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Drupady [299]
3 years ago
8

A/An __________ is a small, raised red lesion that is less than 0.5 cm in diameter and does not contain pus.

Physics
1 answer:
CaHeK987 [17]3 years ago
6 0

Answer:

See Explanation

Explanation:

Papule is a small, raised red lesion that is less than 0.5 cm in diameter and does not contain pus. Small pimples are examples

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A mass moves back and forth in simple harmonic motion with amplitude A and period T.
Sever21 [200]

a. 0.5 T

- The amplitude A of a simple harmonic motion is the maximum displacement of the system with respect to the equilibrium position

- The period T is the time the system takes to complete one oscillation

During a full time period T, the mass on the spring oscillates back and forth, returning to its original position. This means that the total distance covered by the mass during a period T is 4 times the amplitude (4A), because the amplitude is just half the distance between the maximum and the minimum position, and during a time period the mass goes from the maximum to the minimum, and then back to the maximum.

So, the time t that the mass takes to move through a distance of 2 A can be found by using the proportion

1 T : 4 A = t : 2 A

and solving for t we find

t=\frac{(1T)(2 A)}{4A}=0.5 T

b. 1.25T

Now we want to know the time t that the mass takes to move through a total distance of 5 A. SInce we know that

- the mass takes a time of 1 T to cover a distance of 4A

we can set the following proportion:

1 T : 4 A = t : 5 A

And by solving for t, we find

t=\frac{(1T)(5 A)}{4A}=\frac{5}{4} T=1.25 T

6 0
3 years ago
Two identical bowling balls are rolling on a horizontal floor without slipping. The initial speed of both balls is v = 10 m/s. B
Lapatulllka [165]

Answer:

The difference between frictionless ramp and a regular ramp is that on a frictionless ramp the ball cannot roll it can only slide, but on a regular ramp the ball can roll without slipping.

We will use conversation of energy.

K_A_1 + U_A_1 = K_A_2 + U_A_2\\\frac{1}{2}I\omega^2 + \frac{1}{2}mv^2 + 0 = 0 + mgH_A

Note that initial potential energy is zero because the ball is on the bottom, and the final kinetic energy is zero because the ball reaches its maximum vertical distance and stops.

For the ball B;

K_B_1 + U_B_1 = K_B_2 + U_B_2

\frac{1}{2}I_B\omega^2 + \frac{1}{2}mv^2 + 0 = 0 + mgH_B

The initial velocities of the balls are equal. Their maximum climbing point will be proportional to their final potential energy. Since their initial kinetic energies are equal, their final potential energies must be equal as well.

Hence, both balls climb the same point.

Explanation:

4 0
3 years ago
What are the constituent metals of stainless steel?
bogdanovich [222]

<span>Stainless steel is a metal alloy that made up mainly of carbon and chromium. In combination with low carbon contents, chromium is highly reactive element that imparts remarkable resistance to corrosion and heat.</span>

Moreover, stainless steel is mixed up with sufficient nickel, which is an essential allying element in the series of stainless steel grades. Other components are manganese, molybdenum, silicon, titanium, aluminum, niobium, copper, nitrogen, and sulfur.

 

6 0
3 years ago
Which change will always result in an increase in the gravitational force between 2 objects
Nat2105 [25]
<span>Reducing the distance between them. In theory, also increasing the mass; but you can't really change the mass of an object. However, you can compare the forces if you replace an object by a different object, which has a different mass.
</span>
i hope this will work..
4 0
3 years ago
Read 2 more answers
When light of wavelength 240 nm falls on a potassium surface, electrons having a maximum kinetic energy of 2.93 eV are emitted.
olchik [2.2K]

A) We want to find the work function of the potassium. Apply this equation:

E = 1243/λ - Φ

E = energy of photoelectron, λ = incoming light wavelength, Φ = potassium work function

Given values:

E = 2.93eV, λ = 240nm

Plug in and solve for Φ:

2.93 = 1243/240 - Φ

Φ = 2.25eV

B) We want to find the threshold wavelength, i.e. find the wavelength such that the energy E of the photoelectrons is 0eV. Plug in E = 0eV and Φ = 2.25eV and solve for the threshold wavelength λ:

E = 1243/λ - Φ

0 = 1243/λ - Φ

0 = 1243/λ - 2.25

λ = 552nm

C) We want to find the frequency associated with the threshold wavelength. Apply this equation:

c = fλ

c = speed of light in a vacuum, f = frequency, λ = wavelength

Given values:

c = 3×10⁸m/s, λ = 5.52×10⁻⁷m

Plug in and solve for f:

3×10⁸ = f(5.52×10⁻⁷)

f = 5.43×10¹⁴Hz

7 0
3 years ago
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