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DiKsa [7]
3 years ago
6

The frequency of a wave is 560 Hz. What is it’s period

Physics
1 answer:
Crazy boy [7]3 years ago
4 0
The answer would be, "1/560 seconds".
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A stone is dropped from the upper observation deck of a tower, "500" m above the ground. (Assume g = 9.8 m/s2.) (a) Find the dis
Yuri [45]

Answer:

h₍₁₎ = 495,1 meters

h₍₂₎ = 480,4 m

h₍₃₎ = 455,9 m

...

..

Explanation:

The exercise is "free fall". t = \sqrt{\frac{2h}{g} }

Solving with this formula you find the time it takes for the stone to reach the ground (T) = 102,04 s

The heights (h) according to his time (t) are found according to the formula:

h(t) = 500 - 1/2 * g * t²

Remplacing "t" with the desired time.

4 0
3 years ago
Help me with this question please
vichka [17]

Answer:

Its true i'm pretty sure

Explanation:

7 0
3 years ago
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The term "ideal gas" refers to a gas for which certain assumptions have been made. Which of the following is not such an assumpt
Jobisdone [24]

Answer:

(A) Consists of a small number of tiny particles that are far apart- relative in their size.

Explanation:

An <em>ideal gas</em> is defined as a simplification of a real gas, with punctual particles, in which all collisions are elastic, with random displacements and with no attractive force between them.

The assumption of the particles being punctual make clear that they do not have size at all. So if they were far apart-relative in their size, they can not collide each other, that is why assumption (B) can not be possible (<u><em>for that particular case</em></u>).

It is clear that (A) is not an assumption for an ideal gas, because do not fit in any of its properties.

Elastic collision: It is a case in which the energy is conserved (Kinetic Energy).

Kinetic Energy: It is the energy that will have an object as a consequence of its movement.

6 0
3 years ago
The current in a wire varies with time according to the relationship 1=55A−(0.65A/s2)t2. (a) How many coulombs of charge pass a
mario62 [17]

Answer:

(A) Q = 321.1C (B) I = 42.8A

Explanation:

(a)Given I = 55A−(0.65A/s2)t²

I = dQ/dt

dQ = I×dt

To get an expression for Q we integrate with respect to t.

So Q = ∫I×dt =∫[55−(0.65)t²]dt

Q = [55t – 0.65/3×t³]

Q between t=0 and t= 7.5s

Q = [55×(7.5 – 0) – 0.65/3(7.5³– 0³)]

Q = 321.1C

(b) For a constant current I in the same time interval

I = Q/t = 321.1/7.5 = 42.8A.

3 0
3 years ago
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According to Newton's Law of Cooling, if a body with temperature T 1 is placed in surroundings with temperature T 0, different f
Mila [183]

We can substitute the given values into the equation for T, given the surrounding temperature T0 = 0, initial temperature T1 = 140, constant k = -0.0815, and time t = 15 minutes.

T = 0 + (140 - 0)e^(-0.0815*15) = 140e^(-1.2225) = 41.23°F

3 0
3 years ago
Read 2 more answers
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