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2 years ago

A charged object traveling 7 m in a uniform electric field of 5 N/C experiences a 4 J increase in Kinetic Energy.

1 answer:

3 0

To solve this problem it is necessary to apply the principles of conservation of Energy in order to obtain the final work done.

The electric field in terms of the Force can be expressed as

$E = \frac{F}{q} \rightarrow F=Eq$

Where,

F = Force

E= Electric Field

q = Charge

Puesto que el trabajo realizado es equivalente al cambio en la energía cinetica entonces tenemos que

KE = W

KE = F*d

In the First Case,

$4 = (qE)d\\q = \frac{4}{Ed}\\q = \frac{4}{5*7}\\q = 0.1142C$

In Second Case,

$KE = q E'd$

$KE = (0.1142)(40)(7)$

$KE = 31.976J$

The total energy change would be subject to,

$\Delta KE = 31.976-4$

$\Delta KE = 27.976J$

Therefore the Kinetic Energy change of the charged object is 27.976J

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If an exit sign is hanging from the ceiling by three chains each of which has a tension of 15N what is the weight of the sign

kvasek [131]

The weight is 45 N, because the three chains hold the sign, and each contributes 15 N.

Notice that the mass would be the weight/acceleration of gravity, m = 45/9.8 kg. But they ask the weight (force, so Newtons)

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3 years ago

A 57-kg woman holds a 6-kg package as she stands within an elevator which briefly accelerates upward at a rate of 0.15g. Determi

Temka [501]

Answer:

R = 710.7N

L = 67.689 N

During gravity fall L = R = 0 N

Explanation:

So the acceleration that the elevator is acting on the woman (and the package) in order to result in a net acceleration of 0.15g is

g + 0.15g = 1.15g

The force R that the elevator exerts on her feet would be product of acceleration and total mass (Newton's 2nd law):

a(m + M) = 1.15g(57 + 6) = 1.15*9.81*63 = 710.7N

The force L that she exerts on the package would be:

am = 1.15g *6 = 1.15*9.81*6 = 67.689N

When the system is falling, all have a net acceleration of g. So the acceleration that the elevator exerts on the woman (and the package) is 0, and so are the forces L and R.

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2 years ago

The surface charge density on an infinite charged plane is - 2.10 ×10−6C/m2. A proton is shot straight away from the plane at 2.

inn [45]

Explanation:

Formula to calculate electric field because of the plate is as follows.

E = $\frac{\sigma}{2 \times \epsilon_{o}}$