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frozen [14]
3 years ago
14

HOW MANY KILOMETERS IS EARTH FROM THE SUN?

Physics
2 answers:
maria [59]3 years ago
7 0
Earth is 150 million kilometers away for the sun 

dlinn [17]3 years ago
3 0
The Sun, the star of our Galaxy, the reason of our life, is at about 150 million kilometers from us: The Earth!
But in January, it gets a little more closer to 146 Million kilometers (that's the closest possible), and the furthest is 152 million kilometers in July.

Hope this Helps! :)
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When light strikes an opaque material, which of the following accurately describes what happens to the light rays? Some of the l
ExtremeBDS [4]
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Describe five things that you do to make yourself feel warmer or cooler.
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Buy a fan or heater, get a blanket, take a shower, go outside, go by the fireplace
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A friend claims that there are just two forms of energy: kinetic energy and gravitational potential energy. What is wrong with y
Oksanka [162]

Answer:

D

Explanation:

There is a variety of many types of energies. Some of which are mechanical, thermal, electromagnetic, nuclear, chemical, sonic, and more.

7 0
3 years ago
Pete Zaria works on weekends at Barnaby's Pizza Parlor. His primary responsibility is to fill drink orders for customers. He fil
laila [671]

Answer:

W_n_e_t=7.648512 \approx 7.6J

K.E=0.8J

v=0.7844645406 \approx 0.78m/s

Explanation:

From the question we are told that

Mass of pitcher   M= 2.6kg

Force on pitcher f=8.8N

Distance traveled 48cm=>0.48m

Coefficient of friction \mu=0.28

a)Generally frictional force is mathematically given by

F=\mu N

F=0.28*2.6*9.8

F=7.1344N

Generally work done on the pitcher is mathematically given as

W_n_e_t=W_f+W_F

W_f=8.8*0.48=>4.224N\\W_F=7.1344*0.48=>3.424512N

W_n_e_t=4.224-3.424512

W_n_e_t=0.799488\approx 0.8J

b)Generally K.E can be given mathematically as

K.E= W_n_e_t

Therefore

K.E=0.8J

c)Generally the equation for kinetic energy is mathematically represented by

K.E=1/2mv^2

0.8=1/2mv^2

Velocity as subject

v=\sqrt{\frac{K.E*2}{m} }

v=\sqrt{\frac{0.8*2}{2.6} }

v=0.7844645406 \approx 0.78m/s

6 0
3 years ago
An 89 kg man drops from rest on a diving board −3.1 m above the surface of the water and comes to rest 0.5 s after reaching the
OLga [1]

To solve this problem we will use the linear motion kinematic equations, for which the change of speed squared with the acceleration and the change of position. The acceleration in this case will be the same given by gravity, so our values would be given as,

m= 89 kg\\x = 3.1 m\\t = 0.5s\\a = g = 9.8m/s^2

Through the aforementioned formula we will have to

v_f^2-v_i^2 = 2ax

The particulate part of the rest, so the final speed would be

v_f^2 = 2gx

v_f=\sqrt{2(9.8)(3.1)}

v_f = 7.79m/s

Now from Newton's second law we know that

F = ma

Here,

m = mass

a = acceleration, which can also be written as a function of velocity and time, then

F = m\frac{dv}{dt}

Replacing we have that,

F = (89)\frac{7.79}{0.5}

F = 1386.62N

Therefore the force that the water exert on the man is 1386.62

3 0
3 years ago
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