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Lilit [14]
3 years ago
13

An object that travels around another object in space is called a(n)

Physics
2 answers:
HACTEHA [7]3 years ago
3 0
Satellite. think of the moon.

Hope this helps!
Vote me Brainliest!
user100 [1]3 years ago
3 0
It's definitely D, satellite
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An astronaunt takes am object to the moon where there is less gravity. Explain how the mass amd weight of an object on the moon
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Even tho the object is big it will still Float around like a balloon cause of the gravity in space that's why they hook the rocket-ship into the ground of space 
6 0
3 years ago
A virtual image produced by a lens is always
JulijaS [17]
The Answer Is D Because When Uu Magnify Large Items The Image Is Reflected Back To The Magnifying Glass Which Makes The Image Appear In The Back
5 0
3 years ago
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A person kicks a ball, giving it an initial velocity of 20.0 m/s up a wooden ramp. When the ball reaches the top, it becomes air
Alex Ar [27]

Answer:

(a) Height is 4.47 m

(b) Height is 4.37 m

Solution:

As per the question:

Initial velocity of teh ball, v_{o} = 20.0 m/s

Angle made by the ramp, \theta = 22.0^{\circ}

Distance traveled by the ball on the ramp, d = 5.00 m

Now,

(a) At any point on the projectile before attaining maximum height, the velocity can be given by the eqn-3 of motion:

v^{2} = v_{o}^{2} - 2gH

where

H = dsin22^{\circ} = 5sin22^{\circ}

g = 9.8 m/s^{2}

v^{2} = 20^{2} - 2\times 9.8\times 5sin22^{\circ}

v = \sqrt{400 - 19.6\times 5sin22^{\circ}} = 19.06 m/s

Now, maximum height attained is given by:

h = \frac{(vsin\theta)^{2}}{2g}

h = \frac{(19sin(22^{\circ}))^{2}}{2\times 9.8} = 2.60 m

Height from the ground = 5sin22^{circ} + 2.86 = 1.87 + 2.60 = 4.47m

(b) now, considering the coefficient of friction bhetween ramp and the ball, \mu = 0.150:

velocity can be given by the eqn-3 of motion:

v^{2} = v_{o}^{2} - 2gH - \mu gd

v^{2} = 20^{2} - 2\times 9.8\times 5sin22^{\circ} - 0.150\times 9.8\times 5

v = \sqrt{400 - 19.6\times 5sin22^{\circ} - 0.150\times 9.8\times 5} = 18.7 m/s

Now, maximum height attained is given by:

h = \frac{(vsin\theta)^{2}}{2g}

h = \frac{(18.7sin(22^{\circ}))^{2}}{2\times 9.8} = 2.50 m

Height from the ground = 5sin22^{circ} + 2.86 = 1.87 + 2.50 = 4.37 m

6 0
2 years ago
An observer is approaching at stationary source at 17.0 m/s. Assuming the speed of sound is 343 m/s, what is the frequency heard
UNO [17]

Answer:

the frequency heard by the observer is equal to 2677 Hz

Explanation:

given,      

velocity of the observer = 17 m/s

speed of the sound = 343 m/s    

velocity of the source = 0 m/s    

frequency emitted from the source  = 2550 Hz              

f = f_0(\dfrac{v-v_0}{v-v_s})              

f = 2550\times (\dfrac{343+17}{343-0})

velocity of observer is negative as it is approaching the source.                   f = 2676.38 Hz ≈ 2677 Hz                    

hence, the frequency heard by the observer is equal to 2677 Hz

7 0
3 years ago
A current of 0. 82 a flows through a light bulb. how much charge passes through the light bulb during 94 s?
gladu [14]

A current of 0. 82A flows through a light bulb. The charge passed through the light bulb during 94 s is 77.08C

The amount of charge flown for a given period of time determines the current passed through a bulb or electrical body.

The relation between the charge, current and time is given as:

Q = I × t

where, Q is the charge flown through bulb

I is the current passed through bulb

t is the time for which charge passes through bulb

Given,

I = 0.82A

t = 94s

Q = ?

Substituting the values in the above formula:

Q = I × t

Q = 0.82 × 94

Q = 77.08C

Hence, The charge passed through the light bulb during 94 s is 77.08C

Learn more about Current here, brainly.com/question/2264542

#SPJ4

4 0
1 year ago
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