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RSB [31]
2 years ago
8

An astronaut weighs 200 lb at sea level. The radius of the earth is 3960 miles. What force is exerted on the astronaut if he is

floating 125 miles above the earth's surface?
Physics
1 answer:
LiRa [457]2 years ago
6 0

Answer:

85.31 N

Explanation:

Given,

Radius of Earth = 3960 miles = 6373 Km

Weight of Astronaut at Sea level = 200lb = 90.72 Kg

Altitude of Astronaut above Earth = 125 miles = 201.17 km

We know that,

F = m\times g --------------------------  (1)

where,

F = force on the object due to gravity also called the weight

m = mass of the object = 200 lb = 90.71 kg

g = acceleration due to gravity = 9.8 m/s²

Also,

F = \frac{GMm}{r^{2}} -------------------(2)

where,

F = force due to gravity

G = Gravitational constant = 6.67\times 10^{-11} Nm²/kg²

M = mass of the Earth = 5.97\times 10^{24} kg

r = distance between the two objects

here, r = (6373+201.17)km = 6574170 m

From equation (1),

m\times g = 90.71\\m=\frac{90.71}{g} \\m=\frac{90.71}{9.8}\\m=9.26 kg\\

Putting value of m in equation (2)

F = \frac{6.67\times 10^{-11}\times 5.97\times 10^{24}\times 9.26}{6574170^{2}}\\

F=85.31N

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The front 1.20 m of a 1,600-kg car is designed as a "crumple zone" that collapses to absorb the shock of a collision. (a) If a c
eimsori [14]

To develop the problem it is necessary to apply the kinematic equations for the description of the position, speed and acceleration.

In turn, we will resort to the application of Newton's second law.

PART A) For the first part we look for the time, in a constant acceleration, knowing the speeds and the displacement therefore we know that,

X_f = X_i +\frac{1}{2}(V_i+V_f)t

Where,

X = Desplazamiento

V = Velocity

t = Time

In this case there is no initial displacement or initial velocity, therefore

X_f = \frac{1}{2} (V_i+V_f)t

Clearing for time,

t = \frac{2X_f}{(V_i+V_f)}

t = \frac{2*1.2}{24+0}

t = 0.1s

PART B) This is a question about the impulse of bodies, where we turn to Newton's second law, because:

F = ma

Where,

m=mass

a = acceleration

Acceleration can also be written as,

a= \frac{\Delta V}{t}

Then

F = m\frac{\Delta V}{t}

F = m\frac{V_f-V_i}{t}

F = m\frac{-V_i}{t}

F = \frac{(1600kg)(-24m/s)}{(0.1s)}

F = -384000N

Negative symbol is because the force is opposite of the direction of moton.

PART C) Acceleration through kinematics equation is defined as

V_f^2=V_i^2-2ax

0 = (24m/s)^2-2*a(1.2m)

a = \frac{(24m/s)^2}{1.2m}

a=480m/s^2

The gravity is equal to 0.8, then the acceleration is

a = 480*\frac{g}{9.8}

a = 53.3g

3 0
2 years ago
An overhead projector lens is 32.0 cm from a slide (the object) and has a focal length of 30.1 cm. What is the magnification of
puteri [66]

Answer: 15.8

Explanation:

You are given that the

Object distance U = 32 cm

Focal length F = 30.1 cm

First calculate the image distance V by using the formula

1/F = 1/U + 1/V

Substitute F and V into the formula

1/30.1 = 1/32 + 1/V

1/V = 1/30.1 - 1/32

1/V = 0.00197259

Reciprocate both sides

V = 506.94 cm

Magnification M is the ratio of image distance to object distance.

M = V/U

substitute the values of V and U into the formula

M = 506.94/32

M = 15.8

Therefore, the magnification of the image is 15.8 or approximately 16.

6 0
2 years ago
Which three methods are valid for preventing further environmental damage from CO2?
vredina [299]
By afforestation.
by following recycle rule
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3 0
2 years ago
Read 2 more answers
A 1 530-kg automobile has a wheel base (the distance between the axles) of 2.70 m. The automobile's center of mass is on the cen
NeTakaya

Answer:

Force on front axle = 6392.85 N

Force on rear axle = 8616.45 N

Explanation:

As we know that the weight of the car is balanced by the normal force on the front wheel and rear wheels

Now we know that

F_1 + F_2 = W

F_1 + F_2 = (1530\times 9.81)

F_1 + F_2 = 15009.3 N

now we know that distance between the axis is 2.70 m and centre of mass is 1.15 m behind front axle

so we can write torque balance about its center of mass

F_1(1.15) = F_2(2.70 - 1.15)

F_1 = 1.35 F_2

now from above equation

F_2 + 1.35F_2 = 15009.3

now we have

F_2 = 6392.85 N

now the other force is given as

F_1 = 8616.45 N

4 0
2 years ago
A 0.20 kg mass on a horizontal spring is pulled back 2.0 cm and released. If, instead, a 0.40 kg mass were used in this same exp
KIM [24]

Answer:

The total mechanical energy does not change if the value of the mass is changed. That is, remain the same

Explanation:

The total mechanical energy of a spring-mass system is equal to the elastic potential energy where the object is at the amplitude of the motion. That is:

E=U=\frac{1}{2}kA^2       (1)

k: spring constant

A: amplitude of the motion = 2.0cm

As you can notice in the equation (1), the total mechanical energy of the system does not depend of the mass of the object. It only depends of the amplitude A and the spring constant.

Hence, if you use a mass of 0.40kg the total mechanical energy is the same as the obtained with a mas 0.20kg

Remain the same

8 0
2 years ago
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