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3 years ago

If the mass of the earth and all objects on it were suddenly doubled, then the ________

Physics

1 answer:

krok68 [10]3 years ago

3 0

Answer:

The force of universal gravitation between earth and all objects in it will be quadrupled

Explanation:

Newtons law of universal gravitation tell us the force of attraction between two bodies on the earth surface . This force is directly proportional to the product of the masses of the bodies and inversely proportional to their distance apart.

f is directly proportional to m1*m2

doubling these masses will be 2m1*2m2 = 4m1m2

if mas of earth is Me and mass of all objects is Mo

the f is proportional to Me*Mo

doubling the masses of the earth and all objects we have, 2Me*2Mo= 4MeMo

This means that doubling the masses of the earth and all objects on it will cause the force of gravitational attraction to be quadrupled.

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Find the equivalent resistance, the current supplied by the battery and the current through each resistor when the specified res

VLD [36.1K]

Answer:

$R_{eq}$ = 19 Ω, I = 0.105 A, V1 = 1.05 V and V2 = 0.95 V

Explanation:

The correct way to solve this type of problem is to find the current or voltage values for the equivalent resistance and from here find the other values.

For a series circuit the equivalent resistance is the sum of the resistance

$R_{eq}$ = R1 + R2

$R_{eq}$ = 10 +9

$R_{eq}$ = 19 Ω

Let's use the equation for the voltage

V = I $R_{eq}$

I = V / $R_{eq}$

I = 2/19

I = 0.105 A

In a series circuit the current is constant, so let's use the voltage equation for each resistor

V1 = I R1

V1 = 0.105 10

V1 = 1.05 V

V2 = 0.105 9

V2 = 0.95 V

Note that the sum of this voltage is the total voltage applied.

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3 years ago

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dedylja [7]

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3 years ago

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The electric potential at a point equidistant from two particles that have charges +Q and –Q is larger than zero. a. smaller tha

Marizza181 [45]

Answer:

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4 years ago

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gregori [183]

Answer:

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What is the escape speed for an electron initially at rest on the surface of a sphere with a radius of 1.3 cm and a uniformly di

Vlad [161]

Answer:

2.37 * 10^4 m/s

Explanation:

Constants :

Mass of electron = 9.11 * 10^(-31) kg

Electric charge of an electron = 1.602 * 10^(-19) C

Parameters given:

Radius of sphere = 1.3cm = 0.013m

Charge of sphere = 2.3 * 10^(−15) C

Using the law of conservation of energy, we have that:

K. E.(initial) + P. E.(initial) = K. E.(final) + P. E.(final)

K. E.(final) = 0, since final velocity is zero and P. E.(final) = 0 since the electron reaches a final distance of infinity.

Hence,

K. E.(initial) = P. E.(initial)

0.5mv^2 = (kqQ)/r

Where k = Coulumbs constant

Q = charge of the sphere.

r = radius of the sphere.

=> 0.5*m*v^2 = (kqQ)/r

0.5 * 9.11 * 10^(-31) * v^2 = (9 * 10^9 * 1.602 * 10^(-19) * 2.3 * 10^(-15))/0.013

4.555 * 10^(-31) * v^2 = 2550.88 * 10^(-25)

=> v^2 = 2550.88 * 10^(-25) / 4.555 * 10^(-31)

v^2 = 560 * 10^6 = 5.60 * 10^8

=> v = 2.37 * 10^4 m/s

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3 years ago