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NARA [144]
3 years ago
6

A gymnast is swinging on a high bar. The distance between his waist and the bar is 0.829 m, as the drawing shows. At the top of

the swing his speed is momentarily zero. Ignoring friction and treating the gymnast as if all of his mass is located at his waist, find his speed at the bottom of the swing.
Physics
1 answer:
vivado [14]3 years ago
6 0

To find the speed with the given values we can apply the energy conservation equations for which we have that the increase in potential energy is compensated in the decrease of the kinetic energy or vice versa.

Since there is conservation and part of the balance we have to

KE = PE

Where,

KE = Kinetic Energy

PE = Potential Energy

The values for this energy are given as

\frac{1}{2}mv^2 = mgh

\frac{1}{2} v^2 = gh

v = \sqrt{2gh}

v = \sqrt{2(9.8)(0.829)}

v = 4.03m/s

Therefore the speed at the bottom of the swing is 4.03m/s

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Multiply the following numbers, using scientific notation and the correct amount of significant digits. 1.003 m⋅3.09 = _____ 3.0
lozanna [386]

Answer

correct answer is 3.10

Explanation:

in this question we have to multiply  two numbers 1.003 and 3.09.

1.003 has 4 significant digits and 3.09 has 3 significant digits so answer must have 3 significant digits.

1.003\times 3.09=3.09927

1.003\times 3.09=3.10              


Hope it will help you

3 0
2 years ago
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Which of the following is an example of the Doppler effect?
Oliga [24]

b is your answers in this thread

3 0
2 years ago
All fungi are
sertanlavr [38]

Answer:

multicellular

Explanation:

all fungi are multicellular

4 0
2 years ago
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A gang of robbers is escaping across city roofs at night. They come to the edge of one building and need to drop down to their g
REY [17]

Answer:

a) They will hit the ground with a speed of 19.6 m/s.

b) They are at a height of 20 m.

c) It is not a safe jump.

Explanation:

Hi there!

a) The equations of height and velocity in function of time of a free falling body are the following:

h = h0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

h = height of the object at time t.

h0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.8 m/s² considering downward as negative direction).

v = velocity of the object at time t.

Using the equation of velocity, let's find the velocity at which they will hit the ground. The pebble is dropped (initial velocity = 0) and it takes 2 s to reach the ground:

v = v0 + g · t     (v0 = 0)

v = g · t

v = -9.8 m/s² · 2.0 s

v = -19.6 m/s

They will hit the ground with a speed of 19.6 m/s.

b)Now, we have to use the equation of height:

h = h0 + v0 · t + 1/2 · g · t²

If we place the origin of the frame of reference on the ground, we have to find the initial height (h0) knowing that at t = 2.0 s, h = 0 m

0 m = h0 - 1/2 · 9.8 m/s² · (2.0 s)²

h0 = 1/2 · 9.8 m/s² · (2.0 s)²

h0 = 20 m

They are at a height of 20 m.

c)According to a NASA paper (Issues on Human Acceleration Tolerance After Long-Duration Space Flights, figure 10), if you fall with a vertical velocity greater than 17 m/s it is unlikely that you will survive. So, it is not a safe jump.  

3 0
3 years ago
Consider two ideal gases, A & B, at the same temperature. The rms speed of the molecules of gas A is twice that of gas B. Ho
uysha [10]

Answer:

option (d)

Explanation:

The relation between the rms velocity and the molecular mass is given by

v   proportional to  \frac{1}{\sqrt{M}} keeping the temperature constant

So for two gases

\frac{v_{A}}{v_{B}}=\sqrt{\frac{M_{B}}{M_{A}}}

\frac{2v_{B}}{v_{B}}=\sqrt{\frac{M_{B}}{M_{A}}}

{\frac{M_{B}}{M_{A}}} = 4

{\frac{M_{B}}{4}} = M_{A}

7 0
3 years ago
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