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77julia77 [94]
2 years ago
12

Use Hooke's Law to determine the work done by the variable force in the spring problem. A force of 450 newtons stretches a sprin

g 30 centimeters. How much work is done in stretching the spring from 50 centimeters to 80 centimeters
Physics
1 answer:
natima [27]2 years ago
5 0

The work done in stretching the spring from 50 cm to 80 cm is 67.5 J.

<h3>Hooke's Law</h3>

Hooke's law states that the force applied to an elastic material is directly proportional to its extension, provided its elastic limit is not exceeded.

To calculate the amount of work done by Hooke's law, first, we need to find the force constant of the spring.

Formula:

  • F = ke................. Equation 1

Where:

  • F = Force applied
  • k = Spring constant
  • e = extension

make k the subject of the equation

  • k = F/e................ Equation 2

From the question,

Given:

  • F = 450 N
  • e = 30 cm = 0.3 m

Substitute these values into equation 2

  • k = 450/0.3
  • k = 1500 N/m.

Finally, To find the work done in stretching the spring from 50 cm to 80 cm, we use the formula below.

  • W = ke²/2........... Equation 3

Where:

  • W = Work done
  • k = spring constant
  • e = extension

Also, From the question,

Given:

  • e = (80-50) = 30 cm = 0.3 m
  • k = 1500 N/m

Substitute these values into equation 3

  • W = 1500(0.3²)/2
  • W = 67.5 J.

Hence, The work done in stretching the spring from 50 cm to 80 cm is 67.5 J.

Learn more about Hooke's law here: brainly.com/question/12253978

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Yuri [45]

The frictional force required is 9000 N

Explanation:

In order to keep the car in the turn in circular motion without sliding, the frictional force must provide the centripetal force necessary for the circular motion.

Therefore, we can write:

F_f = m \frac{v^2}{r}

where the term on the left is the frictional force while the term on the right is the centripetal force, and where:

m is the mass of the car

v is its speed

r is the radius of the curve

For the car in this turn, we have

m = 1000 kg

v = 30 m/s

r=\frac{0.20 km}{2}=0.10 km = 100 m (since the diameter is 0.20 km, the radius is half that value)

And substituting, we find

F_f = (1000) \frac{30^2}{100}=9000 N

Learn more about friction:

brainly.com/question/6217246

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8 0
4 years ago
Light is incident on a piece of glass in air at an angle of 33 degrees from the normal. If the index of refraction of the glass
lilavasa [31]
About 21 to 22 degrees 

as below

4 0
3 years ago
What is the net charge of a metal ball if there are 21,749 extra electrons in it?
pickupchik [31]

Answer:

Q=3.47\times 10^{-15}\ C

Explanation:

Given that,

Number of extra electrons, n = 21749

We need to find the net charge on the metal ball. Let Q is the net charge.

We know that the charge on an electron is q=1.6\times 10^{-19}\ C

To find the net charge if there are n number of extra electrons is :

Q = n × q

Q=21749\times 1.6\times 10^{-19}\ C

Q=3.47\times 10^{-15}\ C

So, the net charge on the metal ball is 3.47\times 10^{-15}\ C. Hence, this is the required solution.

6 0
3 years ago
The current through two identical light bulbs connected in series is 0.25 A. The voltage across both bulbs is 110 V. The resista
Rus_ich [418]

The resistance of a single light bulb is 220 ohms per bulb.

<h3>What is Ohm's Law?</h3>

Ohm's Law is a formula used to determine how voltage, current, and resistance in an electrical circuit relate to one another.

Ohm's Law (E = IR) is as basic to students of electronics as Einstein's Relativity equation (E = mc2) is to physicists.

E = I x R

The formula reads voltage = current x resistance, or V = A xΩ., or volts = amps x ohms.

110volts divided by .25amps = 440 ohms. 440 divided by 2 =220 ohms per bulb.

R = 110/(2*0.25) = 220 ohms

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4 0
2 years ago
At a certain elevation, the pilot of a balloon has a mass of 120 lb and a weight of 119 lbf. What is the local acceleration of g
Strike441 [17]

Answer:

31.905 ft/s²

Explanation:

Given that

Mass of the pilot, m = 120 lb

Weight of the pilot, w = 119 lbf

Acceleration due to gravity, g = 32.05 ft/s²

Local acceleration of gravity of found by using the relation

Weight in lbf = Mass in lb * (local acceleration/32.174 lbft/s²)

119 = 120 * a/32. 174

119 * 32.174 = 120a

a = 3828.706 / 120

a = 31.905 ft/s²

Therefore, the local acceleration due to gravity at that elevation is 31.905 ft/s²

3 0
3 years ago
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