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emmainna [20.7K]
3 years ago
9

Helllp me will give brainlist

Physics
1 answer:
Mandarinka [93]3 years ago
7 0

The top row of boxes is " F O R C E " .

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the answer is a because they start on the north side

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Which one of the following accurately describes a proper use of eyeglasses?
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A box has a length of 18.1 cm and a width of 19.2cm tall ?
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Calculate the change in velocity of a 0.070kg tennis ball hit by Serena with a force of 140 N over 0.020 s
Brilliant_brown [7]
V=at and a=F/m

140/.070 = 2000m/s^2

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The ball’s velocity increased by 40m/s.
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A 0.750 kg block is attached to a spring with spring constant 13.0 N/m . While the block is sitting at rest, a student hits it w
trapecia [35]

To solve this problem we will apply the concepts related to energy conservation. From this conservation we will find the magnitude of the amplitude. Later for the second part, we will need to find the period, from which it will be possible to obtain the speed of the body.

A) Conservation of Energy,

KE = PE

\frac{1}{2} mv ^2 = \frac{1}{2} k A^2

Here,

m = Mass

v = Velocity

k = Spring constant

A = Amplitude

Rearranging to find the Amplitude we have,

A = \sqrt{\frac{mv^2}{k}}

Replacing,

A = \sqrt{\frac{(0.750)(31*10^{-2})^2}{13}}

A = 0.0744m

(B) For this part we will begin by applying the concept of Period, this in order to find the speed defined in the mass-spring systems.

The Period is defined as

T = 2\pi \sqrt{\frac{m}{k}}

Replacing,

T = 2\pi \sqrt{\frac{0.750}{13}}

T= 1.509s

Now the velocity is described as,

v = \frac{2\pi}{T} * \sqrt{A^2-x^2}

v = \frac{2\pi}{T} * \sqrt{A^2-0.75A^2}

We have all the values, then replacing,

v = \frac{2\pi}{1.509}\sqrt{(0.0744)^2-(0.750(0.0744))^2}

v = 0.2049m/s

7 0
3 years ago
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