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2 years ago

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7. All electromagnetic waves travel at the same speed in empty space, the

2 answers:

Vilka [71]2 years ago

8 0

Answer:

TRUE

Explanation:

Ghella [55]2 years ago

3 0

Answer:

True

Explanation:

All electromagnetic waves travel at the same speed through empty space. That speed, called the speed of light, is 300 million meters per second (3.0 × 108 m/s).

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JOSEPH JOGS FROM END A TO OTHER END B OF A straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back

zzz [600]

(a) The average speed from A to B would be 1.76 metre per second and the average velocity from A to B would also be 1.76 metre per second

<span>(b) The average speed from A to C would be 1.73 metre per second and the average velocity from A to C would be 0.87 metre per second</span>

3 0

3 years ago

According to Newton's<br> Ist law, what will an<br> object in motion tend<br> to do?

SCORPION-xisa [38]

Answer:

According to <em>Newton's first law of motion:</em>

<u>An object in motion tends to remain in motion unless an external force acts upon it.</u>

<u>It stays in motion with the same speed and goes in the same direction.</u>

<u></u>

<em>Hope this helped </em>

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4 0

3 years ago

Read 2 more answers

An irregular object of mass 3 kg rotates about an axis, about which it has a radius of gyration of 0.2 m, with an angular accele

Citrus2011 [14]

Answer:

The magnitude of the applied torque is $6.0\times10^{-2}\ N-m$

(e) is correct option.

Explanation:

Given that,

Mass of object = 3 kg

Radius of gyration = 0.2 m

Angular acceleration = 0.5 rad/s²

We need to calculate the applied torque

Using formula of torque

$\tau=I\times\alpha$

Here, I = mk²

$\tau=mk^2\times\alpha$

Put the value into the formula

$\tau=3\times(0.2)^2\times0.5$

$\tau=0.06\ N-m$

$\tau=6.0\times10^{-2}\ N-m$

Hence, The magnitude of the applied torque is $6.0\times10^{-2}\ N-m$

7 0

3 years ago

A truck tire rotates at an initial angular speed of 21.5 rad/s. The driver steadily accelerates, and after 3.50 s the tire's ang

erma4kov [3.2K]

Given:

initial angular speed, $\omega _{i}$ = 21.5 rad/s

final angular speed, $\omega _{f}$ = 28.0 rad/s

time, t = 3.50 s

Solution:

Angular acceleration can be defined as the time rate of change of angular velocity and is given by:

$\alpha = \frac{\omega_{f} - \omega _{i}}{t}$

Now, putting the given values in the above formula:

$\alpha = \frac{28.0 - 21.5}{3.50}$

$\alpha = 1.86 m/s^{2}$

Therefore, angular acceleration is:

$\alpha = 1.86 m/s^{2}$

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2 years ago

A positive charge of 3.2 x 10 -5 C is located 0.85 m away from another positive charge of 7.4 x 10 -6 C. What is the electric fo

zmey [24]

by using Coulumbs Law its 2.95N.

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3 years ago