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dangina [55]
3 years ago
8

Types of energy i really need this answer

Physics
2 answers:
anygoal [31]3 years ago
8 0

Answer:

These are the two basic forms of energy. The different types of energy include thermal energy, radiant energy, chemical energy, nuclear energy, electrical energy, motion energy, sound energy, elastic energy and gravitational energy.

Explanation:

Digiron [165]3 years ago
6 0
Thermal and nuclear energy
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pashok25 [27]
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8 0
4 years ago
Jupiter's moon Io has active volcanoes (in fact, it is the most volcanically active body in the solar system) that eject materia
kramer

Answer:

The height reached by the material on Earth is 91 km.

Explanation:

Given that,

Mass M_{Io}=8.93\times10^{22}\ kg

Radius = 1821 km

Height h_{Io}=500\ km

Suppose we need to find that how high would this material go on earth if it were ejected with the same speed as on Io?

We need to calculate the acceleration due to gravity on Io

Using formula of gravity

g =\dfrac{GM_{Io}}{(R_{Io})^2}

Put the value into the formula

g=\dfrac{6.67\times10^{-11}\times8.93\times10^{22}}{(1821\times10^{3})^2}

g=1.79\ m/s^2

Let  v be the speed at which the material is ejected.

We need to calculate the height

Using the formula of height

H=\dfrac{v^2}{2g}

Using ratio of height of earth and height of Io

\dfrac{H_{e}}{H_{Io}}=\dfrac{\dfrac{v^2}{2g_{e}}}{\dfrac{v^2}{2g_{Io}}}

\dfrac{H_{e}}{H_{Io}}=\dfrac{g_{Io}}{g_{e}}

Put the value into the formula

\dfrac{H_{e}}{H_{Io}}=\dfrac{1.79}{9.8}

\dfrac{H_{e}}{H_{Io}}=0.182

H_{e}=0.182\times H_{Io}

H_{e}=0.182\times500

H_{e}=91\ km

Hence, The height reached by the material on Earth is 91 km.

3 0
3 years ago
A car in a roller coaster moves along a track that consists of a sequence of ups and downs. Let the
olga_2 [115]

Answer:

Explanation:

r=A(1 m/s)t^i+A((1 m/s3)t3−6(1 m/s2)t2)^j,

dr / dt = A i  + (3 A t² - 6 x 2 t ) j

At t = 2

dr / dt = A i  + (12A - 24 A ) j  = A i - 12 A j .

Since slope is negative so velocity is downwards , hence it is descending

b ) velocity = dr / dr =  .  A i  + (3 A t² - 6 x 2 t ) j

Vx = A

Vy = 3A t² - 12 t

3 0
3 years ago
How can oxygen enter the lungs?
Oksanka [162]
I just got this question not to long ago but to answer your question the primary function of the respiratory system is to exchange oxygen<span> and carbon dioxide. Inhaled </span>oxygen enters the lungs<span> and reaches the alveoli. </span>Oxygen<span> passes quickly through this air-blood barrier into the blood in the capillaries. Similarly, carbon dioxide passes from the blood into the alveoli and is then exhaled.</span>
6 0
3 years ago
A charged box (m=445 g, ????=+2.50 μC) is placed on a frictionless incline plane. Another charged box (????=+75.0 μC) is fixed i
victus00 [196]

The concept required to perform this exercise is given by the coulomb law.

The force expressed according to this law is given by

F= \frac{kqQ}{r^2}

Where,

k = 8.99 * 10^9 N m^2 / C^2.

q = charges of the objects

r= distance/radius

Our values are previously given, so

q= 2.5*10^{-6}C\\Q= 75*10^{-6}C\\r=0.59

Replacing,

F=\frac{kqQ}{r^2}

F= \frac{(8.99 x 10^9)(2.5*10^{-6})(75*10^{-6})}{0.59^2}

F= 4.8423N

The force acting on the block are given by,

F-mgsin\theta = ma

a = \frac{F-mgsin\theta}{m}

a = \frac{4.8423-(0.445)(9.8)sin(35)}{0.445}a = 10.31m/s^2

Therefore the box is accelerated upward.

3 0
3 years ago
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