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Soloha48 [4]
3 years ago
7

PLEASE HELP........Compare and contrast microwaves with visible light using wavelength frequency and energy.

Physics
2 answers:
Margarita [4]3 years ago
6 0
The microwaves have higher wavelength than visible light, but lower frequency than visible light and it also has lower energy than the visible light and that is the answer so I hope this will help you for whatever it is.
stealth61 [152]3 years ago
5 0
Microwaves heat up food. Ha just kidding!
Microwaves are lower on the spectrum. They emit less energy and have longer wavelengths. Light Waves emit more energy and have shorter wavelengths than Microwaves. Light Waves can be seen by the naked eye. Both of these waves are harmless and emit low energy, compared to waves higher in the spectrum that have smaller wavelengths, therefore more energy. Some examples are Gamma rays and X-rays, emitting high levels of energy and even radiation.

Hope that makes sense!  :V

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A wheel that is rotating at 33.3 rad/s is given an angular acceleration of 2.15 rad/s 2. Through what angle has the wheel turned
Neporo4naja [7]

Answer:

The angle through which the wheel turned is 947.7 rad.

Explanation:

initial angular velocity, \omega _i = 33.3 rad/s

angular acceleration, α = 2.15 rad/s²

final angular velocity, \omega_f = 72 rad/s

angle the wheel turned, θ = ?

The angle through which the wheel turned can be calculated by applying the following kinematic equation;

\omega_f^2 = \omega_i^2 + 2\alpha \theta\\\\\theta = \frac{\omega_f^2\   -\  \omega_i^2}{2\alpha } \\\\\theta = \frac{(72)^2\   -\  (33.3)^2}{2(2.15)}\\\\\theta = 947.7 \ rad

Therefore, the angle through which the wheel turned is 947.7 rad.

5 0
2 years ago
Lonnie pitches a baseball of mass 0.20 kg. The ball arrives at home plate with a speed of 40 m/s and is batted straight back to
suter [353]

Answer:

B) 20N.s is the correct answer

Explanation:

The formula for the impulse is given as:

Impulse = change in momentum

Impulse = mass × change in speed

Impulse = m × ΔV

Given:

initial speed  = 40m/s

Final speed = -60 m/s (Since the the ball will now move in the opposite direction after hitting the bat, the speed is negative)

mass = 0.20 kg

Thus, we have

Impulse = 0.20 × (40m/s - (-60)m/s)

Impulse = 0.20 × 100 = 20 kg-m/s or 20 N.s

4 0
3 years ago
In the writing of ionic chemical formulas, what factor is "crossed over" in the crossover rule?
Aleks04 [339]

In the writing of ionic chemical formulas the value of each ion's charge is crossed over in the crossover rule.

Rules for naming Ionic compounds

  • Frist Rule
    The cation (element with a negative charge) is written first in the name then the anion(element with a positive charge) is written second in the name.
  • Second rule
    When the formula unit contains two or more of the same polyatomic ion, that ion is written in parentheses with the subscript written outside the parentheses.
    Example: Sodium carbonate is written as Na₂CO₃ not Na₂(CO)₃
  • Third rule
    If the cation is a metal ion with a fixed charge then the name of the cation will remain the same as the (neutral) element from which it is derived (Example: Na+ will be sodium).
    If the cation is a metal ion with a variable charge, the charge on the cation is indicated using a Roman numeral, in parentheses, immediately following the name of the cation (example: Fe³⁺ = iron(III)).
  • Fourth rule
    If the anion is a monatomic ion, the anion is named by adding the suffix <em>-ide</em> to the root of the element name (example: F = Fluoride).

The oxidation state of each ion is also important, thus in the crossover rule, the value of each ion's charge is crossed over.

Learn more about chemical formulas here:

<u>brainly.com/question/11995171</u>

#SPJ4

3 0
2 years ago
How much power is used by a car engine if it does 48496 J of work in 4 min?
elena-14-01-66 [18.8K]

<em>Hope this will help u</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>✌</em><em>✌</em><em>✌</em>

8 0
3 years ago
Read 2 more answers
The drawing shows a large cube (mass = 21.0 kg) being accelerated across a horizontal frictionless surface by a horizontal force
MaRussiya [10]

Answer:

The blocks must be pushed with a force higher than 359 Newtons horizontally in order to accomplish this friction levitation feat.

Explanation:

The first step in resolving any physics problem is to draw the given scenario (if possible), see the attached image to have an idea of the objects and forces involved.

The large cube in red is being pushed from the left by a force \vec{P} whose value is to be found. That cube has its own weight \vec{w}_1=m_1\vec{g}, and it is associated with the force of gravity which points downward. Newton's third law stipulates that the response from the floor is an upward pointing force on the cube, and it's called the normal force \vec{N}_1.

A second cube is being pushed by the first, and since the force \vec{P} is strong enough it is able to keep such block suspended as if it were glued to the first one, due to friction. As in the larger cube, the smaller one has a weight \vec{w}_2=m_2\vec{g} pointing downwards, but the normal force in this block doesn't point upwards since its 'floor' isn't below it, but in its side, therefore the normal force directs it to the right as it is shown in the picture. Normal forces are perpendicular to the surface they contact. The final force is the friction between both cubes, that sets a resistance of one moving parallel the other. In this case, the weight of the block its the force pointing parallel to the contact surface, so the friction opposes that force, and thus points upwards. Friction forces can be set as Fr=\mu~N, where \mu is the coefficient of static friction between the cubes.

Now that all forces involved are identified, the following step is to apply Newton's second law and add all the forces for each block that point in the same line, and set it as equal its mass multiplied by its acceleration. The condition over the smaller box is the relevant one so its the first one to be analyzed.

In the vertical component: \Sigma F^2_y=Fr-w_2=m_2 a_y Since the idea is that it doesn't slips downwards, the vertical acceleration should be set to zero a_y=0, and making explicit the other forces: \mu N_2-m_2g=0\quad\Rightarrow (0.710)N_2-(4.5)(10)=0\quad\Rightarrow N_2=(4.5)(10)/(0.710)\approx 63.38 [N]. In the last equation gravity's acceleration was rounded to 10 [m/s^2].

In its horizontal component: \Sigma F^2_x=N_2=m_2 a_x, this time the horizontal acceleration is not zero, because it is constantly being pushed. However, the value of the normal force and the mass of the block are known, so its horizontal acceleration can be determined: 63.38=(4.5) a_x \quad \Rightarrow a_x=(63.38)/(4.5)\approx 14.08 [m/s^2]. Notice that this acceleration is higher than the one of gravity, and it is understandable since you should be able to push it harder than gravity in order for it to not slip.

Now the attention is switched to the larger cube. The vertical forces are not relevant here, since the normal force balances its weight so that there isn't vertical acceleration. The unknown force comes up in the horizontal forces analysis: \Sigma F_x=P=m a_x, since the force \vec{P} is not only pushing the first block but both, the mass involved in this equation is the combined masses of the blocks, the acceleration is the same for both blocks since they move together; P=(21.0+4.5) 14.08\approx 359.04 [N]. The resulting force is quite high but not impossible to make by a human being, this indicates that this feat of friction suspension is difficult but feasable.

4 0
3 years ago
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