Topics related to physics

According to the diagram of the electromagnetic spectrum shown, what would best represent the

iren2701 [21]

The best represent the size of visible light will be C. Protozoa

The electromagnetic spectrum, gives the overall distribution of electromagnetic radiation by the frequency or wavelength. All EM waves travel at the speed of light in a vacuum, but over a wide range of frequencies, wavelengths, and photon energies.

Visible light wavelengths cover the range of approximately 0.4 to 0.7 μm. electromagnetic spectrum that the human eye can see is the Visible light. Visible light is a form of electromagnetic (EM) radiation, along with radio waves, infrared, ultraviolet, X-rays, and microwaves. the wavelengths that are visible to most human eyes is generally known as Visible light

the best represent the size of visible light is Protozoa, According to the diagram of the electromagnetic spectrum shown,

Learn more about electromagnetic spectrum here brainly.com/question/25847009

#SPJ9

6 0

10 months ago

If a wave is traveling at 60cm/second and has a wavelength of 15 cm what is the frequency?

REY [17]

The frequency of the wave is 4 Hz

4 0

3 years ago

7. Two people are pushing a 40.0kg table across the floor. Person 1 pushes with a force of 490N

artcher [175]

Answer:

20.4 $m/s^{2}$

Explanation:

To start doing this problem, first draw a free body diagram of the table. My teacher always tells us to do this, and I find that it is very helpful. I have attached a free body diagram to this answer- take a look at it.

First, let us see if Net force = MA. To do that, we need to determine whether the object is at equilibrium horizontally. For an object to be at equilibrium, it either needs to be moving at a constant velocity or not moving at all. Also, if an object is at equilibrium, there will not be any acceleration. But we know that there IS acceleration horizontally, so it cannot be in equilibrium. If it is not in equilibrium, we can use the formula ∑F= ma.

Let us determine the net force. Since the object is moving horizontally, we can ignore the weight and normal force, because they are vertical forces. The only horizontal forces we need to worry about are the applied force and force of friction.

Applied force = 1055 N (490 + 565)

Friction force= Unknown

To find the friction force, use the kinetic friction formula, Friction = μkN

μk is the coefficient, which the problem includes- it is 0.613.

N is the normal force, which we have to find.

*To find the normal force, we have to determine if the object is at equilibrium VERTICALLY. Since it has no acceleration vertically (it's not moving up/down), it is at equilibrium. Now, when an object is at equilibrium in one direction, it means that all the forces in that direction are equal. What are our vertical forces? Weight (mg) and Normal force (N). So it means that the Normal force is equal to the Weight.

Weight = mg = (40)(9.8) = 392 N

Normal force = 392 N

Now, plug it back into the formula (μkN): (0.613)(392) = 240.296 N

Friction = 240.296 N

Now that we know the friction, we can find the horizontal net force. Just subtract the friction force, 240.296 from the applied force, 1055 N

Horizontal Net Force: 814.704 N

Now that we know the net force, plug in the numbers for the formula

∑F= ma.

814.704 = (40.0)(a)

*Divide on both sides)

a = 20.3676 m/s^2

Round it to 3 significant figures, to get:

20.4 $m/s^{2}$

7 0

3 years ago

What affect does a tripling of the net force have upon the acceleration of the object ? Be quantitative

vovikov84 [41]

The formula of net Force is:
F = ma
where m is the mass of the object
a is the acceleration of the object

so if we triple the net force applied to the object:
3F = ma
a = 3F / m

so the acceleration will also be tripled. because from the equation, the force is directly proportional to the acceleration

5 0

3 years ago

You are driving your car along a country road at a speed of 27.0 m/s. as you come over the crest of a hill, you notice a farm tr

Bezzdna [24]

speed of the car = 27 m/s

speed of truck ahead = 10 m/s

relative speed of car with respect to truck

$v_r = 27 - 10 = 17 m/s$