PHYSICS COLLEGE

What type of electromagnetic radiation is being shown in the picture?A. Gamma rays
B. Ultraviolet radiation
C. X-rays
D. Infrared radiation

Answers

Answer 1

Answer:

Answer:

I think D. Infrared radiation.

Answer 2

Answer:

Answer:

infrared radition

Explanation:

valid

Related Questions

Write a paragraph or two (no more than 1 page) describing the misconception and then explain the correct physics by identifying the forces or physics concepts involved in simple terms, like you are talking to your 3rd grade sibling. Start with a question (will water fall out of the bucket when I stop rotating it?)Give the common answer that is incorrect (what does your intuition tell you should happen?).Explain what happens and why in simple terms (try practicing on an elementary student to make sure they understand).Use diagrams and sketches in your explanation and avoid using physics terms like rotational, centripetal, normal forces, tension, friction, acceleration, velocity, vectors, buoyancy, energy, kinetics, potential, heat capacity, etc.
Which of the following is not a motivation behind taking performance enhancing drugs?A.Having bigger, stronger muscles B.improving recovery time fir injuries C. Pumping
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An element emits light at two nearly equal wavelengths, 577 nm and 579 nm If the light is normally incident on a diffraction grating with 2000 lines/cm., what is the distance between the 3rd order fringes of the two wavelengths on a screen 1 m from the grating?

Answers

Answer:

Explanation:

d = width of slit = 1 / 2000 cm =5 x 10⁻⁶ m

Distance of screen D = 1 m.

wave length λ₁ and λ₂ are 577 x 10⁻⁹ and 579 x 10⁻⁹ m.respectively.

distance of third order bright fringe = 3.5 λ D/d

for 577 nm , this distance = 3.5 x 577 x 10⁻⁹ x 1 /5 x 10⁻⁶

= .403 m = 40.3 cm

For 579 nm , this distance = 3.5 x 579 x 1 / 5 x 10⁻⁶

= 40.5 cm

Distance between these two = 0.2 cm.

A skateboarder with mass ms = 54 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then jumps on his skateboard and descends down the ramp. His speed at the bottom of the ramp is vf = 6.2 m/s.

Answers

The work ( $W_f$ ) done by the friction force between the ramp and the skateboarder is given by $-\mu_k \cdot m_s \cdot g \cdot h_y$ .

The workdone by the friction force ( $W_f$ ) can be calculated using the formula for work, which is the product of the force applied ( $F_f$ ) and the displacement (d) over which the force is applied:

$W_f = F_f \cdot d$

In this scenario, the frictionforce works against the skateboarder's momentum down the ramp, therefore it does no good.

Given:

Mass of skateboarder ( $m_s$ ) = 54 kg

Height of the ramp ( $h_y$ ) = 3.3 m

Final velocity ( $v_f$ ) = 6.2 m/s

Coefficient of kineticfriction ( $\mu_k$ ) between skateboarder and ramp

Acceleration due to gravity (g) = $9.81 \, \text{m/s}^2$

The normal force ( $F_{\text{normal}}$ ) is equal to the weight of the skateboarder:

$F_{\text{normal}} = m_s \cdot g$

The displacement (d) is the vertical distance ( $h_y$ ) that the skateboarder descends down the ramp.

Now we can write the expression for the work done by the friction force ( $W_f$ ):

$W_f = -\mu_k \cdot F_{\text{normal}} \cdot d$

Substitute the expression for the normal force:

$W_f = -\mu_k \cdot (m_s \cdot g) \cdot h_y$

Thus, this expression represents the work done by the friction force between the ramp and the skateboarder in terms of the given variables.

For more details regarding friction force visit:

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Your question seems incomplete, the probable complete question is:

Write an expression for the work, Wf, done by the friction force between the ramp and the skateboarder in terms of the variables given in the problem statement.

Final answer:

The momentum of the box with respect to the floor can be found by multiplying its mass by its velocity. When the box is put down on the frictionless skating surface, its velocity becomes zero and its momentum with respect to the floor is also zero.

Explanation:

To find the momentum of the box, we can use the formula:

Momentum = mass x velocity

a. The momentum of the box with respect to the floor is: 5 kg x 5 m/s = 25 kg·m/s

b. When the box is put down on the frictionless skating surface, its velocity becomes zero. So, the momentum of the box with respect to the floor is also zero.

Learn more about Momentum here:

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Impulse is the______of the force and time of contact

Answers

Answer:

Product

Explanation:

Impulse is defined as the average force acting on an object times the time the force acts:

Impulse = F · Δt

Ezra (m = 20.0 kg) has a tire swing and wants to swing as high as possible. He thinks that his best option is to run as fast as he can and jump onto the tire at full speed. The tire has a mass of 10.0 kg and hangs 3.50 m straight down from a tree branch. Ezra stands back 10.0 m and accelerates to a speed of 3.62 m/s before jumping onto the tire swing. (a) How fast are Ezra and the tire moving immediately after he jumps onto the swing? m/s (b) How high does the tire travel above its initial height?

Answers

Answer:

a) $v=5.6725\,m.s^(-1)$

b) $h= 1.6420\,m$

Explanation:

Given:

mass of the body, $M=20\,kg$

mass of the tyre, $m=10\,kg$

length of hanging of tyre, $l=3.5m$

distance run by the body, $d=10m$

acceleration of the body, $a=3.62m.s^(-2)$

(a)

Using the equation of motion :

$v^2=u^2+2a.d$ ..............................(1)

where:

v=final velocity of the body

u=initial velocity of the body

here, since the body starts from rest state:

$u=0m.s^(-1)$

putting the values in eq. (1)

$v^2=0^2+2* 3.62 * 10$

$v=8.5088\,m.s^(-1)$

Now, the momentum of the body just before the jump onto the tyre will be:

$p=M.v$

$p=20* 8.5088$

$p=170.1764\,kg.m.s^(-1)$

Now using the conservation on momentum, the momentum just before climbing on the tyre will be equal to the momentum just after climbing on it.

$(M+m)* v'=p$

$(20+10)* v'=170.1764$

$v'=5.6725\,m.s^(-1)$

(b)

Now, from the case of a swinging pendulum we know that the kinetic energy which is maximum at the vertical position of the pendulum gets completely converted into the potential energy at the maximum height.

So,

$(1)/(2) (M+m).v'^2=(M+m).g.h$

$(1)/(2) (20+10)* 5.6725^2=(20+10)* 9.8* h$

$h\approx 1.6420\,m$

above the normal hanging position.

A force of 200N acts on a body that moves along a horizontal plane in the same direction of movement. The body moves 30m. What is the work done by that force?

Answers

Work = Force times Distance

Work = 200 x 30

Work = 6000

The work done by a force of 200N on a body that moved 30m is 6000J or 6000 Joules.

Sultan throws a ball horizontally from his window, 12 m above the garden. It reaches the ground afterSelect........seconds.

4.0

5.0

2.4

1.6

Answer and I will give you brainiliest

Answers

Answer:

2.4

Explanation:

Hope it help mark as Brainlist

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What type of electromagnetic radiation is being shown in the picture?A. Gamma rays B. Ultraviolet radiation C. X-rays D. Infrared radiation

Answers

Related Questions

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Answers

Final answer:

Explanation:

Learn more about Momentum here:

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Answers

2.4

Hope it help mark as Brainlist

What type of electromagnetic radiation is being shown in the picture?A. Gamma rays
B. Ultraviolet radiation
C. X-rays
D. Infrared radiation