Answer:
i) 3.514 s, ii) 5.692 m/s
Explanation:
i) We can use Newton's second law of motion to find out how long does it take for the Eagle to touch down.
as the equation says for free-falling
h = ut +0.5gt^2
Here, h = 10 m, g = acceleration due to gravity = 1.62 m/s^2( on moon surface)
initial velocity u = 0
10 = 0.5×1.62t^2
t = 3.514 seconds
Therefore, it takes t = 3.514 seconds for the Eagle to touch down.
ii) use Newton's 1st equation of motion to calculate the velocity of the lunar module when it hits the surface of the moon
v = u + gt
v = 0+ 1.62×3.514
v= 5.692 m/s
Answer:
x = 7.14 meters
Explanation:
It is given that,
Current in wire 1, 
Current in wire 2, 
Distance between parallel wires, r = 25 cm
Let at P point the net magnetic field equal to 0. The magnetic field at a point midway between the is given by :
Let the distance is x from wire 1. So,
x = 7.14 meters
So, the magnetic field will be 0 at a distance of 7.14 meters from wire 1. Hence, this is the required solution.
Displacement (between time 0 and time 25) is the area under the velocity time curve, i.e. ∫ vdt.
Here, v(0)=10, v(25)=34 (approx.)
Therefore
displacement = (1/2)(10+34 m/s)*(25-0) s [ trapezoid area ]
=550 m
Answer:
1) the new power coming from the amplifier is 19.02 W
2) The distance away from the amplifier now is 5.50 m
3) u₁ = 69.24 m
Therefore have to move u₁ - u ( 69.24 - 5.50) = 63.74 farther
Explanation:
Lets say that I am at a distance "u" from the TV,
Let I₁ be the corresponding intensity of the sound at my location when sound level is 125dB
SO
S(indB) = 10log (I₁/1₀)
we substitute
125 = 10(I₁/10⁻¹²)
12.5 = log (I₁/10⁻¹²)
10^12.5 = I₁/10^-12
I₁ = 10^12.5 × 10^-12
I₁ = 10^0.5 W/m²
Now I₂ will be intensity of sound when corresponding sound level is 107 dB
107 = 10log(I₂/10⁻²)
10.7 = log(I₂/10⁻¹²)
10^10.7 = I₂ / 10^-12
I₂ = 10^10.7 × 10^-12
I₂ = 10^-1.3 W/m²
Now since we know that
I = P/4πu² ⇒ p = 4πu²I
THEN P₁ = 4πu²I₁ and P₂ =4πu²I₂
Therefore
P₁/P₂ = I₁/I₂
WE substitute
P₂ = P₁(I₂/I₁) = 1200 × ( 10^-1.3 / 10^0.5)
P₂ = 19.02 W
the new power coming from the amplifier is 19.02 W
2)
P₁ = 4πu²I₁
u =√(p₁/4πI₁)
u = √(1200/4π × 10^0.5)
u = 5.50 m
The distance away from the amplifier now is 5.50 m
3)
Let I₃ be the intensity corresponding to required sound level 85 dB
85 = 10log(I₃/10⁻¹²)
8.5 = log (I₃/10⁻¹²)
10^8.5 = I₃ / 10^-12
I₃ = 10^8.5 × 10^-12
I₃ = 10^-3.5 w/m²
Now, I ∝ 1/u²
so I₂/I₃ = u₁²/u²
u₁ = √(I₂/I₃) × u
u₁ = √(10^-1.3 / 10^-3.5) × 5.50
u₁ = 69.24 m
Therefore have to move u₁ - u ( 69.24 - 5.50) = 63.74 farther
Answer:
Constructive interference
Explanation:
- This is an example of a standing wave produced when two ends of a string are oscillated in the same plane. The displacement of of point on two ends oscillates vertically.
- We are given that two pulses move along the string each coming towards each other and meet at a common point ( P ).
- Each pulse have their own magnitude or displacement in the vertical plane. If the pulses are to meet at a common point at the same instant, then they interfere with each other constructively.
- Where constructive interference of two pulses is the addition of magnitudes of induvidual pulses and form a single puls of the constructed magnitude.
magnitude ( New pulse ) = magnitude (Pulse 1) + magnitude (Pulse 2)
