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3 years ago

if you drop a stone from height of 2.5m. what is the speed of the stone right before it hits the ground?

1 answer:

6 0

Since the stone was dropped from height, its initial velocity = 0 m/s

Using v² = u² + 2gs.

Where g ≈ 10 m/s², u = initial velocity = 0 m/s, s = height from drop = 2.5 m

v² = u² + 2gs

v² = 0² + 2*10*2.5

v² = 0 + 50

v² = 50

v = √50

v ≈ 7.07 m/s

Hence velocity just before hitting the ground is ≈ 7.07 m/s

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Answer:

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Explanation:

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3 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

7) Three resistors having resistances of 4.0 Ω, 6.0 Ω, and 10.0 Ω are connected in parallel. If the combination is connected in

viva [34]

Answer:

A, 0.59A

Explanation:

The total resistance in the circuit is the resistances in parallel plus that in series.

Total resistance for those in parallel is;

1/(1/4 +1/6 +1/10) = 1/ (15+10+6 /60)

1/(31/60)= 60/31 ohms

Hence total resistance of the circuit is;

60/31 + 2 = (60+62)/31 = 122/31=3.94 ohms

To calculate the current flowing through the 10ohm resistance we need to know the voltage drop by subtracting the voltage drop in the 2ohm resistance from the total voltage drop.

Voltage drop on the 2 ohm resistance is;

Current on the 2 ohm resistor × 2 ohms

V = I ×R ; I - current

R - resistance

Current drop on the 2ohm resistance is;

Total voltage in the circuit/ total resistance in the circuit

12/3.94= 3.05A

Voltage drop on the 2 ohm resistance;

3.05 × 2 = 6.10volts

Hence voltage drop on the parallel resistance would be ;

12-6.10= 5.90V

Now voltage drop in a parallel circuit is the same hence 5.90v is dropped in each of the parallel resistance.

That said, the current drop on the 10 ohm resistor would be;

5.90/10 = 0.59A

Remember V= I× R so that I = V/R

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3 years ago

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pickupchik [31]

Answer:

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Explanation:

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3 years ago