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Slav-nsk [51]
3 years ago
6

An astronaut weighs 8.00 × 102 newtons on the sur- face of Earth. What is the weight of the astronaut 6.37 × 106 meters above th

e surface of Earth?
Physics
1 answer:
kolbaska11 [484]3 years ago
5 0

Answer:

mg=200.4 N.

Explanation:

This problem can be solved using Newton's law of universal gravitation: F=G\frac{m_{1}m_{2}}{r^{2}},

where F is the gravitational force between two masses m_{1} and m_{2}, r is the distance between the masses (their center of mass), and G=6.674*10^{-11}(m^{3}kg^{-1}s^{-2}) is the gravitational constant.

We know the weight of the astronout on the surface, with this we can find his mass. Letting w_{s} be the weight on the surface:

w_{s}=mg,

mg=8*10^{2},

m=(8*10^{2})/g,

since we now that g=9.8m/s^{2} we get that the mass is

m=81.6kg.

Now we can use Newton's law of universal gravitation

F=G\frac{Mm}{r^{2}},  

where m is the mass of the astronaut and M is the mass of the earth. From Newton's second law we know that

F=ma,

in this case the acceleration is the gravity so

F=mg, (<u>becarefull, gravity at this point is no longer</u> 9.8m/s^{2} <u>because we are not in the surface anymore</u>)

and this get us to

mg=G\frac{Mm}{r^{2}}, where mg is his new weight.

We need to remember that the mass of the earth is M=5.972*10^{24}kg and its radius is 6.37*10^{6}m.

The total distance between the astronaut and the earth is

r=(6.37*10^{6}+6.37*10^{6})=2(6.37*10^{6})=12.74*10^{6} meters.

Now we can compute his weigh:

mg=G\frac{Mm}{r^{2}},

mg=(6.674*10^{-11})\frac{(5.972*10^{24})(81.6)}{(12.74*10^{6})^{2}},

mg=200.4 N.

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3 years ago
To test the quality of a tennis ball, you drop it onto the floor from a height of 4.00 m. It rebounds to a height of 2.00 m. If
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Answer:

Part a)

a = 1260.3 m/s^2

Part b)

Direction = upwards

Explanation:

When ball is dropped from height h = 4.0 m

then the speed of the ball just before it will strike the ground is given as

v_f^2 - v_i^2 = 2 a d

v_1^2 - 0^2 = 2(9.81)(4.0)

v_1 = 8.86 m/s

Now ball will rebound to height h = 2.00 m

so the velocity of ball just after it will rebound is given as

v_f^2 - v_i^2 = 2 a d

0 - v_2^2 = 2(-9.81)(2.00)

v_2 = 6.26 m/s

Part a)

Average acceleration is given as

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a = \frac{6.26 - (-8.86)}{12.0 \times 10^{-3}}

a = 1260.35 m/s^2

Part B)

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Read 2 more answers
Pls help
belka [17]

Answer:

2156 J

Explanation:

From the question,

Work done = Combined mass of the bucket and water×height×gravity.

W = (M+m)hg............................. Equation 1

Where M = mass of water, m = mass of the bucket, h = height, g = acceleration due to gravity.

Given: M = 20 kg, m = 2 kg, h = 10 m

Constant: g = 9.8 m/s²

Substitute these  value into equation 1

W = (20+2)×10×9.8

W = 22×98

W = 2156 J

4 0
2 years ago
A person is standing on and facing the front of a stationary skateboard while holding a construction brick. The mass of the pers
Inessa [10]

Answer:

    v₁ = -0.8087 m / s

Explanation:

To solve this problem we can use the conservation of momentum, for this we define a system formed by the man, the skateboard and the brick, therefore the force during the separation is internal and the momentum is conserved

Initial instant. When they are united

        p₀ = 0

Final moment. After throwing the brick

        p_{f} = (m_man + m_skate) v1 + m_brick v2

the moment is preserved

        p₀ = p_{f}

        0 = (m_man + m_skate) v₁ + m_brick v₂

        v₁ = -  \frac{ m_{brick}   }{m_{man} + m_{skate}   }  v_{2}

the negative sign indicates that the two speeds are in the opposite direction

let's calculate

        v₁ = - \frac{2.5}{67 + 4.10}  23.0

        v₁ = -0.8087 m / s

6 0
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