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3 years ago

The weight of an astronaut on the moon is the same as on Earth. True False

Physics

2 answers:

Vaselesa [24]3 years ago

5 0

Answer:

False

Explanation:

This is due to the gravitational pull, since the moon does not have the same force or gravity like Earth, your weight would change.

☆anvipatel77☆

•Expert•

Brainly Community Contributor

tangare [24]3 years ago

5 0

False. Looking at the equation g=GM/r^2, the mass of the object creating gravity (the moon) is directly proportional to gravitational field strength. This means that the lower the mass of the object, the weaker the gravitational field strength will be. Since the moon is nowhere near as massive as earth, the gravitational field strength is weaker than it would be on earth, meaning your weight will be different.

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A 70.0 kg astronaut is training for accelerations that he will experience upon reentry. He is placed in a centrifuge (r = 15.0 m

Levart [38]

Answer:

1.3823 rad/s

20.7345 m/s

28.66129935 m/s²

$a=2.92164g$

2006.29095 N radially outward

Explanation:

r = Radius = 15 m

m = Mass of person = 70 kg

g = Acceleration due to gravity = 9.81 m/s²

Angular velocity is given by

$\omega=13.2\times \dfrac{2\pi}{60}\\\Rightarrow \omega=1.3823\ rad/s$

Angular velocity is 1.3823 rad/s

Linear velocity is given by

$v=r\omega\\\Rightarrow v=15\times 1.3823\\\Rightarrow v=20.7345\ m/s$

The linear velocity is 20.7345 m/s

Centripetal acceleration is given by

$a_c=r\omega^2\\\Rightarrow a_c=15\times 1.3823^2\\\Rightarrow a_c=28.66129935\ m/s^2$

The centripetal acceleration is 28.66129935 m/s²

Acceleration in terms of g

$\dfrac{a}{g}=\dfrac{28.66129935}{9.81}\\\Rightarrow a=2.92164g$

$a=2.92164g$

Centripetal force is given by

$F_c=ma_c\\\Rightarrow F_c=70\times 28.66129935\\\Rightarrow F_c=2006.29095\ N$

The centripetal force is 2006.29095 N radially outward

The torque will be experienced when the centrifuge is speeding up of slowing down i.e., when it is accelerating and decelerating.

3 0

4 years ago

For a moon orbiting its planet, rp is the shortest distance between the moon and its planet andra is the longest distance betwee

Natasha2012 [34]

Answer: D. 0.57

Explanation:

The formula to calculate the eccentricity $e$ of an ellipse is (assuming the moon's orbit in the shape of an ellipse):

$e=\frac{r_{a}-r_{p}}{r_{a}+r_{p}}$

Where:

$r_{a}$ is the apoapsis (the longest distance between the moon and its planet)

$r_{p}=0.27 r_{a}$ is the periapsis (the shortest distance between the moon and its planet)

Then:

$e=\frac{r_{a}-0.27 r_{a}}{r_{a}+0.27 r_{a}}$

$e=\frac{0.73 r_{a}}{1.27 r_{a}}$

$e=0.57$ This is the moon's orbital eccentricity

3 0

4 years ago

Find the speed of a satellite in a circular orbit around the Earth with a radius 3.57 times the mean radius of the Earth. (Radiu

madam [21]

The speed of the satellite in a circular orbit around the Earth is 1.32 x 10⁵ m/s.

<h3>Speed of the satellite</h3>

v = √(GM/r)

where;

G is universal gravitation constant
M is mass of Earth
r is radius of the satellite

v = √(6.67 x 10⁻¹¹ x 5.98 x 10²⁴/3.57 x 6.37x 10³)

v = 1.32 x 10⁵ m/s

Thus, the speed of the satellite in a circular orbit around the Earth is 1.32 x 10⁵ m/s.

Learn more about speed of satellite here: brainly.com/question/22247460

#SPJ1

7 0

2 years ago

A resistor, an inductor, and a switch are all connected in series to an ideal battery of constant terminal voltage. Suppose at f

Airida [17]

Answer:

c. The steady-state value of the current depends on the resistance of the resistor.

Explanation:

Since all the components are connected in series, when the switch is at first open, current will not flow round the circuit. As current needs to flow through from the positive terminal of the battery through the resistor, inductor, and switch to the negative terminal of the battery.

But the moment the switch is closed, at the initial time t = 0, the current flow through from the positive terminal of the battery through the resistor, inductor, and switch to the negative terminal of the battery. It then begins to increase at a rate that depends upon the value of the inductance of the inductor.

6 0

3 years ago

A 1500 kg car drives around a flat 200-m-diameter circular track at 25m/s. What are the magnitude and direction of the net force

MakcuM [25]

The correct answer to the question is : 9375 N.

CALCULATION:

As per the question, the mass of the car m = 1500 Kg.

The diametre of the circular track D = 200 m.

Hence, the radius of the circular path R = $\frac{D}{2}$

= $\frac{200}{2}\ m$

= 100 m.

The velocity of the truck v = 25 m/s.

When a body moves in a circular path, the body needs a centripetal force which helps the body stick to the orbit. It acts along the radius and towards the centre.

Hence, the force acting on the car is centripetal force.

The magnitude of the centripetal force is calculated as -

Force F = $\frac{mv^2}{R}$

= $\frac{1500\times (25)^2}{100}\ N$

= 9375 N. [ANS}

The centripetal force is provided to the car in two ways. It is the friction which provides the necessary centripetal force. Sometimes friction is not sufficient. At that time, the road is banked to some extent which provides the necessary centripetal force.

6 0

3 years ago