PHYSICS COLLEGE

Hoover Dam on the Colorado River is the highest dam in the United States at 221m, with a power output of 680 MW. The dam generates electricity by flowing water down to a point 150 m below the stop, at an average flow rate of 650 m3/s.

Answers

Answer 1

Answer:

Answer:

The power in this flow is $9.56*10^(8)\ W$

Explanation:

Given that,

Distance = 221 m

Power output = 680 MW

Height =150 m

Average flow rate = 650 m³/s

Suppose we need to calculate the power in this flow in watt

We need to calculate the pressure

Using formula of pressure

$Pressure=\rho g h$

Where, $\rho$ = density

h = height

g = acceleration due to gravity

Put the value into the formula

$Pressure=1000*9.8*150$

$Pressure=1470000\ Nm^2$

We need to calculate the power

Using formula of power

$P=Pressure* flow\ rate$

Put the value into the formula

$P=1470000*650$

$P=9.56*10^(8)\ W$

Hence, The power in this flow is $9.56*10^(8)\ W$

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Answers

Answer:

4611.58 ft/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 32.174 ft/s²

Equation of motion

$v^2-u^2=2as\n\Rightarrow v=√(2as+u^2)\n\Rightarrow v=√(2* 32.174* 516+0^2)\n\Rightarrow v=182.218\ ft/s$

$v^2-u^2=2as\n\Rightarrow a=(v^2-u^2)/(2s)\n\Rightarrow a=(0^2-182.218^2)/(2* 3.6)\n\Rightarrow a=-4611.58\ ft/s^2$

Magnitude of acceleration while stopping is 4611.58 ft/s²

*PLEASE HELP*A baseball is pitched with a horizontal velocity of 25.21 m/s. Mike Trout hits the ball, sending it in the opposite direction (back toward the pitcher) at a speed of -50.67 m/s. The ball is in contact with the bat for 0.0014 seconds. What is the
acceleration of the ball?

Answers

Answer:

-54,200 m/s^2

Explanation:

a=(vf-vi)/t

As the Moon revolves around the Earth, it also rotates on its axis. Why is it that the same side of the Moon is always visible from Earth?

Answers

Answer: The speed of the moon's rotation keeps the same side always facing Earth.

Explanation: Please mark me brainiest

Answer:

The speed of the Moon's rotation keeps the same side always facing Earth.

Explanation:

got it right on study island :)

A rocket accelerates upwards at 6.20 ft/s/s. How far will the rocket travel in 2 minutes?

Answers

Answer:

44,640 ft

Explanation:

assuming the rocket started from rest, then v₀ = 0

2 min = 120 s

Δx = v₀t + 1/2at²

Δx = 0 + 1/2(6.2 ft/s²)(120 s)² = 44,640 ft ≈ 8.45 mi

A string that passes over a pulley has a 0.341 kg mass attached to one end and a 0.625 kg mass attached to the other end. The pulley, which is a disk of radius 9.00 cm , has friction in its axle.What is the magnitude of the frictional torque that must be exerted by the axle if the system is to be in static equilibrium? (Answer should be in N m)

Answers

Answer:

The frictional torque is $\tau = 0.2505 \ N \cdot m$

Explanation:

From the question we are told that

The mass attached to one end the string is $m_1 = 0.341 \ kg$

The mass attached to the other end of the string is $m_2 = 0.625 \ kg$

The radius of the disk is $r = 9.00 \ cm = 0.09 \ m$

At equilibrium the tension on the string due to the first mass is mathematically represented as

$T_1 = m_1 * g$

substituting values

$T_1 = 0.341 * 9.8$

$T_1 = 3.342 \ N$

At equilibrium the tension on the string due to the mass is mathematically represented as

$T_2 = m_2 * g$

$T_2 = 0.625 * 9.8$

$T_2 = 6.125 \ N$

The frictional torque that must be exerted is mathematically represented as

$\tau = (T_2 * r ) - (T_1 * r )$

substituting values

$\tau = ( 6.125 * 0.09 ) - (3.342 * 0.09 )$

$\tau = 0.2505 \ N \cdot m$

Answer:here to earn points

Explanation:

Certain insects can achieve seemingly impossible accelerations while jumping. the click beetle accelerates at an astonishing 400g over a distance of 0.52 cm as it rapidly bends its thorax, making the "click" that gives it its name. part a assuming the beetle jumps straight up, at what speed does it leave the ground? part b how much time is required for the beetle to reach this speed? part c ignoring air resistance, how high would it go?

Answers

(a) The launching velocity of the beetle is 6.4 m/s

(b) The time taken to achieve the speed for launch is 1.63 ms

(a) The beetle starts from rest and accelerates with an upward acceleration of 400 g and reaches its launching speed in a distance 0.53 cm. Here g is the acceleration due to gravity.

Use the equation of motion,

$v^2=u^2+2as$

Here, the initial velocity of the beetle is u, its final velocity is v, the acceleration of the beetle is a, and the beetle accelerates over a distance s.

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 0.52×10⁻²m for s.

$v^2=u^2+2as\n = (0 m/s)^2+2 (400)(9.8 m/s^2)(0.52*10^-^2 m)\n =40.768 (m/s)^2\n v=6.385 m/s$

The launching speed of the beetle is 6.4 m/s.

(b) To determine the time t taken by the beetle for launching itself upwards is determined by using the equation of motion,

$v=u+at$

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 6.385 m/s for v.

$v=u+at\n 6.385 m/s = (0 m/s) +400(9.8 m/s^2)t\n t = (6.385 m/s)/(3920 m/s^2) = 1.63*10^-^3s=1.63 ms$

The time taken by the beetle to launch itself upwards is 1.62 ms.

(c) After the beetle launches itself upwards, it is acted upon by the earth's gravitational force, which pulls it downwards towards the earth with an acceleration equal to the acceleration due to gravity g. Its velocity reduces and when it reaches the maximum height in its path upwards, its final velocity becomes equal to zero.

Use the equation of motion,

$v^2=u^2+2as$

Substitute 6.385 m/s for u, -9.8 m/s² for g and 0 m/s for v.

$v^2=u^2+2as\n (0m/s)^2=(6.385 m/s)^2+2(-9.8m/s^2)s\n s=((6.385 m/s)^2)/(2(9.8m/s^2)) =2.08 m$

The beetle can jump to a height of 2.1 m

We have that for the Question the Speed,Time and Height are

u=6.38m/s
T=13sec
h=2m

From the question we are told

Certain insects can achieve seemingly impossible accelerations while jumping.

the click beetle accelerates at an astonishing 400g over a distance of 0.52 cm as it rapidly bends its thorax, making the "click" that gives it its name.