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Olegator [25]
3 years ago
6

select all that apply the force of gravity a. changes slightly with the location on the earth b. decreases with height above sea

level c. is unaffected by altitude
Physics
1 answer:
Reptile [31]3 years ago
7 0
The answer to your question is C
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Dolphins communicate using compression waves (longitudinal waves). Some of the sounds dolphins make are outside the range of hum
Sphinxa [80]
Water is a really good conductor of sound so I would have to say that it would be to send the message underwater because a more dense medium produces a louder sound
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3 years ago
 Why does a child in a wagon seem to fall backward when you give the wagon a sharp pull forward? 15. What force is needed to acc
slavikrds [6]

Answer:

Force, F = 77 N

Explanation:

A child in a wagon seem to fall backward when you give the wagon a sharp pull forward. It is due to Newton's third law of motion. The forward pull on wagon is called action force and the backward force is called reaction force. These two forces are equal in magnitude but they acts in opposite direction.

We need to calculate the force is needed to accelerate a sled. It can be calculated using the formula as :

F = m × a

Where

m = mass = 55 kg

a = acceleration = 1.4 m/s²

F=55\ kg\times 1.4\ m/s^2

F = 77 N

So, the force needed to accelerate a sled is 77 N. Hence, this is the required solution.

3 0
3 years ago
Based on its orbit, which planet behaves the least like the others?
Gekata [30.6K]
I believe it's Mercury, because the only other option would be Pluto and it's not even considered a planet anymore
Hope this helps
3 0
3 years ago
Tom rides his motorcycle at a speed of 15 meters/second for an hour.
Dahasolnce [82]

Answer:

1 hour to ride his motorcycle

4 0
3 years ago
A stone with a mass of 0.100kg rests on a frictionless, horizontal surface. A bullet of mass 2.50g traveling horizontally at 500
jolli1 [7]

Answer:

Explanation:

Given that:

mass of stone (M) = 0.100 kg

mass of bullet (m) = 2.50 g = 2.5  ×10 ⁻³ kg

initial velocity of stone (u_{stone}) = 0 m/s

Initial velocity of bullet (u_{bullet}) = (500 m/s)i

Speed of the bullet after collision (v_{bullet}) = (300 m/s) j

Suppose we represent (v_{stone}) to be the velocity of the stone after the truck, then:

From linear momentum, the law of conservation can be applied which is expressed as:

m*u_{bullet} + M*{u_{stone}}= mv_{bullet}+Mv_{stone}

(2.50*10^{-3} \ kg) (500)i+0 = (2.50*10^{-3} \ kg)(300 \ m/s)j + (0.100 \ kg)v_{stone}

(2.50*10^{-3} \ kg) (500)i- (2.50*10^{-3} \ kg)(300 \ m/s)j=  (0.100 \ kg)v_{stone}

v_{stone}= (1.25\  kg.m/s)i-(0.75\ kg m/s)j

v_{stone}= (12.5\  m/s)i-(7.5\ m/s)j

∴

The magnitude now is:

v_{stone}=\sqrt{ (12.5\  m/s)^2-(7.5\ m/s)^2}

\mathbf{v_{stone}= 14.6 \ m/s}

Using the tangent of an angle to determine the direction of the velocity after the struck;

Let θ represent the direction:

\theta = tan^{-1} (\dfrac{-7.5}{12.5})

\mathbf{\theta = 30.96^0 \ below \ the \ horizontal\ level}

5 0
3 years ago
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