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2 years ago

Rearrange the equation = KE -1 2 - my? to solve for v. Show your work.

Physics

1 answer:

vesna_86 [32]2 years ago

3 0

Answer:

See below

Explanation:

KE = 1/2 m v^2 multiply both sides by 2

2 (KE) = mv^2 divide both sides by m

2(KE) / m = v^2 sqrt both sides

√ [(2KE)/m ] = v

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A vertical cylindrical tank 10 ft in diameter, has an inflow line of 0.3 ft inside diameter and an outflow line of 0.4 ft inside

neonofarm [45]

Answer:

$\frac{dh}{dt} = 1.3 \times 10^{-3} \frac{ft}{s}$ , level is rising.

Explanation:

Since liquid water is a incompresible fluid, density can be eliminated of the equation of Mass Conservation, which is simplified as follows:

$\dot V_{in} - \dot V_{out} = \frac{dV_{tank}}{dt}$

$\frac{\pi}{4}\cdot D_{in}^2 \cdot v_{in}-\frac{\pi}{4}\cdot D_{out}^2 \cdot v_{out}= \frac{\pi}{4}\cdot D_{tank}^{2} \cdot \frac{dh}{dt} \\D_{in}^2 \cdot v_{in} - D_{out}^2 \cdot v_{out} = D_{tank}^{2} \cdot \frac{dh}{dt} \\\frac{dh}{dt} = \frac{D_{in}^2 \cdot v_{in} - D_{out}^2 \cdot v_{out}}{D_{tank}^{2}}$

By replacing all known variables:

$\frac{dh}{dt} = \frac{(0.3 ft)^{2}\cdot (5 \frac{ft}{s} ) - (0.4 ft)^{2} \cdot (2 \frac{ft}{s} )}{(10 ft)^{2}}\\\frac{dh}{dt} = 1.3 \times 10^{-3} \frac{ft}{s}$

The positive sign of the rate of change of the tank level indicates a rising behaviour.

6 0

3 years ago

8. Il An 8.00 kg package in a mail-sorting room slides 2.00 m down a

Vitek1552 [10]

Answer:

See below

Explanation:

Normal force = m g cos 53 = 8 kg * 9.8 m/s^2 * cos 53 = 47.1823 N

no work is done by this force

Force friction = coeff friction * force normal = .4 * 47.1823 = 7.55 N

work of friction = 7.55 * 2 m = 15.1 j

Force Downplane = mg sin 53 = 62.61 N

work = 62.61 * 2 = 125.22 j

Net Force downplane = force downplane - force friction = 55.06 N

net Work = force * distance = 55.06 N * 2 M = 110.12 j

3 0

2 years ago

Raising 100 grams of water from 40 to 60 °C (the specific heat capacity of water is 1

ANTONII [103]

C. 2000 calories.

Explanation/calculation:

Specific heat capacity = calories / mass * (final temperature - initial temperature)

1 = calories / 100 * (60 - 40)
1 = calories / 100 * 20
1 * (100 * 20) = calories
1 * 2000 = calories
2000 = calories

5 0

3 years ago

Read 2 more answers

n a downhill ski race, surprisingly, little advantage is gained by getting a running start. (This is because the initial kinetic

Annette [7]

Answer:

Explanation:

a ) starting from rest , so u = o and initial kinetic energy = 0 .

Let mass of the skier = m

Kinetic energy gained = potential energy lost

= mgh = mg l sinθ

= m x 9.8 x 70 x sin 30

= 343 m

Total kinetic energy at the base = 343 m + 0 = 343 m .

b )

In this case initial kinetic energy = 1/2 m v²

= .5 x m x 2.5²

= 3.125 m

Total kinetic energy at the base

= 3.125 m + 343 m

= 346.125 m

c ) It is not surprising as energy gained due to gravitational force by the earth is enormous . So component of energy gained due to gravitational force far exceeds the initial kinetic energy . Still in a competitive event , the fractional initial kinetic energy may be the deciding factor .

7 0

3 years ago

Two teams of nine members each engage in tug-of-war. Each of the first team's members has an average mass of 68 kg and exerts an

diamong [38]

Answer:

(a) Acceleration = 0.1063 m/s^2 (Second team wins)

(b) Tension in rope = 65.106 N

Explanation:

Total mass of first team = 68 * 9 = 612 kg

Total force of first team = 1350 * 9 = 12150 N

Total mass of second team = 73 * 9 = 657 kg

Total force of seconds team = 1365 * 9 = 12285 N

Difference in force = 12285 - 12150 = 135 N (towards the second team as it has more force)

(a) For acceleration we get:

F = m * a

135 = (mass of both teams) * a

a = 135 / (612 + 657)

acceleration = 0.1063 m/s^2 (Second team wins)

(b) Since we know the acceleration of the first team (pulling being pulled towards the second team at an acceleration of 0.1063 m/s^2) , we can find out the force required to move them:

Force required for first team = mass of first team * acceleration

Force required = 612 * 0.1063

Force required = 65.106 N

This is the force exerted on the first team through the rope, so the tension in the rope will also be 65.106 N.

7 0

3 years ago