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Sladkaya [172]
3 years ago
11

W=4j, d=2m,and F=10N find cosine

Physics
1 answer:
ruslelena [56]3 years ago
4 0
Correct your answer to the amount of significant figures the questions wants

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A man on the Moon observes two spaceships coming toward him from opposite directions at speeds of 0.600c and 0.600c. What is the
vaieri [72.5K]

Answer:

If we use the equation for the transformation of velocities for moving frames:

v' = (v - u) / (1 - u * v / c^2) where we measure the speed of v' approaching from the left where v is in a frame moving at -u towards v'

v' = (.6 c - (-.6 c)) / (1 - (-.6 c) * .6 c / c^2) = 1.2 c / (1 + .6 * .6)

or v' = 1.2 c / (1 + .36) = .88 c

v is approaching from the left at .6 c in the reference frame and the other frame approaches from the right at -.6 c with speed u  (-.6 c) and we measure the speed of v as seen in the frame moving to the left

5 0
3 years ago
On a cloudless day, the sunlight that reaches the surface of the earth has an intensity of about W/m². What is the electromagnet
Otrada [13]

Answer:

1.907 x 10⁻⁵ J.

Explanation:

Given,

Volume of space, V = 5.20 m³

Assuming the intensity of sunlight(S) be equal to 1.1 x 10³ W/m².

Electromagnetic energy = ?

E = \mu V

E = (\dfrac{S}{c})\times V

where c is the speed of light.

E = (\dfrac{1.1\times 10^3}{3\times 10^8})\times 5.20

E = 1.907\times 10^{-5}\ J

Hence, Electromagnetic energy is equal to 1.907 x 10⁻⁵ J.

4 0
3 years ago
List three ways in which the study of science has made modern life different from that of 100 years ago​
stealth61 [152]
Life Expectancy Was Shorter
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8 0
2 years ago
Which conditions are low air pressure systems usually associated with?
Inga [223]

cloudy, wet weather                                          

4 0
3 years ago
Read 2 more answers
Suppose a 48-N sled is resting on packed snow. The coefficient of kinetic friction is 0.10. If a person weighing 660 N sits on t
Annette [7]

Assume the snow is uniform, and horizontal.

Given:

coefficient of kinetic friction = 0.10 = muK

weight of sled = 48 N

weight of rider = 660 N

normal force on of sled with rider = 48+660 N = 708 N = N

Force required to maintain a uniform speed

= coefficient of kinetic friction * normal force

= muK * N

= 0.10 * 708 N

=70.8 N


Note: it takes more than 70.8 N to start the sled in motion, because static friction is in general greater than kinetic friction.


8 0
3 years ago
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