Ask question

Ask question

All categories

musickatia [10]

3 years ago

12

Games in Space: On Earth, an astronaut throws a ball straight upward, and it stays in the air for a total of 3.0s before they ca

tch it at the same height they released it. On the Moon, acceleration due to gravity is ∼ 1/6 the value on Earth. If the astronaut repeats this process on the Moon, giving the ball the same initial speed, how much time would pass before they catch it

1 answer:

Vesna [10]3 years ago

3 0

if an object is thrown upwards and then it will return back into our hand then we can say for the complete motion the displacement of the object must be zero

so here we will have

$\Delta y = vt + \frac{1}{2}at^2$

here we know that

$0 = vt - \frac{1}{2}gt^2$

$t = \frac{2v}{g}$

now we will have this time t = 3 s on the surface of earth

again same experiment is performed on surface of moon with same initial speed

so the time on the surface of moon will be

$t_{moon} = \frac{2v}{g/6}$

so here we have

$t_{moon} = 6\frac{2v}{g} = 6(3) =18 s$

so it will take 18 s on moon

You might be interested in

A resonance tube can be used to measured the speed of sound in air. A tuning fork is held above the opening of the tube and stru

LenaWriter [7]

Answer:

330.24 Hz

Explanation:

Given:

Frequency, f = 320 Hz

L1 = 25.8 cm

L2 = 78.4 cm

L3 = 131.1 cm

Let the wavelength be λ

Then, L1 which is the length of the column of air is λ/4.

λ/4 = 25.8 cm

λ = 25.8 × 4 = 103.2 cm = 1.032 m

Then, speed of sound in air is:

v = λ f

⇒ v = 1.032 × 320 Hz

⇒ v = 330.24 m/s

7 0

3 years ago

When this circuit is closed, which way do the electrons flow?

gizmo_the_mogwai [7]

Answer:

from the positive end of the battery through the capacitor through the resistors to the negative end...

Current flows from higher potential (+) to lower potential (-)..

3 0

3 years ago

Read 2 more answers

A WLAN transmitter that emites a 100 mW signal is connected to a cable with a 3dB loss. If the cable is connected to an antenna

irakobra [83]

Answer: 24 dBm

Explanation:

in the attachment

6 0

3 years ago

Suppose you have two magnets. Magnet A doesn't have its poles labeled, but Magnet B does have a clearly labeled north and south

Elina [12.6K]

Before going to answer this question first we have to know the fundamental principle of magnetism.

A magnet have two poles .The important characteristic of a magnet is that like poles will repel each other while unlike poles will attract each other.

Through this concept the question can be answered as explained below-

A-As per first option the side of magnet A is repelled by the south pole of magnet B. Hence the pole of a must be south .It can't be north as it will lead to attraction.

B-The side of magnet A is repelled by the north pole of magnet B. Hence the side of A must be north pole.It can't be a south pole.

C-The side of magnet A is attracted by the south pole of magnet B .Hence the side of magnet A must be north.Hence this is right

D-The side of magnet A is attracted by the north pole of magnet B. Hence the side of A must south.It can't be north as it will lead to repulsion.

Hence the option C is right.

3 0

3 years ago

Read 2 more answers

A spring gun is made by compressing a spring in a tube and then latching the spring at the compressed position. A 4.97-g pellet

dimaraw [331]

Answer:

$v = 2.8898 \frac{m}{s}$

Explanation:

This is a problem easily solve using energy conservation. As there are no non-conservative forces, we know that the energy is conserved.

When the spring is compressed downward, the spring has elastic potential energy. When the spring is relaxed, there is no elastic potential energy, but the pellet will have gained gravitational potential energy and kinetic energy. Lets see what are the terms for each of this.

<h3>Elastic potential energy</h3>

We know that a spring following Hooke's Law has a elastic potential energy:

$E_{ep} = \frac{1}{2} k (\Delta x)^2$

where $\Delta x$ is the displacement from the relaxed length and k is the spring's constant.

To obtain the spring's constant, we know that Hooke's law states that the force made by the spring is :

$\vec{F} = - k \Delta \vec{x}$

as we need 9.12 N to compress 4.60 cm, this means:

$k = \frac{9.12 \ N}{4.6 \ 10^{-2} \ m}$

$k = 198.26 \ \frac{ N}{m}$

So, the elastic energy of the compressed spring is:

$E_{ep} = \frac{1}{2} 198.26 \ \frac{ N}{m} (4.6 \ 10^{-2} \ m)^2$

$E_{ep} = 0.209759 \ Joules$

And when the spring is relaxed, the elastic potential energy will be zero.

<h3>Gravitational potential energy</h3>

To see how much gravitational potential energy will the pellet win, we can use

$\Delta E_{gp} = m g \Delta h$

where m is the mass of the pellet, g is the acceleration due to gravity and \Delta h is the difference in height.

Taking all this together, the gravitational potential energy when the spring is relaxed will be:

$\Delta E_{gp} = 4.97 \ 10^{-3} kg \ 9.8 \frac{m}{s^2} 4.6 \ 10^{-2} m$

$\Delta E_{gp} = 0.00224 \ Joules$

<h3>Kinetic Energy</h3>

We know that the kinetic energy for a mass m moving at speed v is:

$E_k = \frac{1}{2} m v^2$

so, for the pellet will be

$E_k = \frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2$

<h3>All together</h3>

By conservation of energy, we know:

$E_{ep} = \Delta E_{gp} + E_k$

$0.209759 \ Joules = 0.00224 \ Joules + \frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2$

So

$\frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2 = 0.209759 \ Joules - 0.00224 \ Joules$

$\frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2 = 0.207519 \ Joules$

$v = \sqrt{ \frac{ 0.207519 \ Joules}{ \frac{1}{2} \ 4.97 \ 10^{-3} kg } }$

$v = 2.8898 \frac{m}{s}$

7 0

3 years ago