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navik [9.2K]
2 years ago
11

The gravitational acceleration on the moon is about one-sixth the size of the gravitational acceleration on Earth. According to

Newton’s second law of motion, what happens to an astronaut who goes to the moon? The astronaut’s weight decreases because the moon’s gravitational acceleration is greater than Earth’s. The astronaut’s weight increases because the moon’s gravitational acceleration is greater than Earth’s. The astronaut’s weight decreases because the moon’s gravitational acceleration is less than Earth’s. The astronaut’s weight increases because the moon’s gravitational acceleration is less than Earth’s.
Physics
1 answer:
photoshop1234 [79]2 years ago
5 0

Answer: If the gravitacional acceleration is 1/6 of Earth's gravitational acceleration, it means that moon's gravitational acceleration is less than Earth's. Also, if the gravitational acceleration is less than Earth's, the astronaut's weight decreases since we calculate it multiplying his body mass by the gravity in the place given.

On Earth, an astronaut that is 70kg weights 70kg * 9.8 m/s² = 686N

On the Moon, the same astronaut would weight 70kg * 9.8 m/s² * 1/6 = 114,3 N

So, the astronaut’s weight decreases because the moon’s gravitational acceleration is less than Earth’s.

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Describe the energy transformations that occur from the time a skydiver jumps out of a plane until landing on the ground.
Kisachek [45]
When the Skydiver jump out a plane, his Potential Energy is being converted or transform into Kinetic energy due to gravity. Hope this helps
6 0
2 years ago
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In the Daytona 500 auto race, a Ford Thunderbird and a Mercedes Benz are moving side by side down a straightaway at 71.0 m/s. Th
qaws [65]

Answer:

The distance between both cars is 990 m

Explanation:

The equations for the position and the velocity of an object moving in a straight line are as follows:

x = x0 + v0 * t + 1/2 * a * t²

v = v0 + a * t

where:

x = position of the car at time "t"

x0 = initial position

v0 = initial speed

t = time

a = acceleration

v = velocity

First let´s find how much time it takes the driver to come to stop (v = 0).  We will consider the origin of the reference system as the point at which the driver realizes she must stop. Then x0 = 0

With the equation of velocity, we can obtain the acceleration and replace it in the equation of position, knowing that the position will be 250 m at that time.

v = v0 + a*t

v-v0 / t = a

0 m/s - 71.0 m/s / t =a

-71.0 m/s / t = a

Replacing in the equation for position:

x = v0* t +1/2 * a * t²

250 m = 71.0 m/s * t + 1/2 *(-71.0 m/s / t) * t²

250 m = 71.0 m/s * t - 1/2 * 71.0 m/s * t

250m = 1/2 * 71.0m/s *t

<u>t = 2 * 250 m / 71.0 m/s = 7.04 s</u>

It takes the driver 7.04 s to stop.

Then, we can calculate how much time it took the driver to reach her previous speed. The procedure is the same as before:

v = v0 + a*t

v-v0 / t = a      now v0 = 0 and v = 71.0 m/s

(71.0 m/s - 0 m/s) / t = a

71.0 m/s / t =a

Replacing in the position equation:

x = v0* t +1/2 * a * t²      

390 m = 0 m/s * t + 1/2 * 71.0 m/s / t * t²       (In this case, the initial position is in the pit, then x0 = 0 because it took 390 m from the pit to reach the initial speed).

390m * 2 / 71.0 m/s = t

<u>t = 11.0 s</u>

In total, it took the driver 11.0s + 5.00 s + 7.04 s = 23.0 s to stop and to reach the initial speed again.

In that time, the Mercedes traveled the following distance:

x = v * t = 71.0 m/s * 23.0 s = 1.63 x 10³ m

The Thunderbird traveled in that time 390 m + 250 m = 640 m.

The distance between the two will be then:

<u>distance between both cars = 1.63 x 10³ m - 640 m = 990 m.  </u>

3 0
2 years ago
A 250 GeV beam of protons is fired over a distance of 1 km. If the initial size of the wave packet is 1 mm, find its final size
Margarita [4]

Answer:

The final size is approximately equal to the initial size due to a very small relative increase of 1.055\times 10^{- 7} in its size

Solution:

As per the question:

The energy of the proton beam, E = 250 GeV =250\times 10^{9}\times 1.6\times 10^{- 19} = 4\times 10^{- 8} J

Distance covered by photon, d = 1 km = 1000 m

Mass of proton, m_{p} = 1.67\times 10^{- 27} kg

The initial size of the wave packet, \Delta t_{o} = 1 mm = 1\times 10^{- 3} m

Now,

This is relativistic in nature

The rest mass energy associated with the proton is given by:

E = m_{p}c^{2}

E = 1.67\times 10^{- 27}\times (3\times 10^{8})^{2} = 1.503\times 10^{- 10} J

This energy of proton is \simeq 250 GeV

Thus the speed of the proton, v\simeq c

Now, the time taken to cover 1 km = 1000 m of the distance:

T = \frac{1000}{v}

T = \frac{1000}{c} = \frac{1000}{3\times 10^{8}} = 3.34\times 10^{- 6} s

Now, in accordance to the dispersion factor;

\frac{\delta t_{o}}{\Delta t_{o}} = \frac{ht_{o}}{2\pi m_{p}\Delta t_{o}^{2}}

\frac{\delta t_{o}}{\Delta t_{o}} = \frac{6.626\times 10^{- 34}\times 3.34\times 10^{- 6}}{2\pi 1.67\times 10^{- 27}\times (10^{- 3})^{2} = 1.055\times 10^{- 7}

Thus the increase in wave packet's width is relatively quite small.

Hence, we can say that:

\Delta t_{o} = \Delta t

where

\Delta t = final width

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2 years ago
Whale sharks swim forward while ascending or descending. They swim along a straight-line path at a shallow angle as they move fr
Aneli [31]
You did not include the quesetion, but I can help you to understand the problem and how to find the relevant information.

1) The angle of 13° with which the shark ascends meets this:

Vertical ascending velocity = 0.85m/s * sin(13°)

Horizontal velocity = 0.85m/s * cos(13°)

2) The length swan by the shark ascending meets this

Vertical ascending length = 50 m

Horizontal length, y:

\frac{y}{50} = \frac{0.85sin(13)}{0.85cos(13)}

From that y = 50 * tan(13°)

=> y = 11.54 m.

3) Conclusions:

1) The shark run 50 m vertically upward and 11.54 m horizontally.

2) The length of the path run by the shark may be calculated using Pythagoras' theorem:

hypotenuse^2 = (50m)^2 + (11.54m)^2 = 2633.25m^2

hypotenuse = 51.35m

So, the shark swan 51.35 m to reach the surface.

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2 years ago
A person's height will increase from birth until about age 25, and it may decrease starting at about age 70. This is an example
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