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3 years ago

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A car moving with a velocity 15m/s and accelerating by 5m/s² attempts to reach a car moving with 30 m/s velocity.What distance s

hould it cover to reach the velocity of other car?
A)30 m
B) 90 m
C) 67 m
D) 70m

1 answer:

Firlakuza [10]3 years ago

4 0

I think A is correct

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seraphim [82]

Answer:

yes

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3 years ago

Two parallel plates have equal and opposite charges. When the space between the plates is evacuated, the electric field is E= 3.

svetlana [45]

Explanation:

The given data is as follows.

Electric field between plates without dielectric, $E_{1} = 3.50 \times 10^{5} V/m$

Electric field between the plates with dielectric, $E_{2} = 2.40 \times 10^{5} V/m$ .

Permittivity of free space, $\epsilon_{o}$ = $8.85 \times 10^{-12} C^{2}/Nm^{2}$

Now, we will determine the charge density as follows.

$\sigma_{i} = \epsilon_{o}(E_{1} - E_{2})$

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= $9.735 \times 10^{-7} C/m^{2}$

Thus, we can conclude that the charge density on each surface of the dielectric is $9.735 \times 10^{-7} C/m^{2}$ .

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3 years ago

Why it is advise to clean the ends of connecting wires before connecting them?

Svetradugi [14.3K]

If dirt and grease were good conductors of electrical current, then we could make wire
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3 years ago

Read 2 more answers

Discuss the effects of breaking a food chain.

Tom [10]

Explanation:

Food chain is linear sequence of the organisms through which the nutrients and the energy pass as one organism eats the another.

<u>If one species in the food chain ceases to exist, the other members in rest of chain could cease to exist too or may increase in population which can affect other food chains. (Breaking of food chain).</u>

This is because the animals in the food chain are dependent on the nutrient present in the another.

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3 years ago

A charge of -3.02 μC is fixed in place. From a horizontal distance of 0.0377 m, a particle of mass 9.43 x 10^-3 kg and charge -9

Andreyy89

Answer:

$d = 0.0306 m$

Explanation:

Here we know that for the given system of charge we have no loss of energy as there is no friction force on it

So we will have

$U + K = constant$

$\frac{kq_1q_2}{r_1} + \frac{1}{2}mv_1^2 = \frac{kq_1q_2}{r_2} + \frac{1}{2}mv_2^2$

now we know when particle will reach the closest distance then due to electrostatic repulsion the speed will become zero.

So we have

$\frac{(9 \times 10^9)(3.02 \mu C)(9.78 \mu C)}{0.0377} + \frac{1}{2}(9.43 \times 10^{-3})(80.4)^2 = \frac{(9 \times 10^9)(3.02 \mu C)(9.78 \mu C)}{r} + 0$

$7.05 + 30.5 = \frac{0.266}{r}$

$r = 7.08 \times 10^{-3} m$

so distance moved by the particle is given as

$d = r_1 - r_2$

$d = 0.0377 - 0.00708$

$d = 0.0306 m$

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3 years ago