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alexdok [17]
3 years ago
14

Tara prepared a report to show how the amplitude of waves affects the energy of waves. Is her graphical representation correct?

Physics
2 answers:
irga5000 [103]3 years ago
8 0

No, it is not correct because as the amplitude of a wave increases, the energy it carries also increases. This graph shows that as the amplitude increases, the energy decreases.                  

Pls mark as brain-list

Gelneren [198K]3 years ago
6 0

Answer:

It is not correct because the amplitude of the waves can be bigger than others and the graph can be going up and down

Explanation: I got the question right

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A mass m attached to a spring of constant k is oscillating on a frictionless surface. A second mass of mass m is dropped on top
elixir [45]

Answer:

E. The period of oscillation increases.

Explanation:

The period of oscillation is:

T = 2π√(m/k)

Frequency is the inverse of period (f = 1/T), so as period increases, frequency decreases.

Increasing the mass will increase the period and decrease the frequency.

8 0
2 years ago
What are the effects of global warming in simple words?
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3 0
1 year ago
How would you compare and contrast the 3 types of friction?
vlabodo [156]
Rolling friction .<span> the force that slows down the movement of a rolling object</span>
sliding friction.
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7 0
3 years ago
Read 2 more answers
A 50 Q resistor in a circuit has a current flowing through it of 2.0 A. What is
Alex17521 [72]

Hello!

We can use the following equation for calculating power dissipated by a resistor:
P = i^2R

P = Power (? W)
i = Current through resistor (2.0 A)
R = Resistance of resistor (50Ω)

Plug in the known values and solve.

P = (2.0^2)(50) = \boxed{\text{ B. }200 W}

7 0
1 year ago
Suppose a person pushes thumbtack that is 1/5 centimeter long into a bulletin board, and the force (in dynes) exerted when the d
mario62 [17]

Answer:

W = 290.7 dynes*cm

Explanation:

d = 1/5 cm = 0.2 cm

The force is in function of the depth x:

F(x) = 1000 * (1 + 2*x)^2

We can expand that as:

F(x) = 1000 * (1 + 4*x + 4x^2)

F(x) = 1000 + 4000*x + 4000*x^2

Work is defined as

W = F * d

Since we have non constant force we integrate

W = \int\limits^{0.2}_{0} {(1000 + 4000*x + 4000*x^2)} \, dx

W = [1000*x + 2000*x^2 + 1333*X^3] evaluated between 0 and 0.2

W = 1000*0.2 + 2000*0.2^2 + 1333*0.2^3 - 1000*0 - 2000*0^2 - 1333*0^3

W = 200 + 80 + 10.7 = 290.7 dynes*cm

3 0
3 years ago
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