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antoniya [11.8K]
3 years ago
13

Use Hooke's Law to determine the work done by the variable force in the spring problem. Nine joules of work is required to stret

ch a spring 0.5 meter from its natural length. Find the work required to stretch the spring an additional 0.40 meter.
Physics
1 answer:
natima [27]3 years ago
6 0

Answer:

29.16 J

Explanation:

From Hook's law,

W = 1/2(ke²)..................... Equation 1

Where W = work done, k = Spring constant, e = extension.

Given: W = 9 J, e = 0.5 m.

Substitute into equation 1

9 = 1/2(k×0.5²)

Solve for k

k = 18/0.5²

k = 72 N/m.

The work done required to stretch the spring by additional 0.4 m is

W = 1/2(72)(0.4+0.5)²

W = 36(0.9²)

W = 29.16 J.

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At room temperature, an oxygen molecule, with mass of 5.31x10^-26kg , typically has a kinetic energy of about 6.21x10^-21J. How
den301095 [7]

Answer:

V=483.63 m/s

Explanation:

Given that

mass ,m= 5.31 x 10⁻²⁶ kg

Kinetic energy KE= 6.21 x 10⁻²¹ J

As we know that the kinetic energy of the mass m with moving velocity V given as

KE=\dfrac{1}{2}mV^2  

Now by putting the values

6.21\times 10^{-21}=\dfrac{1}{2}\times 5.31\times 10^{-26}\times V^2

V^2=\dfrac{2\times 6.21\times 10^{-21}}{5.31\times 10^{-26}}

V^2=233898.30

V=\sqrt{233898.30}\ m/s

V=483.63 m/s

Therefore the velocity of the molecule at the room temperature will be 483.63 m/s.

6 0
4 years ago
What is the power of a machine that transfers 15,000J of energy in 1 minute?
____ [38]

Answer:

answer is 250 W

Explanation:

W=P * t

t=60s

W= 15000J

15000= P * 60

P= 15000/60

P= 250W. (SI unit of power Watt)

5 0
3 years ago
A long, thin rod parallel to the y-axis is located at x = -1.0 cm and carries a uniform linear charge density of +1.0 nC/m. A se
Bad White [126]

Answer:

Hhere is the other part of the question ; What is the net electric field due to these rods at the origin? (epsilon not = 8.85*10^-12)

net electric field = 3600N/C

Explanation:

The detailed calculation is as shwon below

8 0
3 years ago
Refrigerators are usually kept at about 5∘C, while room temperature is about 20∘C. If you were to take an "empty" sealed 2-liter
expeople1 [14]

Answer:

The volume will not contract to one fourth of the original.

Explanation:

Applying Charles's Law as:

\frac {V_1}{T_1}=\frac {V_2}{T_2}

Given ,  

V₁ = ?

V₂ = 2 L

T₁ = 5 °C

T₂ = 20 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,

T₁ = (5 + 273.15) K = 278.15 K  

T₂ = (20 + 273.15) K = 293.15 K  

Using Charles law as:

V_1=\frac {V_2}{T_2} \times {T_1}

V_1=\frac {2\ L}{293.15 K} \times {278.15 K}

V_1=1.8977 L

Thus, the volume will not contract to one fourth of the original. (1/4 of 2 L is 0.5 L).

8 0
4 years ago
A 75.0-kg person is riding in a car moving at 20.0 m/s when the car runs into a bridge abutment. (a) calculate the average force
iragen [17]

(a) -1.5\cdot 10^6 N

First of all, we need to calculate the acceleration of the person, by using the following SUVAT equation:

v^2 -u ^2 = 2ad

where

v = 0 is the final velocity

u = 20.0 m/s is the initial velocity

a is the acceleration

d = 1.00 cm = 0.01 m is the displacement of the person

Solving for a,

a=\frac{v^2-u^2}{2d}=\frac{0-20.0^2}{2(0.01)}=-20000 m/s^2

And the average force on the person is given by

F=ma

with m = 75.0 kg being the mass of the person. Substituting,

F=(75)(-20000)=-1.5\cdot 10^6 N

where the negative sign means the force is opposite to the direction of motion of the person.

b) -1.0\cdot 10^5 N

In this case,

v = 0 is the final velocity

u = 20.0 m/s is the initial velocity

a is the acceleration

d  = 15.00 cm = 0.15 m is the displacement of the person with the air bag

So the acceleration is

a=\frac{v^2-u^2}{2d}=\frac{0-20.0^2}{2(0.15)}=-1333 m/s^2

So the average force on the person is

F=ma=(75)(-1333)=-1.0\cdot 10^5 N

7 0
4 years ago
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