Answer:
a) 0.0625 I_1
b) 3.16 m
Explanation:
<u>Concepts and Principles </u>
The intensity at a distance r from a point source that emits waves of power P is given as:
I=P/4π*r^2 (1)
<u>Given Data</u>
f (frequency of the tuning fork) = 250 Hz
I_1 is the intensity at the source a distance r_1 = I m from the source.
<u>Required Data</u>
- In part (a), we are asked to determine the intensity I_2 a distance r_2 = 4 in from the source.
- In part (b), we are asked to determine the distance from the tuning fork at which the intensity is a tenth of the intensity at the source.
<u>solution:</u>
(a)
According to Equation (1), the intensity a distance r is inversely proportional to the distance from the source squared:
I∝1/r^2
Set the proportionality:
I_1/I_2=(r_2/r_1)^2 (2)
Solve for I_2 :
I_2=I_1(r_2/r_1)^2
I_2=0.0625 I_1
(b)
Solve Equation (2) for r_2:
r_2=(√I_1/I_2)*r_1
where I_2 = (1/10)*I_1:
r_2=(√I_1/1/10*I_1)*r_1
=3.16 m
Global warming is what will happen
Answer:
The magnitude of the torque on the loop due to the magnetic field is
.
Explanation:
Given that,
Diameter = 10 cm
Current = 0.20 A
Magnetic field = 0.30 T
Unit vector
We need to calculate the torque on the loop
Using formula of torque

Where, N = number of turns
A = area
I = current
B = magnetic field
Put the value into the formula


Hence, The magnitude of the torque on the loop due to the magnetic field is
.
Mathematically it can be expressed as: I = FpΔt. Where F p is the average magnitude of the acting force and Δt = t 2 - t 1, the time lapse in the force acts.
The magnitude of the impulse applied to stop the cart is
I = FpΔt = (10N) * (5s) = 50 N.s
Refer to the diagram shown below.
Neglect wind resistance, and use g = 9.8 m/s².
The pole vaulter falls with an initial vertical velocity of u = 0.
If the velocity upon hitting the pad is v, then
v² = 2*(9.8 m/s²)*(4.2 m) = 82.32 (m/s)²
v = 9.037 m/s
The pole vaulter comes to res after the pad compresses by 50 cm (or 0.5 m).
If the average acceleration (actually deceleration) is (a m/s²), then
0 = (9.037 m/s)² + 2*(a m/s²)*(0.5 m)
a = - 82.32/(2*0.5) = - 82 m/s²
Answer: - 82 m/s² (or a deceleration of 82 m/s²)