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Nutka1998 [239]
3 years ago
13

which is not part of maturity? excising self control, using good judgement, making impulsive decisions, being dependable

Physics
2 answers:
IgorLugansk [536]3 years ago
7 0

Answer:

using good judgement

Explanation:

cricket20 [7]3 years ago
4 0
Using good judgment hope this helps
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Alona [7]

Answer:

Explanation:

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5 0
3 years ago
A block of mass 200g is oscillating on the end of a horizontal spring of spring constant 100 N/m and natural length 12 cm. When
malfutka [58]

In order to determine the acceleration of the block, use the following formula:

F=ma

Moreover, remind that for an object attached to a spring the magnitude of the force acting over a mass is given by:

F=kx

Then, you have:

ma=kx

by solving for a, you obtain:

a=\frac{kx}{m}

In this case, you have:

k: spring constant = 100N/m

m: mass of the block = 200g = 0.2kg

x: distance related to the equilibrium position = 14cm - 12cm = 2cm = 0.02m

Replace the previous values of the parameters into the expression for a:

a=\frac{(\frac{100N}{m})(0.02m)}{0.2\operatorname{kg}}=10\frac{m}{s^2}

Hence, the acceleration of the block is 10 m/s^2

8 0
1 year ago
The apple became weightless when it fell...
Tems11 [23]
The answer I believe would be a sorry if I’m wrong
3 0
3 years ago
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Impact of electricity in the society
Travka [436]

Answer:

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5 0
2 years ago
Letting D D represent the maximum displacement, the extremes of the block's motion are at position A, where x = − D x=−D, and at
Ksju [112]

Answer:

The answer is at x = 0, which represents position B

Explanation:

The full question is:

"A block is attached to a horizontal spring and set in a

simple harmonic motion, as shown from above in the figure. When the spring is relaxed, the block is a position B, where the displacement x from the equilibrium position is 0. Letting D represent the maximum displacement, the extremes of the block's motion are at position A, where x= -D, and at position C, where x= D.

At what point in the motion is the speed of the block at its maximum?"

And you can see the figure on the attached file.

Simple Harmonic motion equations

We can start from the equation that describes the position that is

x(t)=D \sin\left(\omega t)

Here D stands for the amplitude which is the maximum displacement, and \omega is the angular velocity, thus we can find the derivative to find the velocity equation, so we get

v(t)=D \omega \cos (\omega t)

And we can find the derivative again to find the acceleration.

a(t) = -D\omega^2 \sin (\omega t)

Maximum speed

We reach the maximum speed when the acceleration equation is equal to 0,

a(t) =0\\-D\omega^2 \sin (\omega t)=0

Thus it happens when

\sin (\omega t)=0

So if we replace that on the position equation we get

x(t)=D \sin(\omega t) \\x(t)=D(0)\\x(t)=0

Thus the position where the speed of the block is at at its maximum is when it is going back to the origin, that is x = 0, so point b.

7 0
3 years ago
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