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romanna [79]
3 years ago
7

Nicholas is at The Noah's Ark Amusement Park and preparing to ride on The Point of No Return racing slide. At the top of the sli

de, Nicholas (m=50 kg) is 28.5 m above the ground. Determine Nicholas' speed as he arrives at the bottom of the slide.
Physics
2 answers:
Kipish [7]3 years ago
6 0
You can not determine the speed based on the information given
Naya [18.7K]3 years ago
6 0

Answer:

Explanation:

Recall that from the principle of conservation of energy,

P.E = K.E -------- eqn i

Solution

P.E = mgh

K.E = \frac{1}{2} mv^{2}

i.e mgh = \frac{1}{2} mv^{2} --------- eqn ii

m = 50 kg

v = ?

g = 9.8 ms^{-2}

h = 28.5 m

substitute for all values in eqn ii

50 X 9.8 X 28.5 = \frac{1}{2} X 50 X v^{2}

13965 = 25 X v^{2}

v^{2}  = \frac{13965}{25}

v^{2} = 558.6

v = \sqrt{558.6}

v = 23.635 m/s

<em>Therefore, Nicholas arrives the bottom of the slide at a speed of 23.635 m/s</em>

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A 65.0 kg ice skater standing on frictionless ice throws a 0.15 kg snowball horizontally at a speed of 32.0 m/s. What is the vel
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Answer:

(d) 0.07 m/s

Explanation:

Given Data

Snowball mass m₁=0.15 kg

Ice skater mass m₂=65.0 kg

Snowball velocity v₁=32.0 m/s

To find

Velocity of Skater v₂=?

Solution

From law of conservation of momentum

m_{1}v_{1}=m_{2}v_{2}\\  v_{2}=\frac{m_{1}v_{1}}{m_{2}}\\ v_{2}=\frac{(0.15kg)(32.0m/s)}{65.0kg}\\ v_{2}=0.0738m/s\\or\\v_{2}=0.07 m/s

So Option d is correct one

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You roll a bowling ball down the lane and it takes 23 seconds to reach the pins. What is the total time it takes the ball to mov
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Answer:

B. 23s

Explanation:

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3 years ago
Read 2 more answers
A harmonic wave on a string with a mass per unit length of 0.050 kg/m and a tension of 60 N has an amplitude of 5.0 cm. Each sec
Dennis_Churaev [7]

Answer:

Power of the string wave will be equal to 5.464 watt

Explanation:

We have given mass per unit length is 0.050 kg/m

Tension in the string T = 60 N

Amplitude of the wave A = 5 cm = 0.05 m

Frequency f = 8 Hz

So angular frequency \omega =2\pi f=2\times 3.14\times 8=50.24rad/sec

Velocity of the string wave is equal to v=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{60}{0.050}}=34.641m/sec

Power of wave propagation is equal to P=\frac{1}{2}\mu \omega ^2vA^2=\frac{1}{2}\times 0.050\times 50.24^2\times 34.641\times 0.05^2=5.464watt

So power of the wave will be equal to 5.464 watt

6 0
3 years ago
1. A small light bulb is shining light on a basketball (diameter is 23 cm or 9 inches). The light bulb is 3 m from the closest s
siniylev [52]

Answer:

The size (diameter) of the basketball's shadow on the wall is approximately 53.38 cm

Explanation:

The given parameters of the basketball are;

The diameter of the basketball = 23 cm (9 inches)

The distance of the light bulb from the closest side of the basketball = 3 m

The distance from the ball to the wall = 4 m

The distance from the light source to the center of the ball, d = 3 m + 0.23/2 m = 3.115 m

The angle the light ray makes with the edge of the ball, θ = arctan(0.115/3.115)

Therefore, the ratio of the shadow width divided by 2 to the distance from the light from the wall = 0.115/3.115

The distance from the light from the wall = 3 m + 4 m + 0.23 m = 7.23 m

Therefore;

((The width of the shadow)/2)/(The distance from the light from the wall) = 0.115/3.115

∴ ((The width of the shadow)/2)/(7.23 m) = 0.115/3.115

((The width of the shadow)/2) = 7.23 m × 0.115/3.115 = 16629/62300 m ≈ 0.2669 m = 26.69 cm

The width (diameter) of the shadow on the wall = 2 × 16629/62300 m ≈ 0.5338 m = 53.38 cm

The size (diameter) of the basketball's shadow on the wall ≈ 53.38 cm

4 0
3 years ago
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