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3 years ago

The depth of water behind the Hoover Dam is 220 m. Find the water pressure at the base of this dam.

1 answer:

6 0

Pressure under water is called the gauge pressure.
$Pressure= \frac{Force}{Area} \\ F=mg \\ P= \frac{mg}{A}$
Now the next part is difficult to put into an equation because you need the symbol for density, which is ρ, pronounced rho. I cannot type ρ into the equation so I will use p.
$p= \frac{m}{V}$ (m=mass, V=Volume)
$m=pV \\ P= \frac{pVg}{A} \\ V=Ah$
A=Area of the base of the object, or cylinder of water above the object, h= height of object
$P= \frac{p(Ah)g}{A}$
The "A"s cancel, leaving us with
$P=pgh$

You are only looking for the water pressure, so you would not be concerned with atmospheric pressure on top of that.

$P_{gauge} =p_{liquid} g h_{object}$

The density of water is $1000 \frac{kg}{ m^{3} }$
Use 1000 for ρ
Gravity is 9.8
Height is 220

$P=pgh \\ P=1000*9.8*220 \\ P=2.16*10^{6} Pa$
(Pa is Pascals, the unit for pressure)

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(a) 1.49 m/s

The conservation of momentum states that the total initial momentum is equal to the total final momentum:

$p_i = p_f\\m u_b + M u_B = (m+M)v$

where

m = 0.016 kg is the mass of the bullet

$u_b = 280 m/s$ is the initial velocity of the bullet

M = 3 kg is the mass of the block

$u_B = 0$ is the initial velocity of the block

v = ? is the final velocity of the block and the bullet

Solving the equation for v, we find

$v=\frac{m u_b}{m+M}=\frac{(0.016 kg)(280 m/s)}{0.016 kg+3 kg}=1.49 m/s$

(b) Before: 627.2 J, after: 3.3 J

The initial kinetic energy is (it is just the one of the bullet, since the block is at rest):

$K_i = \frac{1}{2}mu_b^2 = \frac{1}{2}(0.016 kg)(280 m/s)^2=627.2 J$

The final kinetic energy is the kinetic energy of the bullet+block system after the collision:

$K_f = \frac{1}{2}(m+M)v^2=\frac{1}{2}(0.016 kg+3 kg)(1.49 m/s)^2=3.3 J$

(c) The Energy Principle isn't valid for an inelastic collision.

In fact, during an inelastic collision, the total momentum of the system is conserved, while the total kinetic energy is not: this means that part of the kinetic energy of the system is losted in the collision. The principle of conservation of energy, however, is still valid: in fact, the energy has not been simply lost, but it has been converted into other forms of energy (thermal energy).

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4 years ago

A U-tube is open to the atmosphere at both ends. Water is poured into the tube until the water column on the vertical sides of t

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Answer:

The value is $\Delta h = 0.003 \ m$

Explanation:

From the question we are told that

The height of the water is $h_1 = 10 \ cm = 0.10 \ m$

The density of oil is $\rho_o = 950 \ kg/m^3$

The height of oil is $h_2 = 6 \ cm = 0.06 \ m$

Given that both arms of the tube are open then the pressure on both side is the same

$P_a = P_b$

=> Here

$P_a = P_z + \rho_w * g * h$

where $\rho_w$ is the density of water with value $\rho_w = 1000 \ kg/m^3$

and $P_z$ is the atmospheric pressure

and

$P_b = P_z + \rho_o * g * h_2$

=> $P_z + \rho_w * g * h = P_z + \rho_o * g * h_2$

=> $\rho_w * h = \rho_o * h_2$

=> $h = \frac{950 * 0.06 }{1000}$

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3 years ago

A boy reaches out of a window and tosses a ball straight up with a speed of 10 m/s. The ball is 20 m above the ground as he rele

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MARK ME BRAINLIEST PLEASE!!!!!

The total energy TE = mgh + 1/2 mU^2; where h = 20 m, g = 9.81 m/sec^2, and U = 10 mps. When the ball reaches max height H, all that TE will be potential energy PE = mgH = TE.

So there you are. TE = mgh + 1/2 mU^2 = mgH = TE from the conservation of energy. Solve for H.

1) H = (gh + 1/2 U^2)/g = h + U^2/2g = ? meters where everything on the RHS is given. You can do the math.

2) As the ball drops from H to h, it picks up KE as the potential energy mgH is converted when the potential energy is diminished to mgh, where h < H. So PE - pe = ke = mg(H - h) = 1/2 mv^2 so solve for v = sqrt(2g(H - h)) and, again, everything is given. You can do the math.

3) Same deal as 2) except now its V = sqrt(2gH) because all the PE = mgH = 1/2 mV^2 = KE when it is about to hit the ground. You can do the math.

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