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deff fn [24]
3 years ago
7

INK]: A car is moving at 200 km/hr east. What is its velocity?

Physics
1 answer:
Greeley [361]3 years ago
5 0

Answer:

200 km\h

or 0.621 mp\h its the same speed just different mesuarements

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The gravitational force would get stronger because the farther the two masses are separated the more gravitational force will be used to pull them together the closer they are the less gravitational pull is used to pull them together
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When a +0.00235 C charge
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Light waves have some similarities with water and sound waves, but they are not exactly the same. Describe all the differences y
makkiz [27]

<u>Answer:</u>


<h2>All the waves are pertubations that propagate (transport) energy.</h2><h2></h2>

Nevertheless, they have some differences:


1. Light waves are<u> electromagnetic waves</u>, while sound and water waves are <u>mechanical waves</u>, this is the first and principal difference.  

2. Electromagnetic waves can<u> propagate in vacuum</u> (they do not need a medium or material), but mechanical waves obligatory need a material to propagate

3. Light waves are always <u>transversal waves</u>, this means <u>the oscillatory movement is in a direction that is perpendicular to the propagation</u>; but mechanical waves may be both: <u>longitudinal waves</u> (the oscillation occurs in the same direction as the propagation) or transversal waves.

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3 0
3 years ago
2. A rock is shot straight up into the air with a slingshot that had been stretched 0.30 m. Assume
Luba_88 [7]

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e

Explanation:

4 0
3 years ago
Read 2 more answers
A body is thrown up with a velocity of 78.4 m per second.How high will it rise and how much time it will take to return to its p
a_sh-v [17]

Answer:

The maximum height reached by the body is 313.6 m

The time to return to its point of projection is 8 s.

Explanation:

Given;

initial velocity of the body, u = 78.4 m/s

at maximum height (h) the final velocity of the body (v) = 0

The following equation is applied to determine the maximum height reached by the body;

v² = u² - 2gh

0 = u² - 2gh

2gh = u²

h = u²/2g

h = (78.4²) / (2 x 9.8)

h = 313.6 m

The time to return to its point of projection is calculated as follows;

at maximum height, the final velocity becomes the initial velocity = 0

h = v + ¹/₂gt²

h = 0 + ¹/₂gt²

h =  ¹/₂gt²

2h = gt²

t² = 2h/g

t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 313.6}{9.8} }\\\\t = 8 \ s

4 0
3 years ago
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