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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:

\dfrac{h_2}{h_1}=\dfrac{1}{2}

Explanation:

Lets take h is height of cylinder immersed in the water

We know that for floating body

h=\dfrac{\rho_cL}{\rho_l}

Where \rho_c density of cylinder

\rho_l density of water

For both cylinder fluid is same also density of cylinders are also same

So

\dfrac{h_1}{L_1}=\dfrac{h_2}{L_2}

\dfrac{h_1}{h_2}=\dfrac{L_1}{L_2}

\dfrac{h_1}{h_2}=\dfrac{20}{10}

\dfrac{h_2}{h_1}=\dfrac{1}{2}

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3 years ago
If you walk a distance of 8 blocks and then 3 blocks south from home, what is your position compared to home? What distance did
Harman [31]

The position compared to that of home is a reference to displacement, I believe.

Displacement = x total - x initial

So I believe the answer is 5 blocks due north (if you’re walking linearly from your home), unless the questions is referring to relative displacement, in which then you’d need to use the Pythagorean theorem to find the hypotenuse between both positions. And then you’d have to find theta for the degrees between the south direction and the other unmentioned direction. But I don’t think that’s the case.

Distance refers to x total and doesn’t care for direction, as this refers to a scalar quantity opposed to a vector. Thus the equation is just

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7 0
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Mathphys Help help help
Keith_Richards [23]

Answer:

1030 mph

Explanation:

The new velocity equals the initial velocity plus the wind velocity.

First, in the x (east) direction:

vₓ = 335 mph + 711 cos 19° mph

vₓ = 1007 mph

And in the y (north) direction:

vᵧ = 0 mph + 711 sin 19° mph

vᵧ = 231 mph

The net speed can be found with Pythagorean theorem:

v² = vₓ² + vᵧ²

v² = (1007 mph)² + (231 mph)²

v ≈ 1030 mph

6 0
3 years ago
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