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Sloan [31]
2 years ago
15

A 50 kg athlete running at speed v grabs a light rope that hangs from a 10-meter-high platform and swings to a maximum of 1.8 m

above the ground. later, a 100 kg athlete, running at the same speed, grabs a similar rope hanging from a 5-meter-high platform. what is the maximum height to which the 100 kg athlete swings?
Physics
1 answer:
balu736 [363]2 years ago
4 0
The answer is: 1.8 meters.

Explanation:
An athlete swinging can be considered a pendulum.

The pendulum's maximum height is the point at which it changes direction, which means that its velocity is equal to zero. In this point, for the mechanical energy conservation, all its kinetic energy is transformed into potential energy. Similarly, when the pendulum is at its resting position (when the athlete grabs the rope), its energy is totally kinetic.

Therefore we can say that:
\frac{1}{2}m v_{1} ^{2}  = mgh_{max}

Solving for h:
h_{max} =  \frac{v^{2}}{g}

As we can see, the maximum height is independent on the mass and on the length of the rope, therefore it will be the same for the 100kg-athlete as it is for the 50kg-athlete, since their initial speeds are the same.

We know that the <span>50kg-athlete reached a height of 1.8 m, h</span>ence, the maximum height reached by the 100kg-athlete will be 1.8 m.
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kiruha [24]

Answer:

5.07\cdot 10^{-7} m (507 nm)

Explanation:

First of all let's calculate the energy of the photon absorbed by the electron, This is given by

E=hf

where

h is the Planck constant

f=8.88\cdot 10^{14} Hz is the frequency of the photon

Substituting,

E=(6.63\cdot 10^{-34}Js)(8.88\cdot 10^{14}Hz)=5.89\cdot 10^{-19} J

The energy of the second photon, the one emitted when the electron drops to the intermediate energy level, is 2/3 of this energy:

E'=\frac{2}{3}E=\frac{2}{3}(5.89\cdot 10^{-19} J)=3.92\cdot 10^{-19} J

The relationship between the energy of the photon and its wavelength \lambda is

E=\frac{hc}{\lambda}

where c is the speed of light. Solving for \lambda, we find the wavelength:

\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{3.92\cdot 10^{-19} J}=5.07\cdot 10^{-7} m

3 0
2 years ago
Read 2 more answers
A slide whistle is an open-closed tube with an adjustable plunger that changes the length. You are playing the slide whistle and
Serjik [45]

Answer:

df / ft = -n 12          n= 1, 3, 5, ...

Explanation:

The speed of sound is

         v = λ f

         

In a whistle that we approach by an open tube at one end and closed at the other, standing waves occur, which has a node in the closed part and a maximum in the open pate, whereby wavelength and the distance of the tube are related, the fundamental wave is

         λ₁ = 4L

   The harmonics are

        λ₃ = 4L / 3

        λ₅ = 4L / 5

The general formula

       λₙ = 4L / n              

with n = 1, 3, 5,…

We substitute and clear in the first equation

           f = v n / 4L                        n = 1, 3, 5,…

Let's use derivatives to find the frequency change

           df / dt = v n /4  dL⁻¹ / dt

          d / dt (1/L) = - 1 / L² dL / dt

Where dL / dt = 8 cm / s

We replace

         df / dt = - n v / L2 dL / dt

Let's calculate

         df / dt = - n 340/152 8

         df / ft = -n 12          n= 1, 3, 5, ...

4 0
2 years ago
The graph indicates Linda’s walk.
Sedaia [141]
I think the right answer is the first one. If she stops moving her Position does not change any more-and the Graph Shows that after 6 seconds she stays at the Position of 5 m. If she Went Back to the start point the Graph would have Developed Back to 0m(decreased).
3 0
2 years ago
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Two students are playing paddle ball with a 5 kg spongy ball. If the ball is thrown at the batter with a speed of 5 m/s and boun
fenix001 [56]

Answer:

75 kgm/s

Explanation:

Impulse: This can be defined as the product of mass and change in velocity. The S.I unit is kgm/s.

From the question,

I = m(v-u)................... Equation 1

Where I = impulse, m = mass, v = final velocity, u = initial velocity.

Let the direction of the initial velocity be the positive direction.

Given: m = 5 kg, v = -10 m/s (bounce off), u = 5 m/s.

Substitute into equation 1

I = 5(-10-5)

I = 5(-15)

I = -75 kgm/s.

The negative sign tells that the impulse act in the same direction as the final velocity of the ball

Hence,

I = 75 kgm/s

3 0
2 years ago
Two identical conducting spheres, A and B, sit atop insulating stands. When they are touched, 1.51 × 1013 electrons flow from sp
erma4kov [3.2K]

Answer:

A = -0.576 μC

B = 4.256 μC

Explanation:

Suppose a single electron charge is 1.6\times10^{-19}C. Then the total charge that is flowing from B to A is:

1.6\times10^{-19} * 1.51 \times 10^{13} = 2.416\times10^{-6}C = 2.416 \mu C

Let A and B be the initial charge of spheres A and B, respectively. Since the net charge is 3.68μC we have the following equation

A + B = 3.68 (1)

When they touch 2.416μC flows from B to A, then they are equal, so we have the following equation

A + 2.416 = B - 2.416

-A + B = 2.416 + 2.416 = 4.832 (2)

Add equation (1) to equation (2) we have

2B = 3.68 + 4.832 = 8.512

B = 8.512 / 2 = 4.256 \mu C

A = 3.68 - B = 3.68 - 4.256 = -0.576 \mu C

6 0
2 years ago
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