Answer:
D) the second at the doorknob
Explanation:
The torque exerted by a force is given by:
where
F is the magnitude of the force
d is the distance between the point of application of the force and the centre of rotation
is the angle between the direction of the force and d
In this problem, we have:
- Two forces of equal magnitude F
- Both forces are perpendicular to the door, so 
- The first force is exerted at the midpoint of the door, while the 2nd force is applied at the doorknob. This means that d is the larger for the 2nd force
--> therefore, the 2nd force exerts a greater torque
Answer:
A) 138.8g
B)73.97 cm/s
Explanation:
K = 15.5 Kn/m
A = 7 cm
N = 37 oscillations
tn = 20 seconds
A) In harmonic motion, we know that;
ω² = k/m and m = k/ω²
Also, angular frequency (ω) = 2π/T
Now, T is the time it takes to complete one oscillation.
So from the question, we can calculate T as;
T = 22/37.
Thus ;
ω = 2π/(22/37) = 10.5672
So,mass of ball (m) = k/ω² = 15.5/10.5672² = 0.1388kg or 138.8g
B) In simple harmonic motion, velocity is given as;
v(t) = vmax Sin (ωt + Φ)
It is from the derivative of;
v(t) = -Aω Sin (ωt + Φ)
So comparing the two equations of v(t), we can see that ;
vmax = Aω
Vmax = 7 x 10.5672 = 73.97 cm/s
Explanation:
O + 6V = 9V, so O = 3V
P = 9V as it is parallel to the 9V power supply.
Frequency of the wave " the loudness"
Answer:
b
Explanation:
Given:
- The ball is fired at a upward initial speed v_yi = 2*v
- The ball in first experiment was fired at upward initial speed v_yi = v
- The ball in first experiment was as at position behind cart = x_1
Find:
How far behind the cart will the ball land, compared to the distance in the original experiment?
Solution:
- Assuming the ball fired follows a projectile path. We will calculate the time it takes for the ball to reach maximum height y. Using first equation of motion:
v_yf = v_yi + a*t
Where, a = -9.81 m/s^2 acceleration due to gravity
v_y,f = 0 m/s max height for both cases:
For experiment 1 case:
0 = v - 9.81*t_1
t_1 = v / 9.81
For experiment 2 case:
0 = 2*v - 9.81*t_2
t_2 = 2*v / 9.81
The total time for the journey is twice that of t for both cases:
For experiment 1 case:
T_1 = 2*t_1
T_1 = 2*v / 9.81
For experiment 2 case:
T_2 = 2*t_2
T_2 = 4*v / 9.81
- Now use 2nd equation of motion in horizontal direction for both cases:
x = v_xi*T
For experiment 1 case:
x_1 = v_x1*T_1
x_1 = v_x1*2*v / 9.81
For experiment 2 case:
x_2 = v_x2*T_2
x_2 = v_x2*4*v / 9.81
- Now the x component of the velocity for each case depends on the horizontal speed of the cart just before launching the ball. Using conservation of momentum we see that both v_x2 = v_x1 after launch. Since the masses of both ball and cart remains the same.
- Hence; take ratio of two distances x_1 and x_2:
x_2 / x_2 = v_x2*4*v / 9.81 * 9.81 / v_x1*2*v
Simplify:
x_1 / x_2 = 2
- Hence, the amount of distance traveled behind the cart in experiment 2 would be twice that of that in experiment 1.
