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Dafna11 [192]
2 years ago
13

A space shuttle was launched from the Earth to the moon. The average, or accepted, value of the distance of the Earth to the moo

n is 238,857 miles, but the shuttle crew recorded that their trip took 226,316 miles.
Calculate the percent error of the distance to the moon.

5.25%
7.15%
3.75%
4.45%
Physics
1 answer:
Allushta [10]2 years ago
7 0
To find the answer it's (theoretical value - experimental value) / theoretical value
(238857-226316)/238857 = 5.25%
The answer is A) 5.25%
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Two satellites revolve around the Earth. Satellite A has mass m and has an orbit of radius r. Satellite B has mass 6m and an orb
melomori [17]

Answer:

aaaaa

Explanation:

M = Mass of the Earth

m = Mass of satellite

r = Radius of satellite

G = Gravitational constant

F=G\frac{Mm}{r^2}

F=G\frac{M6m}{r_b^2}

G\frac{Mm}{r^2}=G\frac{M6m}{r_b^2}\\\Rightarrow \frac{1}{r^2}=\frac{6}{r_b^2}\\\Rightarrow \frac{r_b^2}{r^2}=6\\\Rightarrow \frac{r_b}{r}=\sqrt{6}\\\Rightarrow r_b=2.44948r

r_b=2.44948r

8 0
2 years ago
Find an expression for the acceleration a of the red block after it is released. use mr for the mass of the red block, mg for th
Drupady [299]

<span>Assuming pulley is frictionless. Let the tension be ‘T’. See equation below.</span>

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2 years ago
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A kangaroo jumps to a vertical height of 2.8 m. How long was it in the air before returning to earth
BaLLatris [955]
The answer would be 2.8m height on earth takes 
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8 0
2 years ago
A 5.0-kg crate is resting on a horizontal plank. The coefficient of static friction is 0.50 and the coefficient of kinetic frict
Harlamova29_29 [7]

Answer:

The mass of the crate is 5kg.

We know that the force of friction can be obtained by:

F = N*k

where k is the coefficient of friction, where we use the static one if the object is at rest, and the kinetic one if the object os moving. N is the normal force

If we tilt the base making an angle of 30° with the horizontal, now the normal force against the plank will be equal to the fraction of the weight in the direction normal to the surface of the plank.

Knowing that the angle is 30°, then the fraction of the weight that pushes against the normal is Cos(30°)*W = cos(30°)*5kg*9.8m/s^2 = 42.4N

The fraction of the force in the parallel direction to the plank (the force that would accelerate the crate downwards) is:

F = sin(30°)*5k*9,8m/s = 24.5N

now, the statical friction force is:

Fs = 42.4N*0.5 = 21.2N

The statical force is less than the 24.5N, so the crate will move downwards, then the force that acts on the crate is the kinetic force of friction:

Fk = 42.4N*0.4 = 16.96N

Then, the total force that acts on the crate is:

total force = F - Fk = 24.5N - 16.69N = 7.54N and the direction of this force points downside along the parallel direction of the plank.

3 0
2 years ago
A 56 kg diver runs and dives from the edge of a cliff into the water which is located 4.0 m below. If she is moving at 8.0 m/s t
Reil [10]

Answer:

1) 2197.44 J

2) 0 J

3) 2197.44 J = Constant

4) 2197.44 J

5) Approximately 8.86 m/s

Explanation:

The given parameters are;

The mass of the diver, m = 56 kg

The height of the cliff, h = 4.0 m

The speed with which the diver is moving, vₓ = 8.0 m/s

The gravitational potential energy = Mass, m × Height of the cliff, h × Acceleration due to gravity, g

1) Her gravitational potential energy = 56 × 4.0 × 9.81 = 2197.44 J

2) The kinetic energy = 1/2·m·u²

Where;

u = Her initial velocity = 0 when she just leaves the cliff

Therefore;

Her kinetic energy when she just leaves the cliff = 1/2 × 56 × 0² = 0 J

3) The total mechanical energy = Kinetic energy + Potential energy

The total mechanical energy is constant

Her total mechanical energy relative to the water surface when she leaves the cliff = Her gravitational potential energy = 2197.44 J = Constant

4) Her total mechanical energy relative to the water surface just before she enters the water = 2197.44 J

5) The speed with which she enters the water, v, is given from, v² = u² + 2·g·h

Where;

u = The initial velocity at the top of the cliff before she jumps= 0 m/s

∴ v² = 0² + 2 × 9.81 × 4 = 78.48

v = √78.48 ≈ 8.86 m/s

The speed with which she enters the water, v ≈ 8.86 m/s

7 0
2 years ago
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