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Evgesh-ka [11]
3 years ago
10

An object moving with a speed of 35 m/a and has a kinetic energy of 1500j, what is the mass of the object

Physics
1 answer:
EleoNora [17]3 years ago
5 0

Explanation:

Speed or velocity (V) = 35 m/s

Kinetic energy (K. E) = 1500 Joule

mass (m) = ?

We know

K.E = 1/2 * m * v²

1500 = 1/2 * m * 35²

1500 * 2 = 1225m

m = 3000 / 1225

m = 2.45 kg

The mass of the object is 2.45 kg

Hope it will help :)

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Answer:

i) The pressure acting on the base of <em>B</em> will be half the pressure acting on the base of <em>A</em>

ii) The pressure acting on the base of <em>B</em> will be the same as the pressure acting on the base of <em>A</em>

iii) The pressure on the base of drum <em>A</em> will be slightly less than the pressure on the base of drum <em>B</em>

Explanation:

The pressure acting on the base of the drum, P = h·ρ·g

Where;

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h_{max} = The height of the drums

ρ = The density of the liquid in the drum

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

i) If <em>A</em> is completely filled, we have h_A = h_{max}

Therefore, P_A = h_{max}×\rho_{liquid}×g

If <em>B</em> is half filled, we have, h_B =  (1/2)·h_{max}

P_B = (1/2) × h_{max}×\rho_{liquid}×g

Therefore, P_B = (1/2) × P_A

The pressure acting on the base of <em>B</em> will be half the pressure acting on the base of <em>A</em>

ii) If both <em>A</em> and <em>B</em> are each filled with water (the same liquid), then the pressure on their bases will be P_A = h_{max}×\rho_{water}×g = P_B, the same, given that the acceleration due to gravity, <em>g</em>, is constant and the same in Nepal and India

iii) If <em>A</em> is filled with water, and <em>B</em> is filled with salty water, we have that, the density of salty water is slightly higher than water, therefore, we get;

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Answer:

a)  M = 4,997 10²⁰ kg ,  b)   T = 1.43 10³ s

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          v = v₀ - a t

As it indicates that there is no atmosphere, the friction force is zero and the initial and final velocity have the same module, but the opposite direction

         a = (v₀ - v) / t

         a = (15 - (-15)) /9.00 = 30/9

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Now we use Newton's second law where force is the force of universal attraction

          F = m a

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Let's calculate

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         M = 4,997 10²⁰ kg

b) The period of the ship's orbit

In this case we have a centripetal acceleration

The radius of the orbit is the radius of the plant plus the height of the ship from the surface

         R = R_{m} + h

         R = 1 10⁵ + 2.00 10⁴

         R = 12 10⁴ m

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         a = v² / R

The orbit is circular therefore the velocity module is constant, so we can use the equation of uniform motion, where the distance is the length of the orbit, for a circle

        d = 2π R

        v = d / t

        v = 2π R / T

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        G m M / R² = m (2π R / T)² / R

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Answer:

0.139 rad

Explanation:

We use Snell's law n_1sin\theta_1=n_2sin\theta_2, where if n_1 is the <em>refractive index</em> of the medium containing the <em>incident ray</em>, \theta_1 would be the <em>incident angle</em>, and if n_2 is the <em>refractive index</em> of the medium containing the <em>refracted ray</em>, \theta_2 would be the <em>refraction angle</em>, which we want, so we do:

sin\theta_2=\frac{n_1}{n_2}sin\theta_1

And finally:

\theta_2=arcsin(\frac{n_1}{n_2}sin\theta_1)

We then insert our values:

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