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USPshnik [31]
2 years ago
7

A spring is stretched 2.0 cm. If the same spring is stretched 8.0 cm the ratio of the second and first potential energies of the

spring is
16
9
4
1
Physics
1 answer:
jok3333 [9.3K]2 years ago
6 0
Potential energy of spring equals K times X squared divided by 2 where X is displacement

4 times squared equals 16

choose 1st answer
 
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A bicyclist in the Tour de France crests a mountain pass as he moves at 18 km/h. At the bottom, 4.0 km farther, his speed is at
Allisa [31]

We are given:

v0 = initial velocity = 18 km/h

d = distance = 4 km

v = final velocity = 75 km/h

a =?

<span>
We can solve this problem by using the formula:</span>

v^2 = v0^2 + 2 a d

 

75^2 = 18^2 + 2 (a) * 4

5625 = 324 + 8a

<span>a = 662.625 km/h^2</span>

6 0
3 years ago
When atoms are organized in repeating patterns, they generally make a _________. select one of the options below as your answer:
valkas [14]
A. Solid

Solid structures are actually made of repeating patterns.

4 0
3 years ago
A solenoid has 450 loops each of radius 0.0254 m. The field increases from 0 T to 3.00 T in 1.55 s. What is the EMF generated in
allochka39001 [22]

Answer:

0.175 second

Explanation:

i hope it helps

8 0
2 years ago
The net force acting on a 5 kg discus is 50 n. what is the acceleration of the discus
Lyrx [107]
Answer: 10 m/s^2

Explanation:

1) The second law of Newton gives the definition and formula to calculate the net force:

Net force acting on an object = mass * acceleration.

2) From that, when you know the net force acting of the object and its mass, you can solve for the acceleration:

acceleration = Net force / mass

acceleration = 50 N / 5 kg = 10 m/s^2, which is the answer.
8 0
3 years ago
A solenoid 91.0 cm long has a radius of 1.50 cm and a winding of 1300 turns; it carries a current of 3.60 A. Calculate the magni
irinina [24]

The magnitude of the magnetic field inside the solenoid is 6.46 \times 10^{-3} \ T.

The given parameters;

  • <em>length of the solenoid, L = 91 cm = 0.91 m</em>
  • <em>radius of the solenoid, r = 1.5 cm = 0.015 m</em>
  • <em>number of turns of the solenoid, N = 1300 </em>
  • <em>current in the solenoid, I = 3.6 A</em>

The magnitude of the magnetic field inside the solenoid is calculated as;

B = \mu_0 nI\\\\B = \mu_o(\frac{ N}{L} )I\\\\

where;

\mu_o is the permeability of frees space = 4π x 10⁻⁷ T.m/A

B = (4\pi \times 10^{-7}) \times (\frac{1300}{0.91} ) \times 3.6\\\\B = 6.46 \times 10^{-3} \ T

Thus, the magnitude of the magnetic field inside the solenoid is 6.46 \times 10^{-3} \ T.

Learn more here:brainly.com/question/17137684

7 0
2 years ago
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