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sergey [27]
3 years ago
8

An astronaut lands on a new planet. The new planet’s mass is half that of the

Physics
1 answer:
tensa zangetsu [6.8K]3 years ago
7 0
Ah hah !  Very nice problem.

-- Since the mass is 1/2 of the Earth's mass, the gravitational forces between him
and the new planet would be 1/2 of what they are on Earth.

-- Since the distance between him and the center of the new planet is 1/2 of the
distance between him and the Earth's center, the gravitational forces between him
and the new planet would be (2)² = 4 times what they are on Earth.

-- These conditions combine to make his weight  (1/2) x (4) = 2 times his earth weight.

-- He weighs 1,000 N (about 224.8 pounds) on the new planet.

(Maybe that's why they selected an astronaut for this mission who only weighs
about 112.5 pounds on Earth.)
You might be interested in
What quantity measures the number of complete cycles an oscillation makes per second? A. period B. amplitude C. frequency D. for
Free_Kalibri [48]

Hello, there Jcparris

Your answer is going to be C. Frequency

If my answer helped please leave a thank rate 5 stars and the most important rank me braiinliest thank you and have a great day!

8 0
3 years ago
A rocket is fired at 100 m/s at an angle of 37, what was its speed at the top of its path?
Nadusha1986 [10]

Answer:

0m/s

Explanation:

Since its fired at an angle, at the top there will be a split second where the velocity will be 0, as it has a parabolic shape, so the speed at the top of its path is 0

4 0
2 years ago
4.2 mol of monatomic gas A interacts with 3.2 mol of monatomic gas B. Gas A initially has 9500 J of thermal energy, but in the p
Komok [63]

Answer:

14657.32 J

Explanation:

Given Parameters ;

Number of moles mono atomic gas A ,   n 1  =  4 .2 mol

Number of moles mono atomic gas B ,   n 2  =  3.2mol

Initial energy of gas A ,   K A  =  9500  J

Thermal energy given by gas A to gas B ,   Δ K  =  600 J

Gas constant   R  = 8.314  J / molK

Let  K B  be the initial energy of gas B.

Let T be the equilibrium temperature of the gas after mixing.

Then we can write the energy of gas A after mixing as

(3/2)n1RT = KA - ΔK

⟹ (3/2) x 4.2 x 8.314 x T = 9500 - 600

T = (8900 x 3 )/(2x4.2x8.314) = 382.32 K

Energy of the gas B after mixing can be written as

(3/2)n2RT = KB + ΔK

⟹ (3/2) x 3.2 x 8.314 x 382.32 = KB + 600

⟹ KB = [(3/2) x 3.2 x 8.314 x 382.32] - 600

⟹ KB = 14657.32 J

6 0
3 years ago
Adam drops a ball from rest from the top floor of a building at the same time Bob throws a ball horizontally from the same locat
guapka [62]

Answer:

Both balls hit the ground at the same time

Explanation:

Adam drops the ball from rest, so the ball just "<em>falls</em>" in vertical direction, being gravity its only acceleration, for cinematic movements we use that:

y(t)=y_{0}+v_{0y}t+\frac{1}{2}gt^{2}

In this case we have that gravity is negative, and as Adam drops the ball, v_{0y}=0

Bob throws the ball horizontally, so the movement will be a <em>parabola</em>, we can divide into horizontal direction, and vertical direction.

But we only need to analize the vertical movement, in wich again the only acceleration is gravity, and compare it with Adam's ball. Again we have that gravity is negative, and as the initial throw is horizontal, v_{0y}=0

Finally, we have that

y(t)=h-\frac{1}{2}gt^{2}

where

h=y_{0}

both for Adam's vertical drop, and for Bob's vertical component of the parabolic throw.

Now, if we put y(t)=0 (the origin of the vertical coordinate), we get for both cases that

h=\frac{1}{2}gt^{2}

where we can clear the value for the time t, of the fall, wich will be the same in both cases.

Hence, both balls hit the ground at the same time.

3 0
2 years ago
In a game of pool, the cue ball moves at a speed of 2 m/s toward the eight ball. When the cue ball hits the eight ball, the cue
agasfer [191]

Answer:

a)  p₀ = 1.2 kg m / s,  b) p_f = 1.2 kg m / s,  c)   θ = 12.36, d)  v_{2f} = 1.278 m/s

Explanation:

a system formed by the two balls, which are isolated and the forces during the collision are internal, therefore the moment is conserved

a) the initial impulse is

        p₀ = m v₁₀ + 0

        p₀ = 0.6 2

        p₀ = 1.2 kg m / s

b) as the system is isolated, the moment is conserved so

       p_f = 1.2 kg m / s

we define a reference system where the x-axis coincides with the initial movement of the cue ball

we write the final moment for each axis

X axis

        p₀ₓ = 1.2 kg m / s

        p_{fx} = m v1f cos 20 + m v2f cos θ

        p₀ = p_f

       1.2 = 0.6 (-0.8) cos 20+ 0.6 v_{2f} cos θ

        1.2482 = v_{2f} cos θ

Y axis  

       p_{oy} = 0

       p_{fy} = m v_{1f} sin 20 + m v_{2f} cos θ

       0 = 0.6 (-0.8) sin 20 + 0.6 v_{2f} sin θ

       0.2736 = v_{2f} sin θ

we write our system of equations

        0.2736 = v_{2f} sin θ

        1.2482 = v_{2f} cos θ

divide to solve

        0.219 = tan θ

         θ = tan⁻¹ 0.21919

         θ = 12.36

let's look for speed

           0.2736 = v_{2f} sin θ

            v_{2f} = 0.2736 / sin 12.36

           v_{2f} = 1.278 m / s

7 0
2 years ago
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