Answer:
E=0
Explanation:
Electric field due to each thin sheet of charge=\sigma/2\varepsilon
let us say the right plate has positive charge density \varepsilonand left sheet has a negative charge density -\varepsilon .
In the region between the plates,the electric field due to each plate is in same direction,
E=\sigma/2\varepsilon-(-\sigma/2\varepsilon)
E=\sigma/\varepsilon
in the region outside the plates, the field due to the plates is in opposite directions
E=-\sigma/2\varepsilon-(-\sigma/2\varepsilon)
E=-\sigma/2\varepsilon+\sigma/2\varepsilon
E=0
Answer:
14.7 m/s
Explanation:
a = acceleration experienced by driver's head = 50 g = 50 x 9.8 m/s² = 490 m/s²
v₀ = initial speed of the driver = 0 m/s
v = final speed of the driver after 30 ms
t = time interval for which the acceleration is experienced = 30 ms = 0.030 s
Using the equation
v = v₀ + a t
Inserting the values
v = 0 + (490) (0.030)
v = 14.7 m/s
Answer:
a. The total momentum of the trolleys which are at rest before the separation is zero
b. The total momentum of the trolleys after separation is zero
c. The momentum of the 2 kg trolley after separation is 12 kg·m/s
d. The momentum of the 3 kg trolley is -12 kg·m/s
e. The velocity of the 3 kg trolley = -4 m/s
Explanation:
a. The total momentum of the trolleys which are at rest before the separation is zero
b. By the principle of the conservation of linear momentum, the total momentum of the trolleys after separation = The total momentum of the trolleys before separation = 0
c. The momentum of the 2 kg trolley after separation = Mass × Velocity = 2 kg × 6 m/s = 12 kg·m/s
d. Given that the total momentum of the trolleys after separation is zero, the momentum of the 3 kg trolley is equal and opposite to the momentum of the 2 kg trolley = -12 kg·m/s
e. The momentum of the 3 kg trolley = Mass of the 3 kg Trolley × Velocity of the 3 kg trolley
∴ The momentum of the 3 kg trolley = 3 kg × Velocity of the 3 kg trolley = -12 kg·m/s
The velocity of the 3 kg trolley = -12 kg·m/s/(3 kg) = -4 m/s
The question is incomplete. Here is the entire question.
A jetboat is drifting with a speed of 5.0m/s when the driver turns on the motor. The motor runs for 6.0s causing a constant leftward acceleration of magnitude 4.0m/s². What is the displacement of the boat over the 6.0 seconds time interval?
Answer: Δx = - 42m
Explanation: The jetboat is moving with an acceleration during the time interval, so it is a <u>linear</u> <u>motion</u> <u>with</u> <u>constant</u> <u>acceleration</u>.
For this "type" of motion, displacement (Δx) can be determined by:
is the initial velocity
a is acceleration and can be positive or negative, according to the referential.
For Referential, let's assume rightward is positive.
Calculating displacement:
= - 42
Displacement of the boat for t=6.0s interval is = - 42m, i.e., 42 m to the left.
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
