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2 years ago

E electric- and magnetic-field what is the direction of propagation of the wave?

Physics

1 answer:

castortr0y [4]2 years ago

8 0

As well as the electric and magnetic fields being perpendicular to eachother, a wave moves perpendicularly to both

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Determine the magnitude of the momentum of a ... 107-kg halfback moving eastward at 8 m/s.The halfback's momentum in kgm/s is:

Veronika [31]

Given:

The mass of the halfback is m = 107 kg

The speed of the halfback is v = 8 m/s

To find the momentum.

Explanation:

The momentum of the halfback is

$\begin{gathered} p=\text{ mv} \\ =\text{ 107}\times8 \\ =\text{ 856 kg m/s} \end{gathered}$

Thus, the momentum of the halfback is 856 kg m/s

3 0

10 months ago

A ball with a mass of 3.1 kg is moving in a uniform circular motion upon a horizontal surface. The ball is attached at the cente

inessss [21]

Answer:

3.46 seconds

Explanation:

Since the ball is moving in circular motion thus centripetal force will be acting there along the rope.

The equation for the centripetal force is as follows -

$F=\frac{mv^2}{r}$

Where, $m$ is the mass of the ball, $v$ is the speed and $r$ is the radius of the circular path which will be equal to the length of the rope.

This centripetal force will be equal to the tension in the string and thus we can write,

$20.4 = \frac{3.1\times v^2}{2}$

and, $v^2 = 13.16$

Thus, $v = 3.63$ m/s.

Now, the total length of circular path = circumference of the circle

Thus, total path length = 2πr = 2 × 3.14 × 2 = 12.56 m

Time taken to complete one revolution = $\frac{\text {Path length} }{\text {Speed}}$ = $\frac{12.56}{3.63}$ = 3.46 seconds.

Thus, the mass will complete one revolution in 3.46 seconds.

4 0

2 years ago

What is the advantage of a free market economy??

True [87]

A free market economy<span> has two key </span>advantages<span>. First, it allows for individuals to innovate. Individuals have the freedom to create new ideas, new products, and new services to sell for profit. They are not required to only produce what the government tells them to produce.</span>

5 0

2 years ago

According to Kepler's Third Law, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about

NARA [144]

Answer:

Orbital period, T = 1.00074 years

Explanation:

It is given that,

Orbital radius of a solar system planet, $r=4\ AU=1.496\times 10^{11}\ m$

The orbital period of the planet can be calculated using third law of Kepler's. It is as follows :

$T^2=\dfrac{4\pi^2}{GM}r^3$

M is the mass of the sun

$T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 1.989\times 10^{30}}\times (1.496\times 10^{11})^3$

$T^2=\sqrt{9.96\times 10^{14}}\ s$

T = 31559467.6761 s

T = 1.00074 years

So, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about 1.00074 years.

6 0

2 years ago

the Earth exerts a gravitational force of 500 newtons on an object what is the mass of the object in kilograms

MAXImum [283]

Use Newton's second law F = mass * acceleration
In your problem F = 500, and we know gravity is working on it so use a = 9.81
Substitute into the equation
500 = m * 9.81
m = 50.97 kg

7 0

2 years ago