Answer:
a) y = 16.51 [m]
b) t = 1.83 [s]
Explanation:
To solve this problem we must use two kinematics equations, the first to determine the height to which the ball reaches, and the second equation to determine how long it lasts in the air.
where:
Vf = final velocity = 0
Vi = initial velocity = 18 [m/s]
g = gravity acceleration = 9.81[m/s^2]
t = time [s]
Note: the negative sign of the Equation indicates that the acceleration of gravity acts in the opposite direction to the movement of the ball. The final velocity is zero, since the ball reaches its maximum altitude when the velocity is zero.
Now replacing:
0 = (18)^2 - (2*9.81*y)
y = 16.51 [m]
b)
0 = 18 - (9.81*t)
t = 1.83 [s]
Answer:
A) 443 Hz
B) She has to loosen the string
Explanation:
A) Given;
Beat frequency;f_beat = 3 Hz
Frequency of electronically generated tone; f_e = 440 Hz
We know that formula for beat frequency is given by;
f_beat = |f1 - f2|
Now, applying it to this question, we have;
f_beat = f_v - f_e
Where f_v is frequency of the note played by the violinist
Thus, plugging in the relevant values;
3 = f_v - 440
f_v = 3 + 440
f_v = 443 Hz
B) In the concept of wave travelling in a string, the frequency is directly proportional to the square root of the force acting on the string.
Now, for the violinist to get her violin perfectly tuned to concert A from what it was when she heard the 3-Hz beats, the beat frequency will have to be zero. Which means it has to decrease by 3 Hz. For it to decrease, it means that the force applied has to decrease as we have seen that frequency is directly proportional to the square root of the force acting on the string.
Thus, she would have to loosen the string.
Answer:
V_inside = 36 V
Explanation:
<u>Given </u>
We are given a sphere with a positive charge q with radius R = 0.400 m Also, the potential due to this charge at distance r = 1.20 m is V = 24.0 V.
<u>Required</u>
We are asked to calculate the potential at the centre of the sphere
<u>Solution</u>
The potential energy due to the sphere is given by equation
V = (1/4*π*∈o) × (q/r) (1)
Where r is the distance where the potential is measured, it may be inside the sphere or outside the sphere. As shown by equation (1) the potential inversely proportional to the distance V
V ∝ 1/r
The potential at the centre of the sphere depends on the radius R where the potential is the same for the entire sphere. As the charge q is the same and the term (1/4*π*∈o) is constant we could express a relation between the states , e inside the sphere and outside the sphere as next
V_1/V_2=r_2/r_1
V_inside/V_outside = r/R
V_inside = (r/R)*V_outside (2)
Now we can plug our values for r, R and V_outside into equation (2) to get V_inside
V_inside = (1.2 m )/(0.600)*18
= 36 V
V_inside = 36 V
