}$

where,

$P_{i1}$ = initial momentum of the cue ball = 0.23 N.s

$P_{i2}$ = initial momentum of the eight ball = 0 N.s (since ball is initially at rest)

$P_{f1}$ = final momentum of the cue ball = 0.01 N.s

$P_{f2}$ = final momentum of the eight ball = ?

Therefore,

$0.23\ N.s + 0\N.s = 0.01\ N.s+P_{f2}\\\\P_{f2} = 0.22\ N.s$

Learn more about the law of conservation of momentum here:

brainly.com/question/1113396?referrer=searchResults

3 0

3 years ago

Starting at 1.0 m/s, a cheetah runs with a constant acceleration for 4.8 s reaching a speed of 28 m/s. What is the acceleration

LenaWriter [7]

Answer:

c. $5.6m/s^2$

Explanation:

Initial velocity of cheetah,u=1 m/s

Time taken by cheetah =4.8 s

Final velocity of cheetah,v=28 m/s

We have to find the acceleration of this cheetah.

We know that

Acceleration, $a=\frac{v-u}{t}$

Where v=Final velocity of object

u=Initial velocity of object

t=Time taken by object

Using the formula

Then, we get

Acceleration, a= $\frac{28-1}{4.8}=\frac{27}{4.8} m/s^2$

Acceleration= $a=5.6 m/s^2$

Hence, the acceleration of cheetah= $5.6m/s^2$

5 0

3 years ago