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icang [17]
3 years ago
10

A 2.5 kg block is launched along the ground by a spring with a spring constant of 56 N/m. The spring is initially compressed 0.7

5 m.
Disregarding friction, how fast will the block move after the spring is released all the way and the block slides away from it?

3.5 m/s
4.1 m/s
13 m/s
16 m/s
Physics
2 answers:
Norma-Jean [14]3 years ago
7 0

vf ^2 = kx^2/m = 56(0.75)^2 / 2.5 = 12.6


Therefore, v= 3.5 m/s.

olga55 [171]3 years ago
4 0

Answer: The correct option is 3.5m/s.

Explanation: In the question, conservation of energy is taking place which means that the energy is getting transferred from one form to another form.

Here, elastic potential energy of the spring is getting converted to the kinetic energy of the block.

Mathematically,

\frac{1}{2}kx^2=\frac{1}{2}mv^2

where, k = spring constant = 56N/m

x = compression of extension of spring = 0.75m

m = Mass of the block = 2.5kg

v = velocity of the block = ? m/s

Putting values in above equation, we get:

\frac{1}{2}\times 56\times (0.75)^2=\frac{1}{2}\times 2.5\times v^2\\\\v=3.54m/s

Hence, the block will move at a speed of 3.5m/s

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A car travels 200km in 3.0 hours. Determine the average velocity of the car
Masja [62]

Answer:

<h2>66.67 km/hr</h2>

Explanation:

The average velocity of the car can be found by using the formula

a =  \frac{d}{t }  \\

d is the distance

t is the time taken

From the question we have

a =  \frac{200}{3}  \\  = 66.66666...

We have the final answer as

<h3>66.67 km/hr</h3>

Hope this helps you

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3 years ago
Which of these is a benefit of nuclear energy?
Mademuasel [1]
The answer i think would be D 

4 0
3 years ago
Read 2 more answers
Please help answer question​
nika2105 [10]

Answer:

C = 1.01

Explanation:

Given that,

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The terminal velocity of the mass, v_t=60\ m/s

Area of cross section, A=0.33\ m^2

We need to find the drag coefficient. At terminal velocity, the weight is balanced by the drag on the object. So,

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or

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So,

C=\dfrac{2mg}{\rho Av_t^2}\\\\C=\dfrac{2\times 75\times 9.8}{1.225\times 0.33\times (60)^2}\\\\C=1.01

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4 0
2 years ago
What happens to its kinetic energy when its reaches the maximum height ?​
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7 0
3 years ago
Three monkeys A, B, and C weighing 20, 26, and 25 lb, respectively, are climbing up and down the rope suspended from D. At the i
Marina86 [1]

Answer:2235.2lb-ft/s^2

Explanation:

Given

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acceleration of monkey A=4.2ft/s^2

acceleration of monkey B=0

acceleration of monkey C=-5.4ft/s^2

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Force Due to monkey A\left ( F_C\right )=25\times 5.4 =135 lb-ft/s^2\left ( upward\right )

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T=\left ( 20+26+25\right )32.2+84-135=2235.2 lb-ft/s^2

5 0
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