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Lilit [14]
3 years ago
7

If there is an increase in demand for a product, and that is the only thing that changes, what will that do to the price of the

product?
Physics
2 answers:
Eva8 [605]3 years ago
5 0
I believe that the price will rise.
Hope this helps!
lozanna [386]3 years ago
4 0

Answer:

Price will also increase.

Explanation:

If there is an increase in demand for a product, keeping other factors constant, price of product will also rise. According price demand curve demand of a product and price of product in direct proportional relationship with each other. So as demand of any product increases, its price will also increase simultaneously.

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Exactly the same way that you can photograph a mountain or a skyscraper
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Lenses are used to form a tiny image of a gigantic object.


7 0
2 years ago
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A mass is hung from a spring and set in motion so that it oscillates continually up and down. The velocity v of the weight at ti
otez555 [7]

To solve the problem it is necessary to identify the equation in the manner given above.

This equation corresponds to the displacement of a body under the principle of simple harmonic movement.

Where,

\xi = Acos(\omega t +\phi)

PART A) Our equation corresponds to

y = -5cos(4\pi t)

Therefore the value of omega is equivalent to that of

\omega = 4\pi

From the definition we know that the period as a function of angular velocity is equivalent to

T = \frac{2\pi}{\omega}

T = \frac{2\pi}{4\pi}

T = \frac{1}{2}

This same point is the equivalent of the maximum point of the speed that the body can reach, since the internal expression of the cos\thetaIs equivalent to . So the maximum speed that the body can reach is,

y = -5cos(4\pi t)

y = -5cos(4\pi (1/2))

y = -5*(-1)

y = 5

Therefore the maximum felocity will be 5ft / s

PART B) The period of graph is the time taken to reach from one maximum point to next point maximum point, then

t = \frac{T}{2} = \frac{1}{2}*\frac{1}{2}

t = \frac{1}{4}s

5 0
3 years ago
What is the mass moment of inertia of a 20kg sphere with a radius of 0.2m about a point on the sphere's perimeter
Kobotan [32]

Answer:

I = M R^2 is the moment of inertia about a point that is a distance R from the center of mass (uniform distributed mass).

The moment  of inertia about the center of a sphere is 2 / 5 M R^2.

By the parallel axis theorem the moment of inertia about a point on the rim of the sphere is  I = 2/5 M R^2 + M R^2 = 7/5 M R^2

I = 7/5 * 20 kg * .2^2 m = 1.12 kg m^2

7 0
2 years ago
A uniform string of length 0.50 m is fixed at both ends. Find the
kozerog [31]

Answer:

configuration of string:

Node - Antinode - Node    or N-A-N

This is 1/2 wavelength since a full wavelength is N-A-N-A-N

f (fundamental) = V / wavelength

F0 = 300 m/s / 1 m = 100 / sec

F1 = 300 m/s / .5 m = 600 / sec

Each increase is a multiple of the fundamental since the wavelength

increases by 1/2 wavelength to keep nodes at both ends of the string

4 0
2 years ago
A 0.8 kg object displaces 500 mL of water. What is its specific gravity?
Airida [17]
The specific gravity is how the density of the object compares to the density of water. Water's density is 1gram per milliliter. We just need to figure out the density of the object.

The object is .8 kg and it displaces 500mL of water, so the density is the mass divided by the volume. Since the density of water is given in grams, we have to convert the objects mass from kg to g and then we can get the density.

.8kg * 1000g/kg = 800 grams

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800g/500ml = 1.6grams/mL this is the density.

So divide the density of your object by the density of water, which is 1g/mL, you get 1.6 as the specific gravity. This means the object is 1.6 times more dense than water.
8 0
3 years ago
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