Ask question

Ask question

All categories

3 years ago

14

A man on the moon throws a ball vertically upwards and it is noticed that the ball travels 3.0m less in the fifth second of its

upward motion than it does in the third second. What is the acceleration due to gravity on the moon?

1 answer:

sdas [7]3 years ago

8 0

<h2>Acceleration due to gravity in moon is 1.5 m/s²</h2>

Explanation:

We have equation of motion s = ut + 0.5 at²

Here the ball travels 3 m less distance in fifth second compared to third second.

That is

s₃ = s₅ + 3

Now we have

Distance traveled in third second, s₃ = u x 3 - 0.5 x g x 3² - u x 2 - 0.5 x g x 2²

s₃ = u - 2.5 g

Also

Distance traveled in fifth second, s₅ = u x 5 - 0.5 x g x 5² - u x 4 - 0.5 x g x 4²

s₅ = u - 4.5 g

That is

u - 2.5 g = u - 4.5 g + 3

2 g = 3

g = 1.5 m/s²

Acceleration due to gravity in moon = 1.5 m/s²

You might be interested in

How can we teach our kitty to behave properly

Dmitry_Shevchenko [17]

Answer:

I have no idea.

Explanation:

I just wanted to tell you I absolutely read this wrong.

6 0

3 years ago

Two manned satellites approaching one another at a relative speed of 0.550 m/s intend to dock. The first has a mass of 2.50 ✕ 10

NNADVOKAT [17]

Answer: Their final relative velocity is -0.412 m/s.

Explanation:

According to the law of conservation,

$m_{1}v_{1} + m_{2}v_{2} = (m_{1} + m_{2})v$

Putting the given values into the above formula as follows.

$m_{1}v_{1} + m_{2}v_{2} = (m_{1} + m_{2})v$

$2.50 \times 10^{3} kg \times 0 m/s + 7.50 \times 10^{3} kg \times -0.550 m/s = (2.50 \times 10^{3} kg + 7.50 \times 10^{3} kg)v$

$-4.12 \times 10^{3} kg m/s = (10^{4} kg) v$

v = $\frac{-4.12 \times 10^{3} kg m/s}{10^{4} kg}$

= -0.412 m/s

Thus, we can conclude that their final relative velocity is -0.412 m/s.

8 0

3 years ago

Homeostasis may assist in balancing all of the following drives except __________. A. hunger

jeka94

Answer:

D~

Explanation:

8 0

4 years ago

What provides the energy for the conversion from the open complex to chain elongation?

Marat540 [252]

TP hydrolysis distinct from any incorporation into the chain.

7 0

3 years ago

You have a two-wheel trailer that you pull behind your ATV. Two children with a combined mass of 77.2 kg hop on board for a ride

lions [1.4K]

To solve this problem it is necessary to apply the kinematic equations of motion and Hook's law.

By Hook's law we know that force is defined as,

$F= kx$

Where,

k = spring constant

x = Displacement change

PART A) For the case of the spring constant we can use the above equation and clear k so that

$k= \frac{F}{x}$

$k = \frac{mg}{x}$

$k= \frac{77.2*9.8}{0.0637}$

$k = 11876.92N/m$

Therefore the spring constant for each one is 11876.92/2 = 5933.46N/m

PART B) In the case of speed we can obtain it through the period, which is given by

$T = \frac{2\pi}{\omega}$

Re-arrange to find \omega,

$\omega = \frac{2\pi}{T}$

$\omega = \frac{2\pi}{2.14}$

$\omega = 2.93rad/s$

Then through angular kinematic equations where angular velocity is given as a function of mass and spring constant we have to

$\omega^2 = \frac{k}{m}$

$m = \frac{k}{\omega^2}$

$m = \frac{ 11876.92}{2.93}$

$m = 4093.55Kg$

Therefore the mass of the trailer is 4093.55Kg

PART C) The frequency by definition is inversely to the period therefore

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

$f = \frac{1}{2.14}$

$f = 0.4672 Hz$

Therefore the frequency of the oscillation is 0.4672 Hz

PART D) The time it takes to make the route 10 times would be 10 times the period, that is

$t_T = 10*T$

$t_T = 10 *2.14s$

$t_T = 21.4s$

Therefore the total time it takes for the trailer to bounce up and down 10 times is 21.4s

5 0

3 years ago