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3 years ago

Good Luck! And Happy Weekend!

Physics

1 answer:

Anastaziya [24]3 years ago

7 0

<h3><u>Question</u> :</h3>

Circle the scenario that shows balanced forces and will result in no movement of the gold bar.

<h3><u>Answer</u> :</h3><h3 />

1st one is the scenario that shows balanced forces and will result in no movement of the gold bar. (circled in attachment)

<h3><u>Explanation</u> :</h3>

In the first scenario there are equal forces of 5N acting on both sides of the gold bar, but this is not the case for the second and third scenarios. Equal forces are acting on both sides of the bar in the first scenario, hence showing balanced forces and resulting in no movement of the gold bar.

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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

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

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

4 0

4 years ago

Define inertia, give its classification.

strojnjashka [21]

Answer:

The tendeny of a body to continue its state either motion or rest is called inertia . First law of newton also called law of inertia .

There are three types of inertia

1. Motion inertia

2. Rest inertia

3. Directional inertia

Explanation:

Mark brainliest if you undersand

7 0

2 years ago

The size of the gravitational force between two objects depends on their__. frictional force, inertia, masses and the distance b

Inessa [10]

Answer:

Masses and distance between them

Explanation:

The gravitational force between two objects can be calculated using Newton's Gravitational Law.

However, using logic, we can already dictate what the answer will be, for example. We know that the bigger an object is, the stronger its gravity is. This can be seen with how the moon is much smaller, and also has much less gravity.

Also, the distance between two objects also influences the gravity. This can be seen the further an object gets from Earth, the less of a pull the gravitational field has on it. Another example is that Pluto (being very far from the sun) has less of a gravitational effect from the sun, in comparison to Mercury (the closest plant to the sun).

3 0

3 years ago

A. Draw the electric field lines around a negative charge.

Alborosie

<h2>a. Answer:</h2>

We use Electric field lines for visualizing electric fields, so this helps us to see the problem more real. So an electric field line is an imaginary line or curve drawn through a region of space such that the tangent at any point comes from the direction of the electric-field vector at that point. The electric field lines around a negative charge is shown in the First figure below.

<h2>b. Answer:</h2>

Electric forces can be found by using the Coulomb Law's that states <em>that The magnitude of the electric force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. </em>This can be expressed as follows:

$F=k\frac{\left | q_{1}q_{2} \right |}{r^2} \\ \\ Where: \\ \\ k=9\times 10^9Nm^2/c^2 \\ \\ q_{1}=0.00150 C \ and \ q_{2}=0.00240 C \\ \\ r=0.900 m$

Then:

$F=9\times 10^9\frac{\left | 0.00150 \times 0.00240 \right |}{(0.900)^2} \\ \\ \therefore \boxed{F=40000N}$

This force is repulsive because the two charges are positive and recall that two positive charges or two negative charges repel each other while a positive charge and a negative charge attract each other.

<h2>C. Answer:</h2>

From the statement, we have two charged objects. Let's say that this charges are:

$q_{1} \ and \ q_{2}$

If the amount of charge on one of the objects is tripled, let's say this is the charge $q_{2}$ , then the new charge is:

$q_{N}=3q_{2}$

In the formula of Coulomb:

$F=k\frac{\left | q_{1}q_{N} \right |}{r^2} \\ \\ \therefore F=k\frac{\left | q_{1}(3q_{2}) \right |}{r^2} \\ \\ \therefore \boxed{F=3k\frac{\left | q_{1}q_{2} \right |}{r^2}}$

<em>The conclusion is that if the amount of charge on one of the objects is tripled, the electric force between two charged objects is also tripled</em>

<h2>d. Answer:</h2>

Let's use the Coulomb's Law again to solve this problem. We want to know how the electric force between two charged objects changes if the charges are moved closer together:

$F=k\frac{\left | q_{1}q_{N} \right |}{r^2}$

<em>By saying that the charges are moved closer together, we want to express that r becomes smaller. Since r is in the denominator, this implies that the electric force between these two charged objects becomes greater.</em>

<h2>e. Answer:</h2>

From the figure, we can see a metal sphere on a stand. There we have both positive and negative charges. We can say that the positive charge of this sphere is +10q and the negative and the negative charge is -10q. Since the electric charge is conserved, then the algebraic sum of all the electric charges in any closed system is constant. In conclusion, <em>the sphere has no net charge.</em>

<h2>f. Answer:</h2>

Here we want to know how the negative charges in the same sphere are redistributed when a positively charged rod is brought near it. Therefore, positive charge on rod repels positive charges on the sphere, creating zones of negative and positive charge as indicated in the second Figure.