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3 years ago

How far should the second antenna be moved in order to receive a minimum signal from a station that broadcasts at 99.4 MHz?

Physics

1 answer:

raketka [301]3 years ago

5 0

Answer:

So the distance of the antenna from the station will be 3.018 m

Explanation:

We have given the frequency of the broadcast $f=99.4MHz=99.4\times 10^4Hz$

The speed of light $c=3\times 10^8m/sec$

The distance of the antenna to receive a minimum signal from the station is given by $d=\frac{v}{f}=\frac{3\times 10^{8}}{99.4\times 10^6}=3.018m$

So the distance of the antenna from the station will be 3.018 m

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Please help me with this question

Oksanka [162]

Car at rest:

velocity= 0m/s

Acceleration:

0.2m/s²

Since total time:

3 min = 180s

Formula of acceleration:

acceleration = [final velocity - initial velocity] ÷ [total time]

Velocity at end:

0.2m/s² = [final velocity - 0m/s] ÷ [180s]

0.2m/s² × 180s = [final velocity]

[final velocity] = 36m/s

Distance travelled:

Velocity = displacement(distance) ÷ time

36m/s = displacement(distance) ÷ 180s

displacement(distance) = 36m/s × 180s

displacement(distance) = 6480m

Hey I'm sorry but i do not understand why the answer on your worksheet for distance travelled is 3240m... its half of what my answer is...

6 0

3 years ago

Suppose you want to use this human engine to lift a 2.35 kg box from the floor to a tabletop 1.30 m above the floor. How much mu

morpeh [17]

Answer:

The increase in the gravitational potential energy is 29.93 joules.

Explanation:

Given that,

Mass of the box, m = 2.35 kg

It is lifted from the floor to a tabletop 1.30 m above the floor, h = 1.3 m

We need to find the increase the gravitational potential energy. Initial it will placed at ground i.e. its initial gravitational potential is equal to 0. The increase in the gravitational potential energy is given by :

$U=mgh$

$U=2.35\ kg\times 9.8\ m/s^2\times 1.3\ m$

U = 29.93 Joules

So, the increase in the gravitational potential energy is 29.93 joules. Hence, this is the required solution.

6 0

3 years ago

A simple harmonic oscillator has amplitude 0.43 m and period 3.9 sec. What is the maximum acceleration?

rjkz [21]

Answer:

Maximum acceleration in the simple harmonic motion will be $0.854rad/sec^2$

Explanation:

We have given amplitude of simple harmonic motion is A = 0.43 m

Time period of the oscillation is T = 3.9 sec

We have to find the maximum acceleration

For this we have to find the angular frequency

Angular frequency will be equal to $\omega =\frac{2\pi }{T}=\frac{2\times 3.14}{3.9}=1.61rad/sec$

Maximum acceleration is given by $a_{max}=\omega ^2A=1.61^2\times 0.43=0.854rad/sec^2$

So maximum acceleration in the simple harmonic motion will be $0.854rad/sec^2$

8 0

3 years ago

Read 2 more answers

a soccer player kicks a ball. According to Newton's third law of motion, the player's foot exerts a force on the ball. The ball

Leto [7]

Answer:

The forces are exerted on different objects so they are not balanced forces.

Explanation:

8 0

3 years ago

With 51 gallons of fuel in its tank, the airplane has a weight of 2390.7 pounds. What is the weight of the plane with 81 gallons

Shtirlitz [24]

Answer: 2561.7 pounds

Explanation:

If we assume the total weight of an airplane (in pounds units) as a linear function of the amount of fuel in its tank (in gallons) and we make a Weight vs amount of fuel graph, which resulting slope is 5.7, we can use the slope equation of the line:

$m=\frac{Y-Y_{1}}{X-X_{1}}$ (1)

Where:

$m=5.7$ is the slope of the line

$Y_{1}=2390.7pounds$ is the airplane weight with 51 gallons of fuel in its tank (assuming we chose the Y axis for the airplane weight in the graph)

$X_{1}=51gallons$ is the fuel in airplane's tank for a total weigth of 2390.7 pounds (assuming we chose the X axis for the a,ount of fuel in the tank in the graph)