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2 years ago

5

(Mass vs. Weight) HELP PLZ!!

1 answer:

Lina20 [59]2 years ago

7 0

$\huge \boxed { \sf{Answers}}$

c. The weight of an object on the moon will be the same as its weight on Earth. It is false because the weight of an on the moon will be 1/6 th times its weight on Earth.
d. The weight of an object is its mass multiplied by the force of gravity. The statement is false because the formula of weight is mass × acceleration due to gravity, not force of gravity.
e. The mass and weight of an object are the same thing. The statement is false because mass means a body of matter. While weight of an object is its mass multiplied by the force of gravity.
f. The mass of an object is the force of gravity acting upon an object. It is false because it will be the weight of the object not mass.
So, the answers are c, d, e and f.

Hope you could understand.

If you have any query, feel free to ask.

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An airplane starts at rest and accelerates down the runway for 20 s. At the end of the runway, its velocity is 80 m/s north. Wha

goldenfox [79]

Vi = vf + ( a.t)
0 m/s (rest) = 80 m/s + (a.20s.-1)

a.20.-1= -80, we check if this is true; = 0 m/s= 80 + -80 = 0 ITS TRUE.

so a.20.-1= -80, a= 80/20 , Answers a= 4 m/s

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2 years ago

An astronaut is in equilibrium when he is positioned 140 km from the center of asteroid C and 581 km from the center of asteroid

serg [7]

Answer:B

Explanation:

Given

Distance of astronaut From asteroid x is $r_x=140 km$

Distance of astronaut From asteroid Y is $r_y=581 km$

Suppose M,M_x,M_y be the masses of Astronaut , asteroid X and Y

If the astronaut is in equilibrium then net gravitational force on it is zero

$F_x=F_y$

$\frac{GMM_x}{r_x^2}=\frac{GMM_y}{r_y^2}$

cancel out the common terms we get

$\frac{M_x}{r_x^2}=\frac{M_y}{r_y^2}$

$\frac{M_x}{M_y}=(\frac{r_x}{r_y})^2$

$\frac{M_x}{M_y}=(\frac{140}{581})^2$

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2 years ago

In my trigonometry class, we were assigned a problem on Angular and Linear Velocity.

Rzqust [24]

1) 0.0011 rad/s

2) 7667 m/s

Explanation:

1)

The angular velocity of an object in circular motion is equal to the rate of change of its angular position. Mathematically:

$\omega=\frac{\theta}{t}$

where

$\theta$ is the angular displacement of the object

t is the time elapsed

$\omega$ is the angular velocity

In this problem, the Hubble telescope completes an entire orbit in 95 minutes. The angle covered in one entire orbit is

$\theta=2\pi$ rad

And the time taken is

$t=95 min \cdot 60 =5700 s$

Therefore, the angular velocity of the telescope is

$\omega=\frac{2\pi}{5700}=0.0011 rad/s$

2)

For an object in circular motion, the relationship between angular velocity and linear velocity is given by the equation

$v=\omega r$

where

v is the linear velocity

$\omega$ is the angular velocity

r is the radius of the circular orbit

In this problem:

$\omega=0.0011 rad/s$ is the angular velocity of the Hubble telescope

The telescope is at an altitude of

h = 600 km

over the Earth's surface, which has a radius of

R = 6370 km

So the actual radius of the Hubble's orbit is

$r=R+h=6370+600=6970 km = 6.97\cdot 10^6 m$

Therefore, the linear velocity of the telescope is:

$v=\omega r=(0.0011)(6.97\cdot 10^6)=7667 m/s$

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2 years ago

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aleksandr82 [10.1K]

The greater the mass the greater is inertia.

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2 years ago

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Yanka [14]

Is there information in the previous question which relates to this one?

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