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DedPeter [7]
3 years ago
5

Two 1 MHz radio antennas emitting in-phase are separated by600

Physics
1 answer:
algol [13]3 years ago
6 0

Answer:

1 km

Explanation:

d_0 = Gap between antennas = 600 m

\nu = Frequency = 1 MHz

z_1 = Distance to receiver = 2 km

c = Speed of light = 3\times 10^8\ m/s

Wavelength is given by

\lambda=\dfrac{c}{\nu}\\\Rightarrow \lambda=\dfrac{3\times 10^8}{1\times 10^6}\\\Rightarrow \lambda=300\ m

Distance to be moved is given by

D=\dfrac{z_1\lambda_0}{d_0}\\\Rightarrow D=\dfrac{2\times 300}{600}\\\Rightarrow D=1\ km

The distance to be moved is 1 km north.

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You are at a train station, standing next to the train at the front of the first car. The train starts moving with constant acce
Aleksandr [31]

The time needed for the 7th car to pass is 13.2 s

Explanation:

The motion of the train is a uniformly accelerated motion, therefore we can use suvat equations.

We start by analzying the motion of the first car, by using the equation:

s=ut+\frac{1}{2}at^2

where

s is the distance covered by the first car in a time t, which corresponds to the length of one car

u = 0 is the initial velocity

a is the acceleration

t = 5.0 s is the time

The equation can be rewritten as

a=\frac{2s}{t^2}=\frac{2L}{(5.0)^2}=0.08L[m/s^2]

where L is the length of one car.

The same equation can be written considering the first 7 cars:

7L = ut+\frac{1}{2}at'^2

where

7L is the distance covered by the 7 cars

t' is the time needed

We still have

u = 0

And the acceleration is constant so it is

a=0.08L

Substituting into the equation, we can find t':

7L = \frac{1}{2}(0.08L)t'^2\\7=0.04t'^2\\t'=\sqrt{\frac{7}{0.04}}=13.2 s

In attachment the graph of the distance covered versus the time taken: since the motion is uniformly accelerated, the relationship between the two variables is quadratical.

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

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3 years ago
Select the correct answer. Which of the following is true for balancing equations? A. There must be an equal number of atoms of
Sliva [168]
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3 years ago
A 2-kg wheel rolls down the road with a linear speed of 15m/s. Find its trwansitional and rotational kinetic energies.​
Elza [17]

Answer:

The translational kinetic energy is 225 J

The rotational kinetic energy is 225 J

Explanation:

Given;

mass of the wheel, m = 2-kg

linear speed of the wheel, v = 15 m/s

Transnational kinetic energy is calculated as;

E = ¹/₂MV²

where;

M is mass of the moving object

V is the velocity of the object

E =  ¹/₂ x 2 x (15)²

E = 225 J

Rotational kinetic energy is calculated as;

E = ¹/₂Iω²

where;

I is moment of inertia

ω is angular velocity

E = \frac{1}{2} I \omega^2\\\\E = \frac{1}{2} *mr^2*(\frac{v}{r})^2\\\\E =  \frac{1}{2} *mr^2*\frac{v^2}{r^2} \\\\E =  \frac{1}{2}mv^2

E =  ¹/₂ x 2 x (15)²

E = 225 J

Thus, the translational kinetic energy is equal to rotational kinetic energy

6 0
3 years ago
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