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2 years ago

Which is a result of using a machine?

Physics

1 answer:

kakasveta [241]2 years ago

4 0

Answer:

Yes labor is required therefor jobs like cashier, and other low skill labor jobs I'll not be required

Explanation:

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Gravity on earth is 9.8 m/s squared, and gravity on the moon is 1.6 m/s squared. So if the mass of an object on earth is 40 kilo

garik1379 [7]

The mass of an object on Earth is the same as its mass on the Moon. The weight is different.

Weight = m * g

Weight ( Moon ) = 40 kg * 1.6 m/s² = 64 N

If the mass of an object on Earth is 40 kg, its mass on the Moon is 40 kg and its weight on the Moon is 64 N.

7 0

3 years ago

A block of wood is 4 cm wide, 5 cm long, and 10 cm high. It weighs 100 grams. Calculate its volume. Calculate its density. Will

77julia77 [94]

Answer:

$V=200cm^3\\\\\rho =0.500g/cm^3$

It will float.

Explanation:

Hello.

In this case, given the width, length and height, we can compute the volume as follows:

$V=W*L*H\\\\V=4cm*5cm*10cm\\\\V=200cm^3$

Moreover, since the density is computed via the division of the mass by the volume:

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

We obtain:

$\rho =\frac{100g}{200cm^3} \\\\\rho =0.500g/cm^3$

In such a way, since the solid has a lower density than the water, we infer it will float.

Best regards.

3 0

3 years ago

What's is the kinetic energy of a .235-kg baseball thrown at 50.0m/s

san4es73 [151]

Answer:

34.51

Explanation:

k=1/2mv² is the kenetic energy equation to fill is in

k=[1/2(0.235)×50]²

6 0

2 years ago

Help me with physics

Llana [10]

Answer:

Explanation:

6 0

3 years ago

A ray of light incident in water strikes the surface separating water from air making an angle of 10 ° with the normal to the su

labwork [276]

Answer:

$\theta _2 = 13^o$

$\theta _1 =32.94^o$

$\theta_c = 53.05^o$

Explanation:

From the question we are told that

The angle of incidence is $\theta_1 = 10^o$

The refractive index of water is $n_1 = 1.3$

Generally Snell's law is mathematically represented as

$n_1 sin(\theta_1) = n_2 sin(\theta_ 2)$

Here $n_2$ is the refractive index of air with value $n_2 = 1$

$\theta_2$ is the angle of refraction

$\theta _2 = sin^{-1}[\frac{n_1 * sin(\theta _1)}{n_2} ]$

=> $\theta _2 = sin^{-1}[\frac{1.3 * sin(10)}{1} ]$

=> $\theta _2 = 13^o$

Given that the angle should not be greater than $\theta _2 =45^o$ then the angle of incidence will be

$\theta _1 = sin^{-1}[\frac{n_2 * sin(\theta _2)}{n_1} ]$

=> $\theta _1 = sin^{-1}[\frac{1 * sin(45)}{1.3} ]$

=> $\theta _1 =32.94^o$

Generally for critical angle is mathematically represented as

$\theta_c = sin^{-1}[\frac{n_2}{n_1} ]$

=> $\theta_c = sin^{-1}[\frac{1}{1.3} ]$

=> $\theta_c = 53.05^o$

4 0

2 years ago