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3 years ago

Charles, the 75 kg trampoline artist, lands on a trampoline with a speed of 9.0 m/s.

Physics

1 answer:

Softa [21]3 years ago

7 0

Answer:

The maximum distance Charles will push down the trampoline ≈ 0.342 m

Explanation:

The given parameters are;

The mass of the trampoline artist, m = 75 kg

The speed with which the trampoline artist lands, v = 9.0 m/s

The value of the spring constant of the trampoline, k = 52,000 N/m

Let x represents the maximum distance Charles will push down the trampoline

Therefore, we have;

Kinetic energy = 1/2·m·v²

The kinetic energy with which the trampoline artist lands = 1/2 × 75 × 9.0² = 3037.5

The kinetic energy with which the trampoline artist lands = 3037.5 J

The potential energy stored in a spring = 1/2·k·x² = The kinetic energy with which the trampoline artist lands

∴ 1/2 × 52,000 × x² = 3037.5

∴ x = √(3037.5/(1/2 × 52,000)) ≈ 0.342

The maximum distance Charles will push down the trampoline = x ≈ 0.342 m

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A ball is thrown forward at 5 m/s how would the path of the ball differ on earth than on the moon

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In this problem, we are given that a ball is thrown upward and that it has an initial velocity of 5 m/s. Its path both on Earth and on the moon will be trajectory in nature. This is because while its horizontal velocity is maintained, its vertical velocity increases.

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3 years ago

Read 2 more answers

A boy pushes a rock up a hill in 20 seconds. If he applied a force of 20 N over a distance of 5 m on the rock, calculate the pow

aksik [14]

Here is your answer

$5 watts$

REASON :

Given,

t= 20 secs

F= 20 N

s (displacement)= 5m

Now,

Power= \frac{Work done}{time}

P= W/t ...... (i)

W= F×s×cos(theta)

W= 20×5 × 1 (theta is 0 degree; so cos0= 1)

W= 100 J

Hence,

P= 100/20 (from eq. i)

P= 5 watts

HOPE IT IS USEFUL

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4 years ago

A pulse moving to the right along the x axis is represented by the function of

olasank [31]

Answer:

(a) Peak=2 cm, velocity= 3 cm/s (to the right)

(b) Peak= 2 cm, velocity= -3 cm/s (to the left)

Explanation:

Pulse Movement

The function of a moving wave is given by

$\displaystyle y(x,t)=\frac{2}{(x-3t)^2+1}$

where x,y are measured in cm and t in seconds.

Please check the graph shown in the image provided below. It shows the waveform at two different times, t=0 sec and t=1 sec. The peak value is displaced by 3 cm when t varies by 1 second. It shows a velocity of 3 m/s.

(a) For a given time, say t=0 the expression for the pulse is

$\displaystyle y(x,0)=\frac{2}{x^2+1}$

The maximum value or the pulse amplitude occurs when the denominator has its minimum value, that is when x=0

$\displaystyle A=y(0,0)=\frac{2}{0^2+1}=2$

At t=1 second, the function is

$\displaystyle y(x,1)=\frac{2}{(x-3)^2+1}$

Again, the maximum value of the wave occurs at the minimum value of the denominator, or when

$x-3=0 \rightarrow x=3$

Note the peak has moved 3 cm to the right when t increased by 1 second, this gives us a horizontal velocity of 3 cm/s.

We can corroborate for the general case knowing the peak value moves to the right at the point where

$x-3t=0 \rightarrow x=3t$

Taking the derivative with respect to t gives us the horizontal velocity:

$x'=3\ cm/s$

At t=2 seconds

$\displaystyle y(x,2)=\frac{2}{(x-6)^2+1}$

The amplitude and velocity are the same as determined before. The graph shown in the figure attached shows the pulse waves at t=0 and t=1

b) If the function was

$\displaystyle y(x,t)=\frac{2}{(x+3t)^2+1}$

Then when t increases, the peak value moves to the left on the x-axis. The velocity would be

$x'=-3\ cm/s$

It means the wave is traveling to the left instead of to the right

8 0

3 years ago

Water is traveling through a horizontal pipe with a speed of 1.7 m/s and at a pressure of 205 kPa. This pipe is reduced to a new

Arturiano [62]

Answer:

The velocity is $v_2 = 6.8 \ m/s$

The pressure is $P_2 = 204978 Pa$

Explanation:

From the question we are told that

The speed at which water is travelling through is $v = 1.7 \ m/s$

The pressure is $P_1 = 205 k Pa = 205 *10^{3} \ Pa$

The diameter of the new pipe is $d = \frac{D}{2}$

Where D is the diameter of first pipe

According to the principal of continuity we have that

$A_1 v_1 = A_2 v_2$

Now $A_1$ is the area of the first pipe which is mathematically represented as

$A_1 = \pi \frac{D^2}{4}$

and $A_2$ is the area of the second pipe which is mathematically represented as

$A_2 = \pi \frac{d^2}{4}$

Recall $d = \frac{D}{2}$

$A_2 = \pi \frac{[ D^2]}{4 *4}$

$A_2 = \frac{A_1}{4}$

So $A_1 v_1 = \frac{A_1}{4} v_2$

substituting value

$1.7 = \frac{1}{4} * v_2$

$v_2 = 4 * 1.7$

$v_2 = 6.8 \ m/s$

According to Bernoulli's equation we have that

$P_1 + \rho \frac{v_1 ^2}{2} = P_2 + \rho \frac{v_2 ^2}{2}$

substituting values

$205 *10^{3 }+ \frac{1.7 ^2}{2} = P_2 + \frac{6.8 ^2}{2}$

$P_2 = 204978 Pa$

4 0

3 years ago

While standing on a bridge 40.0 m above the ground, you drop a stone from rest. when the stone has fallen 3.80 m, you throw a se

Norma-Jean [14]

Given: distance 1 d₁ = 40 m; distance 2 d₂ = 3.8 m g = -9.8 m/s²

Initial Velocity Vi = 0 Final Velocity of stone 2 is unknown = ?

Total distance dₓ = d₁ - d₂ = 40 m - 3.8 m = 36.2 m

Formula: a = Vf² - Vi²/2d derive for Final Velocity Vf

acceleration is now due to gravity, therefore a = g

Vf = √2gd Vf = √2(9.8 m/s²)(36.2 m)

Vf = 26.64 m/s

Reason: The second stone will still start from rest.

3 0

4 years ago