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3 years ago

If all objects that have mass also have gravity, why doesn't your pencil get pulled towards you while it sits on your desk?

Physics

1 answer:

Lemur [1.5K]3 years ago

5 0

Answer:

The pencil is not pulled towards a person due to a very small magnitude of force between them, due to lighter masses.

Explanation:

Let us apply Newton's Law of Gravitation between a person and pencil.

Average Mass of a Normal Pencil = m₁ = 10 g = 0.01 kg

Average Mass of a Person = m₂ = 80 kg

Distance between both = r = 1 cm = 0.01 m (Taking minimal distance)

Gravitational Constant = G = 6.67 x 10⁻¹¹ N.m²/kg²

So,

F = Gm₁m₂/r²

F = (6.67 x 10⁻¹¹ N.m²/kg²)(0.01 kg)(80 kg)/(0.01 m)²

This Force is very small in magnitude due to the light masses of both objects.

<u>Therefore, the pencil is not pulled towards a person due to a very small magnitude of force between them, due to lighter masses.</u>

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An RL series circuit is connected to an ac generator with a maximum emf of 20 V. If the maximum potential difference across the

Rama09 [41]

If the maximum emf of the ac generator is 20 V and the maximum potential difference across the resistor is 16 V Then the maximum potential difference across the inductor is 4 V.

Calculation:

Step-1:

It is given that the RL circuit is connected to a 20 V ac generator. The maximum potential difference across the resistor is 16 V. It is required to find the maximum potential drop across the inductor.

Step-2:

The maximum emf of the generator is equal to the sum of the maximum potential difference across the resistor and the maximum potential difference across the inductor.

Therefore,

The maximum potential difference across the inductor + Maximum maximum potential difference across the resistor = Maximum emf of the generator

Thus,

Maximum maximum potential difference across the inductor + 16 V = 20 V

Therefore,

Maximum maximum potential difference across the inductor = 20 V - 16 V = 4 V

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2 years ago

Which is the correct definition for sea floor spreading

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Seafloor spreading is a geologic process in which tectonic plates—large slabs of Earth's lithosphere—split apart from each other.

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2 years ago

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You and a friend work in buildings five equal-length blocks apart, and you plan to meet for lunch. Your friend strolls leisurely

Pavel [41]

Answer:

Your friend is 2.143 blocks from the restaurant.

You are 2.857 blocks from the restaurant.

Explanation:

Let t be the time both you and your friend take to walk to the restaurant.

The distance (m) from your building to the restaurant is your walking time t times your speed v1

$s_1 = v_1t = 1.6t$

Similarly the distance (m) from your friend building to the restaurant:

$s_2 = v_2t = 1.2t$

Let b be the length (in m) of a block, the total distance of 5 blocks is 5b

$s_1 + s_2 = 5b$

$1.6t + 1.2t = 5b$

$2.8t = 5b$

$t = 5b/2.8 = 25b/14$

$s_2 = 1.2t = 1.2(25b/14) = 2.143b$

So your friend are 2.143b meters from the restaurant, since each block is b meters long, 2.143b meters would equals to 2.143b/b = 2.143 blocks. And you are 5 - 2.143 = 2.857 blocks from the restaurant.

7 0

3 years ago

A shopper does 157 J of work pushing a cart with 10.9 N force

Tanzania [10]

The cart travelled a distance of 14.4 m

Explanation:

The work done by a force when pushing an object is given by:

$W=Fd cos \theta$

where:

F is the magnitude of the force

d is the displacement

$\theta$ is the angle between the direction of the force and the displacement

In this problem we have:

W = 157 J is the work done on the cart

F = 10.9 N is the magnitude of the force

$\theta=0^{\circ}$ , assuming the force is applied parallel to the motion of the cart

Therefore we can solve for d to find the distance travelled by the cart:

$d=\frac{W}{F cos \theta}=\frac{157}{(10.9)(cos 0)}=14.4 m$

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2 years ago

When UV light of wavelength 248 nm is shone on aluminum metal, electrons are ejected withmaximum kinetic energy 0.92 eV. What ma

Lina20 [59]

Answer:

The maximum wavelength of light that could liberate electrons from the aluminum metal is 303.7 nm

Explanation:

Given;

wavelength of the UV light, λ = 248 nm = 248 x 10⁻⁹ m

maximum kinetic energy of the ejected electron, K.E = 0.92 eV

let the work function of the aluminum metal = Ф

Apply photoelectric equation:

E = K.E + Ф

Where;

Ф is the minimum energy needed to eject electron the aluminum metal

E is the energy of the incident light

The energy of the incident light is calculated as follows;

$E = hf = h\frac{c}{\lambda} \\\\where;\\\\h \ is \ Planck's \ constant = 6.626 \times 10^{-34} \ Js\\\\c \ is \ speed \ of \ light = 3 \times 10^{8} \ m/s\\\\E = \frac{(6.626\times 10^{-34})\times (3\times 10^8)}{248\times 10^{-9}} \\\\E = 8.02 \times 10^{-19} \ J$

The work function of the aluminum metal is calculated as;

Ф = E - K.E

Ф = 8.02 x 10⁻¹⁹ - (0.92 x 1.602 x 10⁻¹⁹)

Ф = 8.02 x 10⁻¹⁹ J - 1.474 x 10⁻¹⁹ J

Ф = 6.546 x 10⁻¹⁹ J

The maximum wavelength of light that could liberate electrons from the aluminum metal is calculated as;

$\phi = hf = \frac{hc}{\lambda_{max}} \\\\\lambda_{max} = \frac{hc}{\phi} \\\\\lambda_{max} = \frac{(6.626\times 10^{-34}) \times (3 \times 10^8) }{6.546 \times 10^{-19}} \\\\\lambda_{max} = 3.037 \times 10^{-7} m\\\\\lambda_{max} = 303.7 \ nm$

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3 years ago