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nignag [31]
3 years ago
6

Which of the following is true about all electromagnetic waves in a vacuum

Physics
2 answers:
Yakvenalex [24]3 years ago
6 0

They all travel at the same speed!

Helen [10]3 years ago
4 0
Electro waves in a vacuum air is deals with this and electricity when the air and the electricity it  makes electro magnets.
You might be interested in
What net force is required to accelerate a car at a rate of 5 m/s ^ 2 if the car has a mass of 3, 000 kg ?
Olenka [21]

Answer:

The answer is 15000kgms^-2

Explanation:

Actually here we are using the formula F=ma

3 0
3 years ago
A 500kg car is driving through a neighborhood at 20m/s. Then the car hits a speed bump which causes it to slow to 15m/s. What wa
lawyer [7]

Based on the formula for calculating impulse, the impulse of the speed bump to the car is 2500 Ns.

<h3>What is the impulse of the speed bump?</h3>

  • Impulse = change in momentum
  • momentum = mass * velocity

Change in momentum = mu - mv

where u is initial velocity

v is final velocity

Impulse = 500 * 20  - 500 * 15

Impulse = 2500 Ns

Therefore, the impulse of the speed bump to the car is 2500 Ns.

Learn more about impulse at: brainly.com/question/297527

5 0
3 years ago
A 200 g hockey puck is launched up a metal ramp that is inclined at a 30° angle. The coefficients of static and kinetic friction
nikitadnepr [17]

Answer:

71.76 m

Explanation:

We will solve this question using the work energy theorem.

The theorem explains that, the change in kinetic energy of a particle between two points is equal to the workdone in moving the particle from the one point to the other.

ΔK.E = W

In the attached free body diagram for the question, the forces acting on the puck are given.

ΔK.E = (final kinetic energy) - (initial kinetic energy)

Final kinetic energy = 0 J (since the puck comes to a stop)

Initial kinetic energy = (1/2)(m)(v²) = (1/2)(0.2)(26²) = 67.6 J

ΔK.E = 0 - 67.6 = - 67.6 J

W = Workdone between the starting and stopping points = (work done by the force of gravity) + (work done by frictional force)

Work done by the force of gravity = - mgh = - (0.2)(9.8)(h) = - 1.96 h

Workdone by the frictional force = F × d

F = μ N

μ = coefficient of kinetic friction = 0.30 (kinetic frictional force is the only frictional force that moves a distance of d, the static frictional force doesn't move any distance, so it does no work)

N = normal reaction of the plane surface on the puck = mg cos 30° = (0.2)(9.8)(0.866) = 1.697 N

F = μ N = 0.3 × 1.697 = 0.509 N

where d = distance along the incline that the puck travels.

d = h/sin 30° = 2h (from trigonometric relations)

Workdone by the frictional force = F × d = 0.509 × 2h = 1.02 h

ΔK.E = W = (work done by the force of gravity) + (work done by frictional force)

- 67.6 = - 1.96h + 1.02h

-0.942h = - 67.6

h = 71.76 m

6 0
3 years ago
a cepheid variable star is a star whose brightness alternately increases and decreases. suppose that cephei joe is a star for wh
MrRissso [65]

After one day, the rate of increase in Delta Cephei's brightness is;0.46

We are informed that the function has been used to model the brightness of the star known as Delta Cephei at time t, where t is expressed in days;

B(t)=4.0+3.5 sin(2πt/5.4)

Simply said, in order to determine the rate of increase, we must determine the derivative of the function that provides

B'(t)=(2π/5.4)×0.35 cos(2πt/5.4)

Currently, at t = 1, we have;

B'(1)=(2π/5.4)×0.35 cos(2π*1/5.4)

Now that the angle in the bracket is expressed in radians, we can use a radians calculator to determine its cosine, giving us the following results:

B'(1)=(2π/5.4)×0.3961

B'(1)≈0.46

To know more about:

brainly.com/question/17110089

#SPJ4

7 0
1 year ago
I need helppppppp<br> real answers only please
elixir [45]
I d k you figure it out
5 0
2 years ago
Read 2 more answers
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