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Angelina_Jolie [31]
3 years ago
6

a diver of mass 101 kg jumps upward off a diving board into water. Diving board is 6m above water. Diver has a speed of 1.2m/s.

based on speed how high above the board will the diver reach?
Physics
1 answer:
8090 [49]3 years ago
5 0
When the diver reaches maximum height, the upward velocity will be zero.

We shall use the formula
v^2 = u^2 - 2gh
where 
v = 0 (velocity at maximum height)
u = 1.2 m/s, intial upward velocity
g = -9.8 m/s^2, gravitational acceleration (downward)
h = maximum height attained above the diving board.

Therefore
0 = 1.2^2 - 2*9.8*h
h = 1.2^2/(2*9.8) = 0.0735 m

Answer: 0.074 m (nearest thousandth)
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Answer:

The value is   a =  4.0 \ m/s^2

Explanation:

From the question we are told that  

  The velocity  is  a =  120 \  km/h =  [tex]\frac{120 *  1000}{3600} =  33.3 \  m/s[/tex]

  The time taken is t  =  8.28 \  s

    The  time taken for contact is  t_c  =  0.815 \  s

     The  velocity of the of the car after contact is v_c =  71.0 \  km/h = [tex]\frac{71 *1000}{3600} =  19.7 2 \  m/s[/tex]

From the equation of kinematics we have that  

       v =  u  + at

Here   u =  0 \ m/s  since the car is initially motionless

=>    33.3 =  0  + a *  8.28

=>    a =  4.0 \ m/s^2

4 0
3 years ago
Two car horns are sounded creating two sound waves with frequencies that differ by a factor of three. How does the speed of the
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Remains the same

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the speed of sound in a medium is constant  and independent of it's frequency. Moreover, when the frequency changes wavelength changes accordingly, such that their product remains constant.

we know that

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When jumping straight down, you can be seriously injured if you land stiff-legged. One way to avoid injury is to bend your knees
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See the attachment for calculations

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A 3.1 kg ball is dropped from the top of a 38 m tall building. What is the speed of the ball when it is halfway from the buildin
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Answer:

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Explanation:

Use third equation of motion

v^2-u^2=2gh

where v is the velocity at halfway, u is the initial velocity, g is gravity (9.81m/s^2) and h is the height at which you'd want to find the velocity

insert values to get answer

v^2-0^2=2(9.81m/s^2)(38/2)\\v^2=9.81m/s^2 *38\\v^2=372.78\\v=\sqrt[]{372.78} \\v=19.3m/s

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