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allsm [11]
3 years ago
12

Consider an ideal spring, with spring constant k, which is oriented along an x-axis. one end of the spring is fixed, and the fre

e end is located at x = 0 when the spring is neither stretched nor compressed. an object having a mass m is attached to the free end of the spring. consider an action in which the free end of the spring is moved from an initial position xi to a final position xf. the work done by the spring (and on the object attached to the free end) during this action is given by
Physics
1 answer:
Blizzard [7]3 years ago
5 0
The spring is neither stretched nor compressed. an object having a mass m is attached to the free end of the spring. consider an action
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Suppose you take
kati45 [8]

Answer:

a current will be induced.

Explanation:

4 0
3 years ago
Read 2 more answers
A total electric charge of 6.75 nC is distributed uniformly over the surface of a metal sphere with radius 20.0 cm. If the poten
djverab [1.8K]

Answer:

a) 60 V

b) 125 V

c) 125 V

Explanation:

<u>Given</u>

We are given the total electric charge q = 6.75 nC = 6.75x 10^-9 C distributed uniformly over the surface of a metal sphere with a radius of R = 20.0 cm = 0.020 m.  

<u>Required </u>

We are asked to calculate the potential at the distances

(a) r = 10.0 cm

(b) r = 20.0 cm

(c) r = 40.0 cm  

<u>Solution</u>

(a) Here, the distance r > R so, we can get the potential outside the sphere (r > R) where the potential is given by

V = q/4\pi∈_o                       (1)

r is the distance where the potential is measured and the term 1/4\pi∈_o equals  9.0 x 10^9 Nm^2/C^2. Now we can plug our values for q and r into equation (1) to get the potential V where r = 0.10 m  

V= 1*q/4\pi∈_o*r

 =60 V

(b) Here the distance r is the same for the radius R, so we can get the potential inside the sphere (r = R) where the potential is given by  

V = 1*q/4\pi∈_o*R                (2)    

Now we can plug our values for q and R into equation (2) to get the potential V where R = 0.20 m  

V = 1*q/4\pi∈_o*R

    = 125 V

(c) Inside the sphere the electric field is zero therefore, no work is done on a test charge that moves from any point to any other point inside the sphere. Thus the potential is the same at every point inside the sphere and is equal to the potential on the surface. and it will be the same as in part (b)  

V= 125 V

7 0
3 years ago
How do you convert 1.3*10^6cal into joules
ehidna [41]

Answer:

5.4×10⁶J

Explanation:

1 cal = 4.184 J

1.3×10⁶ cal × (4.184 J/cal) = 5.4×10⁶J

3 0
3 years ago
In a hot water heating system, room temperature is controlled by a device called a
erik [133]
<h3><u>Answer</u>;</h3>

Thermostat

<h3><u>Explanation;</u></h3>
  • A thermostat is a device that exerts control by switching heating or cooling devices on or off, or by regulating the flow of a heat transfer fluid as needed, to maintain the correct temperature.
  • Thermostat automatically regulates temperature, or that activates a device when the temperature reaches a certain point.
4 0
3 years ago
Two round rods, one steel andthe other copper, are joined end to end. Each rod is 0.750 mlong and 1.50 cmin diameter. The combin
nikitadnepr [17]

Answer: a) Strain on Steel rod = 0.0001078

b) elongation on the steel rod = 0.00008085m = 0.008085cm = 0.0081mm.

c) strain on Copper Rod = 0.000189

Explanation: a) To obtain the strain of the steel rod, we invoke Hooke's law which states that, provided the elastic limit of A material isn't exceeded, the stress it undergoes is directly proportional to its strain.

(Stress, σ) ∝ (Strain, ε)

The constant of proportionality is called Young's modulus, E.

σ = Eε

For steel, Younger Modulus as obtained from literature = 210GPa.

Strain = Stress/Young's Modulus

Stress = (Force or Load applied)/ Cross sectional Area.

Force applied For the steel = 4000N

Cross sectional Area = (π(D^2))/4

D = 1.50cm = 0.015m

A = 0.0001767 m2

σ = 4000/0.0001767 = 22637238.257 N/m2

ε = σ/E = 22637238.257/(210 × (10^9)) = 0.0001078.

b) To solve for elongation.

Strain, ε = (elongation, dl)/(original length, lo)

Elongation, dl = strain × original length

dl = 0.0001078 × 0.75 = 0.00008085m = 0.008085cm = 0.0081mm.

c) strain in Copper

ε = σ/E; σ = 22637238.257 N/m2

Young's modulus of Copper, from literature, = 120GPa

ε = 22637238.257/(120 × (10^9)) = 0.000189

4 0
3 years ago
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