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IceJOKER [234]
3 years ago
11

Which type of energy is stored in fuels such as gasoline and batteries

Physics
1 answer:
AnnyKZ [126]3 years ago
7 0

Answer:

Chemical Energy

Explanation:

Chemical energy is energy stored in the bonds of atoms and molecules. Batteries, biomass, petroleum, natural gas, and coal are examples of chemical energy.

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A poker is a long thin tool used to move pieces of coal or logs burning in a fire. To be as safe as possible, the poker should b
nata0808 [166]

Answer:

See the answer below

Explanation:

A poker that will effectively and safely function to move pieces of coal or logs in a burning fire must be fireproof itself. Hence, to be as safe as possible, such <u>poker should be made from a material that is fireproof</u> and that does not conduct a lot of heat. Otherwise, the poker will catch fire/becomes too hot during the course of usage.

3 0
2 years ago
Show that rigid body rotation near the Galactic center is consistent with a spherically symmetric mass distribution of constant
irakobra [83]

To solve this problem we will use the concepts related to gravitational acceleration and centripetal acceleration. The equality between these two forces that maintains the balance will allow to determine how the rigid body is consistent with a spherically symmetric mass distribution of constant density. Let's start with the gravitational acceleration of the Star, which is

a_g = \frac{GM}{R^2}

Here

M = \text{Mass inside the Orbit of the star}

R = \text{Orbital radius}

G = \text{Universal Gravitational Constant}

Mass inside the orbit in terms of Volume and Density is

M =V \rho

Where,

V = Volume

\rho =Density

Now considering the volume of the star as a Sphere we have

V = \frac{4}{3} \pi R^3

Replacing at the previous equation we have,

M = (\frac{4}{3}\pi R^3)\rho

Now replacing the mass at the gravitational acceleration formula we have that

a_g = \frac{G}{R^2}(\frac{4}{3}\pi R^3)\rho

a_g = \frac{4}{3} G\pi R\rho

For a rotating star, the centripetal acceleration is caused by this gravitational acceleration.  So centripetal acceleration of the star is

a_c = \frac{4}{3} G\pi R\rho

At the same time the general expression for the centripetal acceleration is

a_c = \frac{\Theta^2}{R}

Where \Theta is the orbital velocity

Using this expression in the left hand side of the equation we have that

\frac{\Theta^2}{R} = \frac{4}{3}G\pi \rho R^2

\Theta = (\frac{4}{3}G\pi \rho R^2)^{1/2}

\Theta = (\frac{4}{3}G\pi \rho)^{1/2}R

Considering the constant values we have that

\Theta = \text{Constant} \times R

\Theta \propto R

As the orbital velocity is proportional to the orbital radius, it shows the rigid body rotation of stars near the galactic center.

So the rigid-body rotation near the galactic center is consistent with a spherically symmetric mass distribution of constant density

6 0
3 years ago
What does Newton's 2nd Law explain?
kow [346]

Answer:

C. Why you must push harder to move a car farther.

Explanation:

Newton's Second Law of Motion states that the acceleration of a physical object is directly proportional to the net force acting on the physical object and inversely proportional to its mass.

Mathematically, it is given by the formula;

Acceleration = \frac {Force}{mass}

Hence, Newton's 2nd Law explains why you must push harder to move a car farther because of its mass. Thus, it is important to increase the force that the engine provides and decrease the mass of the car.

6 0
2 years ago
Will mark BRAINLIEST!!!!
REY [17]

Answer:

Atoms found in nature are either stable or unstable. ... An atom is unstable (radioactive) if these forces are unbalanced; if the nucleus has an excess of internal energy. Instability of an atom's nucleus may result from an excess of either neutrons or protons

7 0
2 years ago
As part of a safety investigation, two 1900 kg cars traveling at 20 m/s are crashed into different barriers. Part A Find the ave
DedPeter [7]

Answer:

-29.2\times 10^{3} N

Explanation:

We are given that

Mass of cars= m=1900 kg

Initial speed of car=u=20 m/s

Final speed of car=v=0

Time=\Delta t=1.3 s

We have to find the average force exerted on the car.

Average force=\frac{change\;in\;momentum}{\Delta t}

F_{avg}=\frac{mv-mu}{1.3}

F_{avg}=\frac{1900(0)-1900(20)}{1.3}

F_{avg}=\frac{-38000}{1.3}=-29.2\times 10^{3} N

Hence, the average force exerted on the car that hits a line of water barrels=-29.2\times 10^{3} N

8 0
3 years ago
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