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Sunny_sXe [5.5K]
3 years ago
9

a marine biologist wants to know the total vertical distance a dolphin travel during a jump with the surface of the water being

zero is started underwater at -35 ft jump up and out of the water the 27 ft and return to the water to swim at -18 feet what is the total vertical distance the dolphin traveled?
Physics
2 answers:
Yuki888 [10]3 years ago
5 0

Answer:

107 feet is the total vertical distance the dolphin traveled.

Explanation:

Negative sign in front of the distance value is just to define the distance covered by dolphin underwater that is beneath the surface of the water

Before jump

Vertical distance covered by dolphin below the surface of water before coming out= 35 feet

Vertical distance covered by dolphin after coming out of the surface of water = 27 feet

After jump

Vertical distance covered by dolphin above the surface of water after jumping = 27 feet

Vertical distance covered by dolphin below the surface of water after jumping = 18 feet

The total vertical distance the dolphin traveled:

35 feet + 27 feet + 27 feet + 18 feet = 107 feet

marin [14]3 years ago
3 0

From the starting depth to the surface, the vertical distance is 35 ft.

From the surface to the peak of the jump, the vertical distance is 27 ft.

From the peak of the jump to the surface, the vertical distance is 27 ft.

From the surface to the ending depth, the vertical distance is 18 ft.

Then the total vertical distance is ...

  35 ft + 27 ft + 27 ft + 18 ft = 107 ft

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If your machine has a mechanical advantage of 2.5, then WHATEVER force you apply to the input, the force at the output will be 2.5 times as great.

If you apply 1 newton to the machine's input, the output force is

                 (2.5 x 1 newton)  =  2.5 newtons.

If you apply 120 newtons to the machine's input, the output force is

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3 years ago
A contestant in a winter games event pulls a 36.0 kg block of ice across a frozen lake with a rope over his shoulder as shown in
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(a) The minimum force F he must exert to get the block moving is 38.9 N.

(b) The acceleration of the block is 0.79 m/s².

<h3>Minimum force to be applied </h3>

The minimum force F he must exert to get the block moving is calculated as follows;

Fcosθ = μ(s)Fₙ

Fcosθ = μ(s)mg

where;

  • μ(s) is coefficient of static friction
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F = [0.1(36)(9.8)] / [(cos(25)]

F = 38.9 N

<h3>Acceleration of the block</h3>

F(net) = 38.9 - (0.03 x 36 x 9.8) = 28.32

a = F(net)/m

a = 28.32/36

a = 0.79 m/s²

Thus, the minimum force F he must exert to get the block moving is 38.9 N.

The acceleration of the block is 0.79 m/s².

Learn more about minimum force here: brainly.com/question/14353320

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Answer:

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Explanation:

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Yf=Yo+Voyt+0.5ayt² (II)

Where Xo, Yo are the initial positions, Xf and Yf are the final positions, Vox and Voy are the initial velocities, ax and ay are the accerelations in x and y directions, t is the time.

The particle starts from rest from the origin, therefore:

Vox=Voy=0

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Replacing Yf=12, Yo=0 and Voy=0 in (I) and solving for t:

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Dividing by 0.5 and extracting thr squareroot both sides:

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