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Setler [38]
4 years ago
8

In a charging process, 4 × 1013 electrons are removed from one small metal sphere and placed on a second identical sphere. Initi

ally, both metal spheres were neutral. After the charging process, the electrical potential energy associated with the two spheres is found to be −0.063 J. What is the distance between the two spheres?
Physics
1 answer:
Alina [70]4 years ago
3 0

Answer:

The distance between the two spheres is 914.41 X 10³ m

Explanation:

Given;

4 X 10¹³ electrons, and its equivalent in coulomb's is calculated as follows;

1 e = 1.602 X 10⁻¹⁹ C

4 X 10¹³ e = 4 X 10¹³ X 1.602 X 10⁻¹⁹ C = 6.408 X 10⁻⁶ C

V = Ed

where;

V is the electrical potential energy between two spheres, J

E is the electric field potential between the two spheres N/C

d is the distance between two charged bodies, m

V = \frac{K*q}{d^2}*d = \frac{K*q}{d}

d = \frac{K*q}{V}

where;

K is coulomb's constant = 8.99 X 10⁹ Nm²/C²

d = (8.99 X 10⁹ X 6.408 X 10⁻⁶)/0.063

d = 914.41 X 10³ m

Therefore, the distance between the two spheres is 914.41 X 10³ m

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3 years ago
A person is standing on and facing the front of a stationary skateboard while holding a construction brick. The mass of the pers
jek_recluse [69]

Answer:

0.74 m/s

Explanation:

From the question,

We apply the law of conservation of momentum,

Total momentum before collision = Total momentum after collision.

Since the skateboard, the person and the brick where stationary, therefore, the total momentum before collision is 0

0 = Total momentum after collision

(m+M)V + m'v = 0

Where m = mass of the  skateboard, M = mass of the person, m' = mass of the brick, V = recoil velocity of the person and the skateboard, v =  velocity of the brick

make V the subject of the equation above

V = -m'v/(m+M)................... Equation 1

Given: m = 4.10 kg, M = 68.0 kg, m' = 2.50 kg, v = 21.0 m/s.

Substitute these values into equation 1

V = -(2.5×21)/(68+2.5)

V = 52.50/70.5

V = 0.74 m/s

4 0
3 years ago
Model rocket engines are sized by thrust, thrust duration, and total impulse, among other characteristics. A size C5 model rocke
umka2103 [35]

Answer:

v_{f} = 115.95 m / s

Explanation:

This is an exercise of a variable mass system, let's form a system formed by the masses of the rocket, the mass of the engines and the masses of the injected gases, in this case the system has a constant mass and can be solved using the conservation the amount of movement. Which can be described by the expressions

        Thrust = v_{e}  \frac{dM}{dt}

        v_{f}-v₀ = v_{e} ln ( \frac{M_{o} }{M_{f}} )

where v_{e} is the velocity of the gases relative to the rocket

let's apply these expressions to our case

the initial mass is the mass of the engines plus the mass of the fuel plus the kill of the rocket, let's work the system in SI units

       M₀ = 25.5 +12.7 + 54.5 = 92.7 g = 0.0927 kg

     

The final mass is the mass of the engines + the mass of the rocket

      M_{f} = 25.5 +54.5 = 80 g = 0.080 kg

thrust and duration of ignition are given

       thrust = 5.26 N

       t = 1.90 s

Let's start by calculating the velocity of the gases relative to the rocket, where we assume that the rate of consumption is linear

          thrust = v_{e} \frac{M_{f} - M_{o}  }{t_{f} - t_{o}  }

          v_{e} = thrust  \frac{\Delta t}{\Delta M}

          v_{e} = 5.26 \frac{1.90}{0.080 -0.0927}

          v_{e} = - 786.93 m / s

the negative sign indicates that the direction of the gases is opposite to the direction of the rocket

now we look for the final speed of the rocket, which as part of rest its initial speed is zero

            v_{f}-0 = v_{e} ln ( \frac{M_{o} }{M_{f} } )

we calculate

            v_{f} = 786.93 ln (0.0927 / 0.080)

            v_{f} = 115.95 m / s

5 0
3 years ago
A 17926-lb truck enters an emergency exit ramp at a speed of 75.6 ft/s. It travels for 6.4 s before its speed is reduced to 30.3
I am Lyosha [343]

Answer:

F_{braking}=337299 pdl

Explanation:

Impulse-Momentum relation:

I=\Delta p\\ F_{total}*t=m(v_{f}-v{o})

F_{total}=-F_{braking}+mgsin{\theta}

We solve the equations in order to find the braking force:

F_{braking}=m(v_{o}-v{f})/t+mgsin{\theta}=17926(75.6-30.3)/6.4+17926*32.17*sin21.4=337299 pdl

6 0
3 years ago
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