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3 years ago

What is sludge dumping?HELP ASAP PLEASE

Physics

1 answer:

yulyashka [42]3 years ago

5 0

Answer:

What is sludge dumping? Sludge is the solid waste in raw sewage. Sludge dumping is discharging that waste into the ocean.

Explanation:

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Which of the following is a transformer used for?

mamaluj [8]

A transformer is used to increase or decrease a voltage. BUT ... it has to be an AC voltage, or the transformer doesn't work.

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3 years ago

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A amusement park moves riders in a circle at a rate of 6.0m/s if the radius is 9.0 meters what is the acceleration of the ride

oksano4ka [1.4K]

Centripetal acceleration is (speed-squared) / (radius)

CA = (6 m/s)² / (9 m)

CA = (36 m²/s²) / (9 m)

CA = (36/9) (m²/m·s²)

<em>Centripetal acceleration = 4 m/s²</em>

4 0

3 years ago

How many ounces is 200 lbs converted

ddd [48]

Answer:

3,200 ounces

Explanation:

1 pound (lb) is equal to 16 ounces (oz):

i.e 1 lb = 16 oz

Given:

200 pounds

To find:

2,000 pounds as ounces.

Steps:

200 (mass) * 16 = 3,200 ounces.

Thank you!

- EE

3 0

2 years ago

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A 4.80 −kg ball is dropped from a height of 15.0 m above one end of a uniform bar that pivots at its center. The bar has mass 7.

Margarita [4]

Answer:

h = 13.3 m

Explanation:

Given:-

- The mass of ball, mb = 4.80 kg

- The mass of bar, ml = 7.0 kg

- The height from which ball dropped, H = 15.0 m

- The length of bar, L = 6.0 m

- The mass at other end of bar, mo = 5.10 kg

Find:-

The dropped ball sticks to the bar after the collision.How high will the other ball go after the collision?

Solution:-

- Consider the three masses ( 2 balls and bar ) as a system. There are no extra unbalanced forces acting on this system. We can isolate the system and apply the principle of conservation of angular momentum. The axis at the center of the bar:

- The angular momentum for ball dropped before collision ( M1 ):

M1 = mb*vb*(L/2)

Where, vb is the speed of the ball on impact:

- The speed of the ball at the point of collision can be determined by using the principle of conservation of energy:

ΔP.E = ΔK.E

mb*g*H = 0.5*mb*vb^2

vb = √2*g*H

vb = ( 2*9.81*15 ) ^0.5

vb = 17.15517 m/s

- The angular momentum of system before collision is:

M1 = ( 4.80 ) * ( 17.15517 ) * ( 6/2)

M1 = 247.034448 kgm^2 /s

- After collision, the momentum is transferred to the other ball. The momentum after collision is:

M2 = mo*vo*(L/2)

- From principle of conservation of angular momentum the initial and final angular momentum remains the same.

M1 = M2

vo = 247.03448 / (5.10*3)

vo = 16.14604 m/s

- The speed of the other ball after collision is (vo), the maximum height can be determined by using the principle of conservation of energy:

ΔP.E = ΔK.E

mo*g*h = 0.5*mo*vo^2

h = vo^2 / 2*g

h = 16.14604^2 / 2*(9.81)

h = 13.3 m

3 0

3 years ago

n the railroad accident, a boxcar weighting 200 kN and traveling at 3 m/s on horizontal track slams into a stationary caboose we

USPshnik [31]

Answer:

ΔK = -6 10⁴ J

Explanation:

This is a crash problem, let's start by defining a system formed by the two trucks, so that the forces during the crash have been internal and the moment is preserved

initial instant. Before the crash

p₀ = m v₁ + M 0

final instant. Right after the crash

p_f = (m + M) v

p₀ = p_f

mv₁ = (m + M) v

v = $\frac{m}{m+M} \ v_1$

we substitute

v = $\frac{20}{20+40}$ 3

v = 1.0 m / s

having the initial and final velocities, let's find the kinetic energy

K₀ = ½ m v₁² + 0

K₀ = ½ 20 10³ 3²

K₀ = 9 10⁴ J

K_f = ½ (m + M) v²

K_f = ½ (20 +40) 10³ 1²

K_f = 3 10⁴ J

the change in energy is

ΔK = K_f - K₀

ΔK = (3 - 9) 10⁴

ΔK = -6 10⁴ J

The negative sign indicates that the energy is ranked in another type of energy

7 0

3 years ago