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timurjin [86]
4 years ago
8

explain how many minimum number of geostationary satellites are required for global coverage of T.V transmission

Physics
1 answer:
kobusy [5.1K]4 years ago
4 0

Answer:I honestly don't know

Explanation:

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PLEASE HELP ME, ILL GIVE BRAINLIEST<br><br> Which image is a correctly labeled prokaryotic cell?
yulyashka [42]

Answer:

I believe it's the bottom left :)

Explanation:

4 0
3 years ago
Read 2 more answers
A box rests on the floor. If you pull sideways on the box with a 300 N force, but don't move it, what is the magnitude of the fr
Dahasolnce [82]

The magnitude of the frictional force is B) 300 N

Explanation:

For this problem, we just have to consider the forces acting on the box in the horizontal direction.

There are only two forces in this direction:

- The pull applied by the man, of magnitude F=300 N, forward

- The frictional force, F_f, acting backward

So the equation of motion for the box is

F-F_f = ma

where

m is the mass of the box

a is its acceleration

However, the box is at rest, so its acceleration is zero:

a = 0

This means therefore that

F-F_f=0

And so the magnitude of the frictional force is equal to the pulling force:

F_f = F = 300 N

Learn more about friction:

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5 0
3 years ago
What can happen to the motion of an object when there is a resultant force on it
MA_775_DIABLO [31]
Then, it would continue the motion in direction of resultant force...
3 0
3 years ago
1. A concave mirror has a focal length of 1.50 meters. What is the radius of curvature of the mirror? An object is placed 4.00 m
matrenka [14]

1) 3.0 m

2) 2.40 m

Explanation:

1)

A concave mirror is a reflecting surface that causes the reflection of the rays of light coming to the mirror, producing an image of the object facing the mirror.

There are two types of mirror:

- Concave mirror: this is curved inward - as a result, the rays of light coming from the object are reflected back into a single point, called focal point

- Convex mirror: this is curved outward - as a result, the rays of light coming from the object are reflected back into diverging direction, not into a single point

For a curved mirror, the radius of curvature is twice the focal length:

R=2f

Where

R is the radius of curvature

f is the focal length

In this problem,

f = 1.50 m

So, the radius of curvature is

R=2(1.50)=3.0 m

2)

The distance of the image from the mirror can be found by using the mirror equation:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where

f is the focal length

p is the distance of the object from the mirror

q is the distance of the image from the mirror

IN this problem we have:

f = 1.50 m is the focal length

p = 4.00 m is the distance of the object from the mirror

Solving for q, we  find:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{1.50}-\frac{1}{4.00}=0.416\\q=\frac{1}{0.416}=2.40 m

8 0
3 years ago
A 5.00μF parallel-plate capacitor is connected to a 12.0 V battery. After the capacitor is fully charged, the battery is disconn
EastWind [94]

(a) 12.0 V

In this problem, the capacitor is connected to the 12.0 V, until it is fully charged. Considering the capacity of the capacitor, C=5.00 \mu F, the charged stored on the capacitor at the end of the process is

Q=CV=(5.00 \mu F)(12.0 V)=60 \mu C

When the battery is disconnected, the charge on the capacitor remains unchanged. But the capacitance, C, also remains unchanged, since it only depends on the properties of the capacitor (area and distance between the plates), which do not change. Therefore, given the relationship

V=\frac{Q}{C}

and since neither Q nor C change, the voltage V remains the same, 12.0 V.

(b) (i) 24.0 V

In this case, the plate separation is doubled. Let's remind the formula for the capacitance of a parallel-plate capacitor:

C=\frac{\epsilon_0 \epsilon_r A}{d}

where:

\epsilon_0 is the permittivity of free space

\epsilon_r is the relative permittivity of the material inside the capacitor

A is the area of the plates

d is the separation between the plates

As we said, in this case the plate separation is doubled: d'=2d. This means that the capacitance is halved: C'=\frac{C}{2}. The new voltage across the plate is given by

V'=\frac{Q}{C'}

and since Q (the charge) does not change (the capacitor is now isolated, so the charge cannot flow anywhere), the new voltage is

V'=\frac{Q}{C'}=\frac{Q}{C/2}=2 \frac{Q}{C}=2V

So, the new voltage is

V'=2 (12.0 V)=24.0 V

(c) (ii) 3.0 V

The area of each plate of the capacitor is given by:

A=\pi r^2

where r is the radius of the plate. In this case, the radius is doubled: r'=2r. Therefore, the new area will be

A'=\pi (2r)^2 = 4 \pi r^2 = 4A

While the separation between the plate was unchanged (d); so, the new capacitance will be

C'=\frac{\epsilon_0 \epsilon_r A'}{d}=4\frac{\epsilon_0 \epsilon_r A}{d}=4C

So, the capacitance has increased by a factor 4; therefore, the new voltage is

V'=\frac{Q}{C'}=\frac{Q}{4C}=\frac{1}{4} \frac{Q}{C}=\frac{V}{4}

which means

V'=\frac{12.0 V}{4}=3.0 V

3 0
3 years ago
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