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Law Incorporation [45]
3 years ago
14

What is the mass of an object that experiences a gravitational force of 510 N near Earth's surface?

Physics
2 answers:
sineoko [7]3 years ago
8 0

Answer:The mass of an object is 52 kg.

Explanation:

Gravitational force on the object ,F=510 N

Acceleration due to gravity = g = 9.8 m/s^2

Mass of the object = m

Force = mass × acceleration

510 N=m\times 9.8 m/s^2

m=52.04 kg\approx 52 kg

The mass of an object is 52 kg.

Ksju [112]3 years ago
3 0

Answer

52.0 kg

The force is due to gravity is given by:

Force = Mass × gravity

Let the acceleration due to gravity(g) on the surface of the earth to be 9.8 m/s²

510 N = m × 9.8

9.8m = 510

m = 510/9.8

    = 52.04 Kg

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Two thin concentric spherical shells of radii r1 and r2 (r1 < r2) contain uniform surface charge densities V1 and V2, respect
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Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

So,

a)  0 < r < r1 :

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

Hence, E = 0 for r < r1

b)  r1 < r < r2:

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E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   } x \frac{r1^{2} }{r1^{2} }

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c) r > r2 :

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E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   } + \frac{Q2}{E_{0} . 4 \pi. r^{2}   }

As we know from above, that:

\frac{Q1}{E_{0} . 4 \pi. r^{2}   } =  (σ1 x r1^{2}) /(E_{0} x r^{2})

Then, Similarly,

\frac{Q2}{E_{0} . 4 \pi. r^{2}   } = (σ2 x r2^{2}) /(E_{0} x r^{2})

So,

E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   } + \frac{Q2}{E_{0} . 4 \pi. r^{2}   }

Replacing the above equations to get E:

E = (σ1 x r1^{2}) /(E_{0} x r^{2}) + (σ2 x r2^{2}) /(E_{0} x r^{2})

Now, for

d) Under what conditions,  E = 0, for r > r2?

For r > r2, E =0 if

σ1 x r1^{2} = - σ2 x r2^{2}

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