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3 years ago

The acceleration due to gravity on Mars is less than that on Earth. On Mars, a person will weigh than on Earth.

2 answers:

4 0

On mars people would way less.
An example of this is that if I weighed 700 pounds (I don't by the way) I would then weigh 500 pounds or less.

Vlad [161]3 years ago

4 0

Answer: On Mars, a person will weigh <u>less</u> than on Earth.

Explanation:

The mass of the planet Mars is less than Earth. Because of this it pulls the objects on it with less force and thus Mars has less value of acceleration due to gravity.

The weight of the object is given by the product of mass and acceleration due to gravity.

W = m g'

W ∝ g'

The mass remains constant irrespective of the planet but weight varies with the value of acceleration due to gravity. Thus, on Mars, a person will weigh less than on Earth.

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The Apollo Lunar Module was used to make the transition from the spacecraft to the moon's surface and back. Consider a similar m

lions [1.4K]

Answer:

Explanation:

a. Landing height is

H=1.3m

Velocity of lander relative to the earth is, i.e this is the initial velocity of the spacecraft

u=1.3m/s

Velocity of lander at impact, i.e final velocity is needed

v=?

The acceleration due to gravity is 0.4 times that of the one on earth,

Then, g on earth is approximately 9.81m/s²

Then, g on Mars is

g=0.4×9.81=3.924m/s²

Then using equation of motion for a free fall body

v²=u²+2gH

v²=1.3²+2×3.924×1.3

v²=1.69+10.2024

v²=11.8924

v=√11.8924

v=3.45m/s

The impact velocity of the spacecraft is 3.45m/s

b. For a lunar module, the safe velocity landing is 3m/s

v=3m/s.

Given that the initial velocity is 1.2m/s²

We already know acceleration due to gravity on Mars is g=3.924m/s²

The we need to know the maximum height to have a safe velocity of 3m/s

Then using equation of motion

v²=u²+2gH

3²=1.2²+2×3.924H

9=1.44+7.848H

9-1.44=7.848H

7.56=7.848H

H=7.56/7.848

H=0.963m

The the maximum safe landing height to obtain a final landing velocity of 3m/s is 0.963m

8 0

3 years ago

If two wires run parallel and the current passes through both wires in the same direction, which happens to the wires?

pashok25 [27]

If two wires run parallel and the current passes through both wires in the same direction, which happens to the wires?

If two wires run parallel and the current passes through both wires in the same direction, the wires wires move together because the current of one wire pulls the other wire closer. The answer is letter C.

7 0

3 years ago

he "escape velocity from Earth (the speed required to escape Earth's gravity) is 2.5 x 10 miles per hour. What is this speed in

Natasha_Volkova [10]

Answer: $11.17\ \text{ m/s}$

Explanation:

Given : The escape velocity : $v=2.5\times10\text{ miles per hour}$

We know that 1 mile = 1609 meters (approx)

and 1 hour= 3600 seconds

To convert escape velocity 2.5 x 10 miles per hour into m/s , we need to multiply it by 1609.34 and divide it by 3600.

Thus, the escape velocity in m/s is given by :-

$v=2.5\times10\times\dfrac{1609}{3600}\\\\=11.1736111111\approx11.17\text{ m/s}$

Hence, the speed in m/s = 11.17

4 0

3 years ago

A disk of radius 25 cm spinning at a rate of 30 rpm slows to a stop over 3 seconds. what is the angular acceleration?

SCORPION-xisa [38]

Answer:

α = -π/3 rad/s²

θ = 1.5π rad ≈ 4.71 rad

θ = 0.75 rev

Explanation:

30 rev/min (2π rad/rev) / (60 s/min) = π rad/s

α = (ωf - ωi) / t = (0 - π) / 3 = -π/3 rad/s²

θ = ½αt² = ½(π/3)3² = 1.5π rad ≈ 4.71 rad

θ = 1.5π rad / 2π rad/rev = 0.75 rev

5 0

2 years ago

The space shuttle releases a satellite into a circular orbit 630 km above the Earth.

solniwko [45]

Answer:

7,539 m/s

Explanation:

Let's use this equation to find the gravitational acceleration of this space shuttle:

$\displaystyle g=\frac{GM}{r^2}$

We know that G is the gravitational constant: 6.67 * 10^(-11) Nm²/kg².

M is the mass of the planet, which is Earth in this case: 5.972 * 10^24 kg.

r is the distance from the center of Earth to the space shuttle: radius of Earth (6.3781 * 10^6 m) + distance above the Earth (630 km → 630,000 m).

Plug these values into the equation:

$\displaystyle g=\frac{(6.67\cdot 10^-^1^1 \ Nm^2kg^-^2)(5.972\cdot 10^2^4 \ kg)}{[(6.3781\cdot 10^6 \ m)+(630000 \ m)]^2}$

Remove units to make the equation easier to read.

$\displaystyle g=\frac{(6.67\cdot 10^-^1^1 )(5.972\cdot 10^2^4 )}{[(6.3781\cdot 10^6)+(630000 )]^2}$

Multiply the numerator out.

$\displaystyle g=\frac{(3.983324\cdot 10^1^4)}{[(6.3781\cdot 10^6)+(630000 )]^2}$

Add the terms in the denominator.

$\displaystyle g=\frac{(3.983324\cdot 10^1^4)}{[(7008100)]^2}$

Simplify this equation.

$\displaystyle g=8.11045189 \ \frac{m}{s^2}$

The acceleration due to gravity g = 8.11045189 m/s². Now we use the equation for acceleration for an object in circular motion which contains v and r.

$\displaystyle a = \frac{v^2}{r}$

a = g, v is the velocity that the space shuttle should be moving (what we are trying to solve for), and r is the radius we had in the previous equation when solving for g.

Plug these values into the equation and solve for v.