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RoseWind [281]
3 years ago
7

While skydiving, your parachute opens and you slow from 50.0 m/s to 8.0 m/s in 0.75 s . Determine the distance you fall while th

e parachute is opening.
Physics
1 answer:
Anestetic [448]3 years ago
8 0

Answer:

21.75 m

Explanation:

t = Time taken for the car to slow down = 0.75 s

u = Initial velocity = 50 m/s

v = Final velocity = 8 m/s

s = Displacement

a = Acceleration

Equation of motion

v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{8-50}{0.75}\\\Rightarrow a=-56\ m/s^2

Acceleration is -56 m/s²

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{8^2-50^2}{2\times -56}\\\Rightarrow s=21.75\ m

The distance covered in the 0.75 seconds is 21.75 m

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Harry and Sally are sitting on opposite sides of a circus tent when an elephant trumpets a loud blast. If Harry experiences a so
FromTheMoon [43]

To solve this problem we will start from the definition of the decibels which refers to the log relationship of the intensities, mathematically each decibel of the people can be described as follows

dB_1 = 10log(\frac{I_1}{I_o})

dB_2 = 10log(\frac{I_2}{I_o})

dB_1 - dB_2 = 10log(\frac{I_1}{I_2})

\frac{67-45}{10} = log(\frac{I_1}{I_2})

log(\frac{I_1}{I_2}) = \frac{11}{5}

For properties of the logarithm then we have that,

\frac{I_1}{I_2} = 10^{11/5}

\frac{I_1}{I_2} = 158.48

We have the relation

\frac{r_2^2}{r_1^2} = 158.48

\frac{r_2}{r_1} = \sqrt{158.48}

\frac{r_2}{r_1} = 12.58

7 0
3 years ago
A stream of water strikes a stationary turbine blade horizontally, as the drawing illustrates. The incident water stream has a v
Umnica [9.8K]

Answer:

Average force exerted by the water on the blades is 1024 N

Explanation:

As we know that the force due to water jet is given by

F = (v_{in} - v_{out})\frac{dm}{dt}

here we know that

v_{in} = 16 m/s

v_{out} = -16 m/s

\frac{dm}{dt} = 32 kg/s

so here we will have

F = (16 + 16)(32)

F = 1024 N

Average force exerted by the water on the blades is 1024 N

6 0
3 years ago
Please tell me if this looks right.
maria [59]
Compression is the best answer. Think about pressing on the toy bus enough to bend the table but not break it
8 0
4 years ago
A tennis ball is released from a height of 4.0 m above the floor. After its third bounce off the floor, it reaches a height of 1
diamong [38]

Answer:

The percentage of its mechanical energy does the ball lose with each bounce is 23 %

Explanation:

Given data,

The tennis ball is released from the height, h = 4 m

After the third bounce it reaches height, h' = 183 cm

                                                                       = 1.83 m

The total mechanical energy of the ball is equal to its maximum P.E

                                      E = mgh

                                          = 4 mg

At height h', the P.E becomes

                                      E' = mgh'

                                           = 1.83 mg

The percentage of change in energy the ball retains to its original energy,

                                 \Delta E\%=\frac{1.83mg}{4mg}\times100\%

                                  ΔE % = 45 %

The ball retains only the 45% of its original energy after 3 bounces.

Therefore, the energy retains in each bounce is

                                   ∛ (0.45) = 0.77

The ball retains only the 77% of its original energy.

The energy lost to the floor is,

                                E = 100 - 77

                                   = 23 %

Hence, the percentage of its mechanical energy does the ball lose with each bounce is 23 %      

5 0
4 years ago
at a particular instant a hot air balloon is 100m in the air and decending at a constant speed of 2m/s at this exact instant a g
labwork [276]
<h2>She will find the ball at a horizontal distance of 86.4 m from landed location</h2>

Explanation:

Consider the vertical motion of ball

We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = 2 m/s

        Acceleration, a = 9.81 m/s²  

        Displacement, s = 100 m      

     Substituting

                      s = ut + 0.5 at²

                      100 = 2 x t + 0.5 x 9.81 xt²

                      4.905t²  + 2t - 100 = 0

                     t = 4.32 s    or   t = -4.72 s

                    After 4.32 seconds the ball reaches ground.

Now we need to find horizontal distance traveled by ball in 4.32 seconds.

We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = 20 m/s

        Acceleration, a = 0 m/s²  

        Time, t = 4.32 s      

     Substituting

                      s = ut + 0.5 at²

                      s = 20 x 4.32 + 0.5 x 0 x 4.32²

                      s = 86.4 m

She will find the ball at a horizontal distance of 86.4 m from landed location

5 0
4 years ago
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