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Ira Lisetskai [31]

2 years ago

6

What happened to the kinetic energy when the velocy of a moving body is doubled?

2 answers:

dem82 [27]2 years ago

8 0

Answer:

If velocity is doubled, Kinetic Energy increases by 4 times. Kinetic energy of a body is the energy possessed by it, by virtue of its motion, i.e. if the body is moving it will always have kinetic energy.Nov 1, 2018

Explanation:

kicyunya [14]2 years ago

5 0

Answer:

If velocity is doubled, Kinetic Energy increases by 4 times. Kinetic energy of a body is the energy possessed by it, by virtue of its motion, i.e. if the body is moving it will always have kinetic energy :)

Explanation:

hope this helps <3

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Compare metamorphic rock and intrusive igneous rock in terms of how and where they form. Then compare sedimentary rock and extru

Mrac [35]

Answer: Igneous rock , formed by the cooling of magma (molten rock) inside the Earth or on the surface. Sedimentary rocks, formed from the products of weathering by cementation or precipitation on the Earth's surface. Metamorphic rocks, formed by temperature and pressure changes inside the Earth.

Explanation:

3 0

2 years ago

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Find electric field at point p which is a distance l away from the both +q and -q

denis-greek [22]

Answer:

$\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }$

Explanation:

As given point p is equidistant from both the charges

It must be in the middle of both the charges

Assuming all 3 points lie on the same line

Electric Field due a charge q at a point ,distance r away

$=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{r^{2} }$

Where

q is the charge
r is the distance
$E$ is the permittivity of medium

Let electric field due to charge q be F1 and -q be F2

I is the distance of P from q and also from charge -q

⇒

F1 $=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }$

F2 $=\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }$

⇒

F1+F2= $\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }$

8 0

3 years ago

A 300-n force acts on a 25-kg object. the acceleration of the object is?

Over [174]

To solve the answer use the equation: a = fnet / m

a = 300 N / 25 kg

300 N / 25 kg = 12m/s

The acceleration of the object is 12m/s

8 0

2 years ago

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4. An airplane maintains a constant acceleration of 4.0 m/s2 [E] as it speeds up from 16 m/s [E] to 28 m/s [E].

ki77a [65]

Answer:

gyvtu jyut

Explanation:

6 0

3 years ago

Air "breaks down" when the electric field strength reaches 3 × 106 n/c, causing a spark. A parallel-plate capacitor is made from

Lera25 [3.4K]

Electric field between the plates of parallel plate capacitor is given as

$E = \frac{Q}{A\epsilon_0}$

here area of plates of capacitor is given as

$A = 0.055 * 0.055$

$A = 3.025 * 10^{-3}$

also the maximum field strength is given as

$E = 3 * 10^6 N/C$

now we will plug in all data to find the maximum possible charge on capacitor plates

$3 * 10^6 = \frac{Q}{3.025 * 10^{-3}*8.85 * 10^{-12}}$

$Q = 8.03 * 10^{-8} C$

so the maximum charge that plate will hold will be given by above

3 0

3 years ago

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