The answer is 570 J. The kinetic energy has the formula of 1/2mV². The total work in this process W= 1/2m(V2²-V1²) = 1/2 * 15.0 * (11.5²-7.50²) = 570 J.
Answer:
45.93°
Explanation:
The angle of incidence is given as 32.7°
The refractive index of the water that is 
Refractive index of the air that is 
(because the refractive index of air is 1 )
We have to find the angle at which the light leave the water means angle of refraction
So according to snell's law 
r =45.93°
So the light leave the water at an angle of 45.93°
Answer:
T = 4.42 10⁴ N
Explanation:
this is a problem of standing waves, let's start with the open tube, to calculate the wavelength
λ = 4L / n n = 1, 3, 5, ... (2n-1)
How the third resonance is excited
m = 3
L = 192 cm = 1.92 m
λ = 4 1.92 / 3
λ = 2.56 m
As in the resonant processes, the frequency is maintained until you look for the frequency in this tube, with the speed ratio
v = λ f
f = v / λ
f = 343 / 2.56
f = 133.98 Hz
Now he works with the rope, which oscillates in its second mode m = 2 and has a length of L = 37 cm = 0.37 m
The expression for standing waves on a string is
λ = 2L / n
λ = 2 0.37 / 2
λ = 0.37 m
The speed of the wave is
v = λ f
As we have some resonance processes between the string and the tube the frequency is the same
v = 0.37 133.98
v = 49.57 m / s
Let's use the relationship of the speed of the wave with the properties of the string
v = √ T /μ
T = v² μ
T = 49.57² 18
T = 4.42 10⁴ N
Answer:
1. 2.67 s
2. 0.1 m/s²
Explanation:
1. Determination of the time taken for the penguin to fall.
Height (h) of cliff = 35 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
h = ½gt²
35 = ½ × 9.8 × t²
35 = 4.9 × t²
Divide both side by 4.9
t² = 35 / 4.9
Take the square root of both side
t = √(35 / 4.9)
t = 2.67 s
Thus, it will take 2.67 s for the penguin to fall onto the head of a napping polar bear.
2. Determination of the acceleration of the penguin.
Initial velocity (u) = 0 m/s.
Final velocity (v) = 2 m/s.
Time (t) = 20 s
Acceleration (a) =?
a = (v – u)/t
a = (2 – 0)/ 20
a = 2 / 20
a = 0.1 m/s²
Thus, the acceleration of the penguin is 0.1 m/s²
