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3 years ago

PHYSICSSS HELP ME PLEASEEEE

Physics

1 answer:

Drupady [299]3 years ago

4 0

Answer:

Volume (of a liquid) = quantities

Pound = Units

Kilogram - meter per second = Units

Weight = quantities

Newton = quantities

Dollar = Units

Impulse = quantities

Temperature = quantities

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A heat pump absorbs heat from the atmosphere at a rate of 30 kW. If work is being done to run this heat pump at a rate of 7.7 kW

Volgvan

Answer:

Option D 3.9

Explanation:

First, you need to use the correct equation which is the following:

COP = Q/W

Where:

Q = heat absorbed

W = work done by the pump

COP = coefficient of perfomance

We have all the data, so, all you need to do is replace in the above expression and you shoould get the correct result:

COP = 30 / 7.7

COP = 3.896

This result you can round it to 3.9. option D.

8 0

3 years ago

What is the kinetic energy of a 0.25kg ball rolling at a speed of 2.5m/s

Lisa [10]

Answer:

Explanation:

Kinetic Energy formula:

KE = $\frac{1}{2}$ mv²

m=mass

v=speed

Given:

m=0.25kg

v=2.5m/s

Plug the values in:

KE = 1/2(0.25kg)(2.5m/s)²

KE = 0.78125 J (Joules)

4 0

3 years ago

A physics student of mass 43.0 kg is standing at the edge of the flat roof of a building, 12.0 m above the sidewalk. An unfriend

Dmitry_Shevchenko [17]

Answer:

The speed of the student just before she lands, v₂ is approximately 8.225 m/s

Explanation:

The given parameters are;

The mass of the physic student, m = 43.0 kg

The height at which the student is standing, h = 12.0 m

The radius of the wheel, r = 0.300 m

The moment of inertia of the wheel, I = 9.60 kg·m²

The initial potential energy of the female student, P.E.₁ = m·g·h₁

Where;

m = 43.0 kg

g = The acceleration due to gravity ≈ 9.81 m/s²

h = 12.0 h

∴ P.E.₁ = 43 kg × 9.81 m/s² × 12.0 m = 5061.96 J

The kinetic rotational energy of the wheel and kinetic energy of the student supporting herself from the rope she grabs and steps off the roof, K₁, is given as follows;

$K_1 = \dfrac{1}{2} \cdot m \cdot v_{1}^2+\dfrac{1}{2} \cdot I \cdot \omega_{1}^2$

The initial kinetic energy, 1/2·m·v₁² and the initial kinetic rotational energy, 1/2·m·ω₁² are 0

∴ K₁ = 0 + 0 = 0

The final potential energy of the student when lands. P.E.₂ = m·g·h₂ = 0

Where;

h₂ = 0 m

The final kinetic energy, K₂, of the wheel and student is give as follows;

$K_2 = \dfrac{1}{2} \cdot m \cdot v_{2}^2+\dfrac{1}{2} \cdot I \cdot \omega_{2}^2$

Where;

v₂ = The speed of the student just before she lands

ω₂ = The angular velocity of the wheel just before she lands

By the conservation of energy, we have;

P.E.₁ + K₁ = P.E.₂ + K₂

∴ m·g·h₁ + $\dfrac{1}{2} \cdot m \cdot v_{1}^2+\dfrac{1}{2} \cdot I \cdot \omega_{1}^2$ = m·g·h₂ + $\dfrac{1}{2} \cdot m \cdot v_{2}^2+\dfrac{1}{2} \cdot I \cdot \omega_{2}^2$

Where;

ω₂ = v₂/r

∴ 5061.96 J + 0 = 0 + $\dfrac{1}{2} \times 43.0 \, kg \times v_{2}^2+\dfrac{1}{2} \times 9.60 \, kg\cdot m^2 \cdot \left (\dfrac{v_2}{0.300 \, m} }\right ) ^2$