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3 years ago

Work occurs when

Physics

2 answers:

azamat3 years ago

8 0

Answer:

The answer is A. an applied force results in movement of an object in the same direction as the applied force.

Explanation:

Work is the product of a force applied to a body and the displacement of the body in the direction of this force, that is, a force performs a work when there is a displacement of its point of application in the direction of that force. The work of the force on that body will be equivalent to the energy needed to move it.

neonofarm [45]3 years ago

6 0

Work occurs when an applied force results in movement of an object in the same direction as the applied force.

Work is done when a force that is applied to an object moves that object. The work is calculated by multiplying the force by the amount of movement of an object (W = F * d). A force of 10 newtons, that moves an object 3 meters, does 30 n-m of work.

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2 years ago

Reception devices pick up the variation in the electric field vector of the electromagnetic wave sent out by the satellite. Give

Keith_Richards [23]

Complete Question

A satellite in geostationary orbit is used to transmit data via electromagnetic radiation. The satellite is at a height of 35,000 km above the surface of the earth, and we assume it has an isotropic power output of 1 kW (although, in practice, satellite antennas transmit signals that are less powerful but more directional).

Reception devices pick up the variation in the electric field vector of the electromagnetic wave sent out by the satellite. Given the satellite specifications listed in the problem introduction, what is the amplitude E0 of the electric field vector of the satellite broadcast as measured at the surface of the earth? Use ϵ0=8.85×10^−12C/(V⋅m) for the permittivity of space and c=3.00×10^8m/s for the speed of light.

Answer:

The electric field vector of the satellite broadcast as measured at the surface of the earth is $E_o = 6.995 *10^{-6} \ V/m$

Explanation:

From the question we are told that

The height of the satellite is $r = 35000 \ km = 3.5*10^{7} \ m$

The power output of the satellite is $P = 1 \ KW = 1000 \ W$

Generally the intensity of the electromagnetic radiation of the satellite at the surface of the earth is mathematically represented as

$I = \frac{P}{4 \pi r^2}$

substituting values

$I = \frac{1000}{4 * 3.142 (3.5*10^{7})^2}$

$I = 6.495*10^{-14} \ W/m^2$

This intensity of the electromagnetic radiation of the satellite at the surface of the earth can also be mathematically represented as

$I = c * \epsilon_o * E_o^2$

Where $E_o$ is the amplitude of the electric field vector of the satellite broadcast so

$E_o = \sqrt{\frac{2 * I}{c * \epsilon _o} }$

substituting values

$E_o = \sqrt{\frac{2 * 6.495 *10^{-14}}{3.0 *10^{8} * 8.85*10^{-12}} }$

$E_o = 6.995 *10^{-6} \ V/m$

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3 years ago

A hot air balloon must be designed to support a basket, cords, and one person for a total payload weight of 1300 N plus the addi

RSB [31]

Answer:

r = 4.44 m

Explanation:

For this exercise we use the Archimedes principle, which states that the buoyant force is equal to the weight of the dislodged fluid

B = ρ g V

Now let's use Newton's equilibrium relationship

B - W = 0

B = W

The weight of the system is the weight of the man and his accessories (W₁) plus the material weight of the ball (W)

σ = W / A

W = σ A

The area of a sphere is

A = 4π r²

W = W₁ + σ 4π r²

The volume of a sphere is

V = 4/3 π r³

Let's replace

ρ g 4/3 π r³ = W₁ + σ 4π r²

If we use the ideal gas equation

P V = n RT

P = ρ RT

ρ = P / RT

P / RT g 4/3 π r³ - σ 4 π r² = W₁

r² 4π (P/3RT r - σ) = W₁

Let's replace the values

r² 4π (1.01 10⁵ / (3 8.314 (70 + 273)) r - 0.060) = 13000

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r² (11.81 r - 0.060) = 1034.51

As the independent term is very small we can despise it, to find the solution

r = 4.44 m

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3 years ago

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zhuklara [117]

Answer:

1:2

Explanation:

It is given that,

Initial RMS AC voltage is 100 V and final RMS AC voltage is 200 V.

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