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3 years ago

How long would it take a 500. W electric motor to do 15010 J of work?

Physics

1 answer:

KIM [24]3 years ago

8 0

Time = energy / power = 15010 / 500 = .... seconds

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How would you set it up?

Firdavs [7]

Answer:

THE LAST OPTION

Explanation:

5 0

4 years ago

The mysterious visitor that appears in the enchanting story The Little Prince was said to come from a planet that "was scarcely

FromTheMoon [43]

Answer:

Part a)

$g = 1 \times 10^{-5} m/s^2$

Part b)

$v = 0.011 m/s$

Explanation:

Part a)

As we know that the acceleration due to gravity is given as

$g = \frac{GM}{R^2}$

$g = \frac{G(\frac{4}{3}\pi R^3 \rho)}{R^2}$

$g = \frac{4}{3}\pi \rho G R$

now we know that

$\rho = \frac{M}{\frac{4}{3}\pi R^3}$

$\rho = \frac{5.98 \times 10^{24}}{\frac{4}{3}\pi (6.37 \times 10^6)^3}$

$\rho = 5523.2 kg/m^3$

now we have

$g = \frac{4}{3}\pi (5523.2) (6.67 \times 10^{-11})(6.5)$

$g = 1 \times 10^{-5} m/s^2$

Part b)

As we know that the escape speed is given as

$v = \sqrt{\frac{2GM}{R}}$

$v = \sqrt{2gR}$

$v = \sqrt{2(1\times 10^{-5})(6.5)}$

$v = 0.011 m/s$

8 0

3 years ago

What is the density of a 2 gallon milk jug that has a mass of 2.0 kg? Answer should be in g/ml.

kirill [66]

volume of milk is given as

$V = 2 Gallon$

we will convert it into mL unit

$V = 2 Gal = 7570.82 ml$

mass of the milk m = 2kg

m = 2000 g

now for the density we can use

$\rho = \frac{m}{V}$

$\rho = \frac{2000}{7570.82}$

$\rho = 0.264 g/mL$

<em>so the density is 0.264 g/mL for above sample</em>

8 0

4 years ago

A student stretches a spring, attaches a 1.20 kg mass to it, and releases the mass from rest on a frictionless surface. The resu

Sphinxa [80]

Answer:

the speed of the mass when it is halfway to the equilibrium position is 1.26 m/s

Explanation:

Given that;

mass of the object m = 1.20 kg

period of oscillation = 0.750 s

Amplitude ( A/x) = 15.0 cm = 0.15 m

now;

a) Determine the oscillation frequency;

oscillation frequency f = 1/T

we substitute

f = 1 / 0.750 s

f = 1.33 Hz

Therefore, the oscillation frequency is 1.33 Hz

b) Determine the spring constant;

we solve for spring constant from the following expression;

T = 2π√(m/k)

k = 4π²m / T²

so we substitute

k = (4π² × 1.20) / (0.750)²

k = 47.3741 / 0.5625

k = 84.22 N/m

Therefore, the spring constant is 84.22 N/m

c) determine the speed of the mass when it is halfway to the equilibrium position

So, at equilibrium, the energy is equal to K.E

such that;

1/2mv² = 1/2kx²

mv² = kx²

v² = kx² / m

v = √( kx²/m)

we substitute

v = √( 84.22×(0.15 m)²/ 1.2 )

v = √( 1.89495 / 1.2 )

v = √ 1.579125

v = 1.26 m/s

Therefore, the speed of the mass when it is halfway to the equilibrium position is 1.26 m/s

3 0

3 years ago

Hi hi someone talk to meeeee

Elina [12.6K]

Hi I’ll talk to you!

4 0

3 years ago