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3 years ago

9

What is the potential energy of a 2,000-kg car parked at the top of a 30-m hill?

2 answers:

babymother [125]3 years ago

5 0

Potential energy is a relative measure, so the answer is dependent on the assumptions we make. The potential energy in the car is going to be gravitational potential energy(PE). PE = mgh, where m is the mass, g is 9.8 m/s^2, and h is the height. So PE = 2000*9.8*h = 19600h. The final answer obviously depends on h. Most likely the problem is assuming that 30 meters under the top of the hill is considered 0 meters. Then h would be 30m and PE would equal 588 kJ.

vampirchik [111]3 years ago

5 0

Answer:

the answer is 588,000 J

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A small sphere has a harge of 9uC and other small sphere has a charge of 4uC.

Helga [31]

Answer:

Electrical force, F = 90 N

Explanation:

It is given that,

Charge on sphere 1, $q_1=9\ \mu C=9\times 10^{-6}\ C$

Charge on sphere 2, $q_1=4\ \mu C=4\times 10^{-6}\ C$

Distance between two spheres, d = 6 cm = 0.06 m

Let F is the electrical force between them. It is given by the formula of electric force which is directly proportional to the product of charges and inversely proportional to the square of distance between them such that,

$F=k\dfrac{q_1q_2}{d^2}$

$F=9\times 10^9\times \dfrac{9\times 10^{-6}\times 4\times 10^{-6}}{(0.06)^2}$

F = 90 N

So, the electrical force between them is 90 N. Hence, this is the required solution.

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3 years ago

A -5.0 μC charge experiences a 11 i^ N electric force in a certain electric field. [Recall that i^ is a unit vector in the x-dir

Pachacha [2.7K]

Answer:

$\vec{F}= -3.52\times 10^{-13}\hat{i}\ N$

Explanation:

given,

charge = -5.0 μC

Electric force, F = 11 i^ N

force would a proton experience = ?

we know

$\vec{F} = q \vec{E}$

$11\hat{i} = -5 \times 10^{-6} \vec{E}$

$\vec{E} =-2.2 \times 10^{6}\hat{i}$

we know charge of proton is equal to 1.6 x 10⁻¹⁹ C

using formula

$\vec{F} = q \vec{E}$

$\vec{F}= 1.6 \times 10^{-19}\times -2.2 \times 10^{6}\hat{i}$

$\vec{F}= -3.52\times 10^{-13}\hat{i}\ N$

Force experienced by the photon in the same field is equal to $\vec{F}= -3.52\times 10^{-13}\hat{i}\ N$

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algol [13]

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nikitadnepr [17]

Answer:

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Explanation:

According to Malu's law the intensity of a polarized light having an initial intensity $I_0$ is mathematically represented as

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