A fuel-filled rocket is at rest. It burns its fuel and expels hot gas. The gas has a momentum of 1,500 kg x m/s backward. What i
1 answer:
You might be interested in
Answer:
a) -238 nC
b) 0.889 N
Explanation:
Concepts and Principles
<u>Particle in Equilibrium:</u> If a particle maintains a constant velocity (so that a = 0), which could include a velocity of zero, the forces on the particle balance and Newton's second law reduces to:
∑F = 0 (1)
<u>Coulomb's Law:</u> the magnitude of the electrostatic force exerted by a point charge q1 on a second point charge q2 separated by a distance r is directly proportional to the product of the two charges and is inversely proportional to the square of the distance between them:
F_12 = k*| q1 |*| q2 |/r^2 (2)
where k = 8.99 x 10^9 N m^2/C^2 is Coulomb constant.
<u>Given Data </u>
<em>mA (mass object A) = (83 g)*(1/1000g)=0.09 kg </em>
<em>qB (charge of object B) = (140 nC)*(1/10^9 nC) = 130 x 10^-9 C </em>
<em>Object A is attracted to object B. </em>
<em>Ф(angle made by object A with the vertical) = 7.2° </em>
<em>( r (distance between the two objects) = (5 cm) * (1 m/ 100 cm) =0.05 m </em>
<em>Object A is in equilibrium. </em>
Required Data
In part (a), we are asked to determine the charge qA of object A.
In part (b), we are asked to determine the tension T in the thread.
(a) The FBD in Figure 1 shows the forms acting on object A; Fe is the electric force exerted on object A by object B, T is the tension force exerted on the thread, and m_a*g is the gravitational force exerted on object A.
Model object A as a particle in equilibrium in the horizontal and vertical direction and apply Equation (1) to it:
∑F_x = F_e-Tsin = 0 F_e=Tsin<em>Ф </em><em>(3)</em>
∑F_y = Tcos<em>Ф - </em>m_a*g= 0 m_a*g=Tsin<em>Ф </em><em>(4)</em>
Divide Equation (3) by Equation (4) to eliminate T:
F_e/m_a*g=tan<em>Ф</em>
F_e=m_a*g*tan<em>Ф</em>
Substitute for F_e by using Coulomb's law from Equation (2):
k*| q_A |*| q_B |/r = m_a*g*tan<em>Ф</em>
Solve for q_A :
| q_A | = m_a*g*tanФ_r/k*| q_B |
Substitute numerical values from given data:
| q_A | = 238 nC
Because object A is attracted to object B. it has an opposite negative charge. Therefore, the charge on object A is | q_A | = -238 nC
(b)
Solve Equation (4) for T:
T = m_a*g/cosФ
Substitute numerical values from given data:
T = (0.09 kg)(9.8 m/s^2) /cos 7.2°
= 0.889 N
