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eimsori [14]
4 years ago
6

Which characteristics is common to the four outer planets in out solar system ?

Physics
2 answers:
Lubov Fominskaja [6]4 years ago
8 0
I think the answer is B
Dvinal [7]4 years ago
6 0
THE ANSWER IS A I AM PRETTY SURE hahaha
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What process produces radiant energy in stars?
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"Nuclear Fusion" <span>produces radiant energy in stars

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true or false: atoms combine in different ways to make up all of the substances you encounter everyday
ioda
Yes, atoms are identified as the particles of an element...these consist of electron,protons and neutrons...that form everything we see and are able to touch
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4 years ago
The Sankey diagrams below show the energy transfers in two light bulbs. What is the efficiency of light bulb 1? 1: 80j Input ene
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7 0
3 years ago
A weightlifter works out at the gym each day. Part of her routine is to lie on her back and lift a 43 kg barbell straight up fro
Tasya [4]

Answer:

A. 231.77 J

B. 5330.71 J

C. 46 donuts

Explanation:

A. To lift the barbell once, she will have to extend it the full length of her arm. The work done will then be:

W = F * d

Where the force is the weight of the barbell.

F = m * g

Hence, the work done in lifting the barbell is:

W = m * g * d

W = 43 * 9.8 * 0.55

W = 231.77 J

B. If she does 23 repetitions, the total energy she expend will be equal to the Potential energy when the barbell is lifted multiplied by 23:

E = 23 * m * g * d

E = 23 * 231.77

E = 5330.71 J

C. 1 Joule = 4.184 calories

5330.71 Joules = 5330.71 * 4.184 = 22303.69

If 1 donut contains 490 calories, the number of donuts she will need will be:

N = 22303.69/490 = 45.5 donuts or 46 donuts

5 0
4 years ago
A cannon shoots an artillery shell with a initial velocity of 400 meters/second at an indirect fire angle of 60 on level ground
notsponge [240]

Answer: 14139.19 m

Explanation:

This situation is related to parabolic motion and can be solved using the following equations:

x=V_{o}cos \theta t (1)

y=y_{o}+V_{o} sin \theta t+\frac{g}{2}t^{2} (2)

Where:

x is the horizontal distance (where the artillery shell lands)

V_{o}=400 m/s is the initial velocity

\theta=60\° is the angle

t is the time

y=0 m is the final height

y_{o}=0 m is the initial height

g=-9.8 m/s^{2} is the acceleration due gravity, always directed downwards

So, let's begin by isolating t from (2):

0=V_{o} sin \theta t+\frac{g}{2}t^{2} (3)

t=-\frac{2 V_{o}sin \theta}{g} (4)

Substituting (4) in (1):

x=V_{o}cos \theta (-\frac{2 V_{o}sin \theta}{g}) (5)

Rewriting (5) and taking into account sin(2\theta)=2 sin \theta cos \theta:

x=-\frac{V_{o}^{2}sin(2\theta)}{g} (6)

x=-\frac{(400 m/s)^{2}sin(2(60\°))}{-9.8 m/s^{2}} (7)

Finally:

x=14139.19 m

6 0
3 years ago
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