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Yuki888 [10]
3 years ago
9

4. Calculate the slope of the line in your graph of the square of the period of the spring vs. mass [slope = (y2 – y1)/(x2 – x1)

]. Robert Hooke figured out the equation that describes the periodic motion of the spring. If you square both sides of the equation, you will find that the slope of the line is related to the spring constant (k). Specifically, slope = 42/k. Use your data to calculate the k of your spring.
Physics
1 answer:
Hunter-Best [27]3 years ago
5 0
You need data for this problem
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12
makkiz [27]

Answer:

i done know sorry for not answering ur question

3 0
3 years ago
A toy car has an initial acceleration of 2 m/s" across a horizontal surface after it is released from rest. After the car travel
Ganezh [65]

Answer:

Closed system, because the speed of the car is as expected in the case where an object has uniform acceleration for a time t

Explanation:

Here in the question it is mentioned that a toy car has an initial acceleration of 2m/s²  across a horizontal surface so we can say that it is acted upon by an external force

Assuming that the acceleration is constant and the reason for this assumption is there at the last

The major difference between an open system and closed system is in case of open system there will be transfer of matter and in case of closed system there will be no change in matter of the system

If acceleration is constant in case of closed system we can expect the speed of the car after a time t by using the formula

  s = u×t + 0·5×a×t²

where s is the distance travelled

t is the time taken to travel that distance

u is the initial velocity

a is the acceleration of that system

But in case of open system as there will be a change of mass there will be a change in velocity of the system so in this case we cannot expect the speed of the car after a time t

And if the acceleration is not constant then we cannot say that the toy car is an open system or closed system, that is why we are assuming that the acceleration of the toy car is constant

3 0
3 years ago
Read 2 more answers
How much heat energy must be added to the gas to expand the cylinder length to 16.0 cm ?
Lapatulllka [165]

This question is incomplete, the complete question is;

A monatomic gas fills the left end of the cylinder in the following figure. At 300 K , the gas cylinder length is 14.0 cm and the spring is compressed by65.0 cm . How much heat energy must be added to the gas to expand the cylinder length to 16.0 cm ?

Answer:

the required heat energy is 16 J

Explanation:

Given the data in the question;

Lets consider the ideal gas equation;

PV = nRT

from the image, we calculate initial pressure;

Pi = ( 2000N/M × 0.06m) / 0.0008 m²

Pi = 15 × 10⁴ Pa

next we find Initial velocity

Vi = (0.0008 m²)(0.14) = 1.1 × 10⁻⁴ m²

now we find the number of moles

n = [(15 × 10⁴ Pa)(1.1 × 10⁻⁴ m²)] / 8.31 J/molK × 300K

N = 6.6 × 10⁻³ mol

next we calculate the final temperature;

Pf = ( 2000N/m × 0.08) / 0.0008 m²

Pf = 2 × 10⁵ Pa

Calculate the final Volume

Vf = (0.0008 m² × 0.16 m = 1.28 × 10⁻⁴ m³

we also determine the final temperature

T_{f} =  (2 × 10⁵ Pa × 1.28 × 10⁻⁴ m³) / 6.6 × 10⁻³ × 8.31 J/molK

T_{f}  = 466.8 K

so change in temperature ΔT

ΔT =  466.8 K - 300K = 166.8 K

we then calculate the change in thermal energy

ΔU = nCΔT

ΔU = ( 6.6 × 10⁻³ mol ) × 12.5 × 166.8K

ΔU = 13.761 J

C is the isochoric molar specific heat which is equal to 3R/2 for monoatomic

now we calculate the work done;

W = 1/2 × K( x_{i\\}² - x_{f\\}² )

W = 1/2 × ( 2000 N/m) ( 0.06² - 0.08² )

= - 2.8 J

and we then calculate the heat energy using the following expression;

Q = ΔU - W

we substitute

Q = 13.761 - (- 2.8 J)

Q = 13.761 + 2.8 J)

Q =  16 J

Therefore, the required heat energy is 16 J

5 0
3 years ago
Need help on these 2 questions please????? Hurry
fredd [130]

Answer:

Question 1 the anwser is 300 meters.  Question 2 the answer is speed.

is Explanation:

3 0
3 years ago
PLEASE ASAP ILL GIVE BRAINLIEST.
taurus [48]

Answer:

applied force

Explanation:

any force where you push or pull is always applied force.

4 0
2 years ago
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