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dedylja [7]
2 years ago
11

An excited hydrogen atom releases an electromagnetic wave to return to its normal state. You use your futuristic dual electric/m

agnetic field tester on the electromagnetic wave to find the directions of the electric field and magnetic field. Your device tells you that the electric field is pointing in the positive y direction and the magnetic field is pointing in the positive x direction. In which direction does the released electromagnetic wave travel?
Physics
1 answer:
aniked [119]2 years ago
7 0

Answer: The released electromagnetic wave will travel in +y direction

Explanation:

It should be noted that, in a situation, whereby an excited hydrogen atom releases an electromagnetic wave to return to its normal state. And it's also evident that the futuristic dual electric/magnetic field tester on the electromagnetic wave to find the directions of the electric field and magnetic field is used. Eventually, your device tells you that the electric field is pointing in the positive y direction and the magnetic field is pointing in the positive x direction. Therefore, the released electromagnetic wave will travel in +y direction.

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a 59kg physics student jumps off the back of her laser sailboat (42kg). after she jumps the laser is found to be travelling at 1
evablogger [386]

From the conservation of linear momentum of closed system,

Initial momentum = final momentum

Mass of the student, M = 59 kg

Mass of the laser boat, m = 42 kg

Initial speed of student + laser boat, u =0

Final speed of laser boat, v = 1.5 m/s

Final speed of the student = V

(M+m) u =M V +m v

0 = (59 kg) V + (42 kg) (1.5m/s)

V = - 1.06 m/s

Thus, the speed of the student is 1.06 m/s in the opposite direction of the motion of boat.

5 0
2 years ago
A box with a mass of 100.0 kg slides down a ramp with a 50 degree angle. What is the weight of the box? N What is the value of t
nadya68 [22]

1) weight of the box: 980 N

The weight of the box is given by:

W=mg

where m=100.0 kg is the mass of the box, and g=9.8 m/s^2 is the acceleration due to gravity. Substituting in the formula, we find

W=(100.0 kg)(9.8 m/s^2)=980 N


2) Normal force: 630 N

The magnitude of the normal force is equal to the component of the weight which is perpendicular to the ramp, which is given by

N=W cos \theta

where W is the weight of the box, calculated in the previous step, and \theta=50^{\circ} is the angle of the ramp. Substituting, we find

N=(980 N)(cos 50^{\circ})=630 N


3) Acceleration: 7.5 m/s^2

The acceleration of the box along the ramp is equal to the component of the acceleration of gravity parallel to the ramp, which is given by

a_p = g sin \theta

Substituting, we find

W_p = (9.8 m/s^2)(sin 50^{\circ})=7.5 m/s^2

5 0
2 years ago
A 202 kg bumper car moving right at 8.50 m/s collides with a 355 kg car at rest. Afterwards, the 355 kg car moves right at 5.80
Sidana [21]

Explanation:

It is given that,

Mass of bumper car, m₁ = 202 kg

Initial speed of the bumper car, u₁ = 8.5 m/s

Mass of the other car, m₂ = 355 kg

Initial velocity of the other car is 0 as it at rest, u₂ = 0

Final velocity of the other car after collision, v₂ = 5.8 m/s

Let p₁ is momentum of of 202 kg car, p₁ = m₁v₁

Using the conservation of linear momentum as :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

202\ kg\times 8.5\ m/s+355\ kg\times 0=m_1v_1+355\ kg\times 5.8\ m/s

p₁ = m₁v₁ = -342 kg-m/s

So, the momentum of the 202 kg car afterwards is 342 kg-m/s. Hence, this is the required solution.

7 0
2 years ago
An object weighs 200 newtons at a distance of 100 kilometers above the center of a small uniform planet. how much will the objec
disa [49]

Since the law of gravitation is an inverse square law if you quadruple the radius the f will drop by a factor of 16 SO the object would weigh 200/16 = 12.5N

In other words, as the distance, or radius, quadruples the weight becomes 1/16 of the original weight. Just plug in 4 for r and when you square it you get 16. The numerator is 1 so that is how the weight becomes 1/16.

7 0
2 years ago
In the equation vx^2=v0x^2+2ax(x-x0) what does the terms vx, v0x, x, and x0 stand for respectively?
tatuchka [14]

B. velocity at position x, velocity at position x=0, position x, and the original position

In the equation

v_{x}^{2} = v_{ox}^{2} +2 a x (x - x₀)

v_{x} = velocity at position "x"

v_{ox} = velocity at position "x = 0 "

x = final position

x_{o} = initial position of the object at the start of the motion

6 0
2 years ago
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