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MatroZZZ [7]
3 years ago
5

HELP PLEASE!! GIVING BRAINLIEST!! If you answer this correctly ill answer some of your questions you have posted! (20pts)

Physics
2 answers:
anyanavicka [17]3 years ago
8 0

Explanation:

potential energy =360800J

mass(m)=?

height (h)=25m

g=9.8m/s²

we have

potential energy =360800J

mgh=360800J

m×9.8×25=360800

m=360800/(9.8×25)=1472.653061kg

Archy [21]3 years ago
8 0

Answer:

HELP PLEASE!! GIVING BRAINLIEST!! If you answer this correctly ill answer some of your questions you have posted!

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What is the law of conservation of energy?
charle [14.2K]
The law of conservation of energy<span>, a fundamental concept of physics, states that the total amount of </span>energy<span> remains constant in an isolated system. It implies that </span>energy<span> can neither be created nor destroyed, but can be change from one form to another.</span>
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2 years ago
An amusement park ride consists of a rotating circular platform 8.26 m in diameter from which 10 kg seats are suspended at the e
VashaNatasha [74]

To solve this problem we will begin by finding the necessary and effective distances that act as components of the centripetal and gravity Forces. Later using the same relationships we will find the speed of the body. The second part of the problem will use the equations previously found to find the tension.

PART A) We will begin by finding the two net distances.

r = \frac{8.26}{2} = 4.13m

And the distance 'd' is

d = lsin\theta

d = 1.14 sin 16.2\°

d = 0.318m

Through the free-body diagram the tension components are given by

Tcos\theta = mg

Tsin\theta = \frac{mv^2}{R}

Here we can watch that,

R = r+d

Dividing both expression we have that,

tan\theta = \frac{v^2}{Rg}

Replacing the values,

tan(16.2) = \frac{v^2}{(4.13+0.318)(9.8)}

v = 4.83371m/s

PART B) Using the vertical component we can find the tension,

Tcos\theta = mg

T = \frac{mg}{cos\theta}

T = \frac{(10+26.2)(9.8)}{cos(16.2)}

T = 369.42N

6 0
3 years ago
What is the acceleration of a 10 kg mass pushed by a 5 N force
hichkok12 [17]

Answer:

The formula is a = F m so in this case a = 5 10 = 0.5 m s 2

Explanation:

3 0
2 years ago
Hey guys, i need some help. I'm having a physics test tmmrow and I understand nothing :(. Can anyone plz explain or give me a br
professor190 [17]

We think of sound as something we hear—something that makes noise. But in pure physics terms, sound is just a vibration going through matter.

The way a vibration “goes through” matter is in the form of a sound wave. When you think of sound waves, you probably think of something like this:1

But that’s not how sound waves work. A wave like that is called a transverse wave, where each individual particle moves up and down to create a snake situation.

A sound wave is more like an earthworm situation:2

Like an earthworm, sound moves by compressing and decompressing. This is called a longitudinal wave. A slinky can do both kinds of waves:13

Sound starts with a vibration of some kind creating a longitudinal wave through matter. Check this out:4

That’s what sound looks like—except picture an expanding ripple of spheres doing that. In this animation, the sound wave is being generated by that vibrating grey bar on the left. The bar might be your vocal chords, a guitar string, or a waterfall continually pounding down into the river below. By looking at the red dots, you can see that even though the wave moves in one direction, each individual particle only moves back and forth, mimicking the vibration of the gray bar.

So instead of a curvy snake wave, sound is a pressure wave, which causes each piece of the air to be at either higher-than-normal pressure or lower-than-normal pressure. So when you see a snake-like illustration of a sound wave, it’s referring to the measure of pressure, not the literal path of movement of the particles:5

6 0
3 years ago
in a softball game, a batter hits the ball at the velocity of 27m/s and angle of 40 shown below. What is the maximum range of th
Nata [24]

Answer:

R = 73.25 m

Explanation:

We have,

Initial speed of the ball is 27 m/s

It is projected at an angle of 40 degrees

The maximum range of the ball is given by :

R=\dfrac{u^2\sin2\theta}{g}

Plugging all the values we get :

R=\dfrac{(27)^2\sin2(40)}{9.8}\\R=73.25\ m

So, the maximum range of the ball is 73.25 m

8 0
3 years ago
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