Yes because Electromagnetic waves differ from mechanical waves in that they do not require a medium to propagate. This means that electromagnetic waves can travel not only through air and solid materials, but also through the vacuum of space.

3 0

3 years ago

samantha a high school teacher wants to spend more time with her children and is looking for a job in her field. what would be a

sp2606 [1]

She could look for an online job or a job she can work at from home.

3 0

3 years ago

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A stuntman drives a car with a mass of 1500 kg on a drawbridge. The car accelerates with a constant force of 10,000 N. While he

Viktor [21]

Answer:

1.77 m/s^2

Explanation:

There are two forces acting on the car along the direction parallel to the incline:

- The driving force of 10,000 N, which pushes forward

- The component of the weigth of the car parallel to the incline, which pulls backward

The component of the weight of the car parallel to the incline is:

$W_p = mg sin \theta=(1500 kg)(9.8 m/s^2)( sin 30^{\circ})=7350 N$

So now we can apply Newton's second law to find the acceleration of the car:

$F-W_p = ma\\a=\frac{F-W_p}{m}=\frac{10000 N-7350 N}{1500 kg}=1.77 m/s^2$

7 0

3 years ago

What must be the distance between point charge q1 = 28.0 μC and point charge q2 = −57.0 μC for the electrostatic force between t

vlabodo [156]

Answer:

1.686 m

Explanation:

From coulomb's law,

F = kq1q2/r² ...................................... Equation 1

Where F = electrostatic force between the two charges, q1 = first charge, q2 = second charge, r = distance between the charges.

making r the subject of the equation,

r = √(kq1q2/F).......................... Equation 2

Given: F = 5.05 N, q1 = 28.0 μC = 28×10⁻⁶ C, q2 = 57.0 μC = 57.0×10⁻⁶ C

Constant: k = 9.0×10⁹ Nm²/C².

Substituting into equation 2

r = √(9.0×10⁹×28×10⁻⁶×57.0×10⁻⁶/5.05)

r = √(14364×10⁻³/5.05)

r = √(14.364/5.05)

r = √2.844

r = 1.686 m

r = 1.686 m.

Thus the distance must be 1.686 m

6 0

3 years ago