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Iteru [2.4K]
3 years ago
5

The spring is compressed a total of 3.0 cm, and used to set a 500 gram cart into motion. Find the speed of the cart at the insta

nt it is released, assuming all the elastic potential energy is converted.
Physics
1 answer:
BartSMP [9]3 years ago
4 0

Answer:

1.15 m/s

Explanation:

Part of the question is missing. Found the missing part on google:

"1. A hanging mass of 1500 grams compresses a spring 2.0 cm.   Find the spring constant in N/m."

Solution:

First of all, we need to find the spring constant. We can use Hooke's law:

F=kx

where

F=mg=(1.5 kg)(9.8 m/s^2)=14.7 N is the force applied to the spring (the weight of the hanging mass)

x = 2.0 cm = 0.02 m is the compression of the spring

Solving for k, we find the spring constant:

k=\frac{F}{x}=\frac{14.7}{0.02}=735 N/m

In the second part of the problem, the spring is compressed by

x = 3.0 cm = 0.03 m

So the elastic potential energy of the spring is

U=\frac{1}{2}kx^2=\frac{1}{2}(735)(0.03)^2=0.33 J

This energy is entirely converted into kinetic energy of the cart, which is:

U=K=\frac{1}{2}mv^2

where

m = 500 g = 0.5 kg is the mass of the cart

v is its speed

Solving for v,

v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(0.33)}{0.5}}=1.15 m/s

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Answer:

Given,

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To find,

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Now,

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Total time taken to display the given number of frames (ie. 25 frames) = 1 second

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8 0
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A 269-turn solenoid is 102 cm long and has a radius of 2.3 cm. It carries a current of 3.9 A. What is the magnetic field inside
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Explanation:

Given;

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length of the solenoid, L = 102 cm = 1.02 m

radius of the solenoid, r = 2.3 cm = 0.023 m

current in the solenoid, I = 3.9 A

Magnitude of the magnetic field inside the solenoid near its centre is calculated as;

B = \frac{\mu_o NI}{l} \\\\

Where;

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B = \frac{4\pi*10^{-7} *269*3.9}{1.02} \\\\B = 1.293 *10^{-3} \ T

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1. Any push or pull is called
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Answer:

b  ok

Explanation:

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