Answer:
λ2 < λ1
Explanation:
- Young's experiment of interference of light when passed through a slit has determined a result. We will use an expression derived from the results that relates wavelength and and the spread of the interference pattern using:
sin (Q) = n*lambda / d
Where,
n is the order of fringe about the central order
lambda is the wavelength of light
d is the space between grating
Q is the angle between the two successive interference fringe
- So in our case using a light of wavelength lambda 2 the pattern got closer together or we could say that angle Q got smaller. According to the relationship given we can see as angle Q decreases when lambda decreases. So we can conclude that λ2 < λ1.
Answer:
Explanation:
Given
acceleration a =2.2 m/s^2
Force F(t) is given by
F(t)=5.40 t N/s
Power supplied by this force at different time t is given by
Power
velocity at any instant t is given by
v=at
Power
Power
at t=1 s
Power =11.88 W
at t=2 s
Power
at =4 s
Power
Answer:
a) 3.43 m/s
Explanation:
Due to the law of conservation of momentum, the total momentum of the bullet - rifle system must be conserved.
The total momentum before the bullet is shot is zero, because they are both at rest, so:
Instead the total momentum of the system after the shot is:
where:
m = 0.006 kg is the mass of the bullet
M = 1.4 kg is the mass of the rifle
v = 800 m/s is the velocity of the bullet
V is the recoil velocity of the rifle
The total momentum is conserved, therefore we can write:
Which means:
Solving for V, we can find the recoil velocity of the rifle:
where the negative sign indicates that the velocity is opposite to direction of the bullet: so the recoil speed is
a) 3.43 m/s
Answer:
1.16cm were cut off the end of the second pipe
Explanation:
The fundamental frequency in the first pipe is,
<em><u>Since the speed of sound is not given in the question, we would assume it to be 340m/s</u></em>
f1 = v/4L, where v is the speed of sound and L is the length of the pipe
266 = 340/4L
L = 0.31954 m = 0.32 m
It is given that the second pipe is identical to the first pipe by cutting off a portion of the open end. So, consider L’ be the length that was cut from the first pipe.
<u>So, the length of the second pipe is L – L’</u>
Then, the fundamental frequency in the second pipe is
f2 = v/4(L - L’)
<u>The beat frequency due to the fundamental frequencies of the first and second pipe is</u>
f2 – f1 = 10hz
[v/4(L - L’)] – 266 = 10
[v/4(L – L’)] = 10 + 266
[v/4(L – L’)] = 276
(L - L’) = v/(4 x 276)
(L – L’) = 340/(4 x 276)
(L – L’) = 0.30797
L’ = 0.31954 – 0.30797
L’ = 0.01157 m = 1.157 cm ≅ 1.16cm
Hence, 1.16 cm were cut from the end of the second pipe
Answer:
75
Explanation:
Power is current times voltage:
P = IV
Voltage is current times resistance:
V = IR
Therefore:
P = I²R
Given I = 0.62 A and R = 195 Ω:
P = (0.62 A)² (195 Ω)
P ≈ 75 W
