Answer:
The average linear velocity (inches/second) of the golf club is 136.01 inches/second
Explanation:
Given;
length of the club, L = 29 inches
rotation angle, θ = 215⁰
time of motion, t = 0.8 s
The angular speed of the club is calculated as follows;
The average linear velocity (inches/second) of the golf club is calculated as;
v = ωr
v = 4.69 rad/s x 29 inches
v = 136.01 inches/second
Therefore, the average linear velocity (inches/second) of the golf club is 136.01 inches/second
Answer:
Force on the object is 20 N
Explanation:
As we know that work done to raise the book from initial position to final position is known as potential energy stored in it
So here we know that
here we know that
U = 30 J
s = displacement = 1.5 m
so we have
Answer: 4,438.96m
Explanation:
(kindly find attachment below)
From the attachment below, it can be seen that the resultant displacement and the other 2 displacements form a right angle triangle, with A+B as the hypotenus, 3.2km as the opposite and the displacement B as the adjacent.
By using phythagoras theorem
H² = O² + A²
(5.38)² = (3.20)² + B²
28.944 = 10.24 + B²
B² = 28.944 - 10.24
B² = 18.7044
B = √18.7044
B = 4.439km to meter is 4.439 * 1000 = 4,438. 96m
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
