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ad-work [718]
3 years ago
14

What would be new resistance if length of conductor is doubled and thickness is halved ​

Physics
1 answer:
nignag [31]3 years ago
6 0

Answer:

The new resistance comes out to be = 4 times of original resistance .

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A machine takes 0.5 seconds to move a brick 1 meter and put 100 Joules of energy into it. (hint: Power is the amount of energy t
Flauer [41]

Answer:

200 Watts.

Explanation:

Power is defines as the amount of work expended per unit time. Mathematically, it is expressed as Power = Workdone/Time

Given parameters

Energy used up 100Joules

Distance moved by brick = 1 meters

Time taken by the machine = 0.5 secs

Power can also be written as Energy/Time

Required

We need to calculate the amount of power used up.

Power = 100J/0.5s

Power = 100/(1/2)

Power = 100 * 2/1

Power = 200Watts.

This shows that the machine would expend 200Watts of power

8 0
3 years ago
A wave carries _____ from one place to another <br><br> Mechanical waves carry energy through ______
gulaghasi [49]

A wave carries <u>energy</u><u> </u>from one place to another.

mechanical waves carry energy through <u>MEDIUM</u><u>.</u>

<u>SO</u><u> </u><u>THIS</u><u> </u><u>IS</u><u> </u><u>MY</u><u> </u><u>ANSWER</u>

7 0
3 years ago
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Which is an example of sliding friction?
Anvisha [2.4K]
I think its D not sure
4 0
3 years ago
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A 1.8-kg object is attached to a spring and placed on frictionless, horizontal surface. A force of 40 N stretches a spring 20 cm
Sergio [31]

Answer:

a) k = 200 N/m

b) E = 4 J

c) Δx = 6.3 cm

Explanation:

a)

  • In order to find force constant of the spring, k, we can use the the Hooke's Law, which reads as follows:

       F = - k * \Delta x (1)

  • where F = 40 N and Δx =- 0.2 m (since the force opposes to the displacement from the equilibrium position, we say that it's a restoring force).
  • Solving for k:

       k =- \frac{F}{\Delta x} =-\frac{40 N}{-0.2m} = 200 N/m (2)

b)

  • Assuming no friction present, total mechanical energy mus keep constant.
  • When the spring is stretched, all the energy is elastic potential, and can be expressed as follows:

        U = \frac{1}{2}* k* (\Delta x)^{2} (3)

  • Replacing k and Δx by their values, we get:

       U = \frac{1}{2}* k* (\Delta x)^{2} = \frac{1}{2}* 200 N/m* (0.2m)^{2} = 4 J (4)

c)

  • When the object is oscillating, at any time, its energy will be part elastic potential, and part kinetic energy.
  • We know that due to the conservation of energy, this sum will be equal to the total energy that we found in b).
  • So, we can write the following expression:

        \frac{1}{2}* k* \Delta x_{1} ^{2} + \frac{1}{2} * m* v^{2}  = \frac{1}{2}*k*\Delta x^{2}   (5)

  • Replacing the right side of (5) with (4), k, m, and v by the givens, and simplifying, we can solve for Δx₁, as follows:

        \frac{1}{2}* 200N/m* \Delta x_{1} ^{2} + \frac{1}{2} * 1.8kg* (-2.0m/s)^{2}  = 4J   (6)

⇒      \frac{1}{2}* 200N/m* \Delta x_{1} ^{2}   = 4J  - 3.6 J = 0.4 J (7)

⇒     \Delta x_{1}   = \sqrt{\frac{0.8J}{200N/m} } = 6.3 cm (8)

6 0
3 years ago
In a standard tensile test a steel rod of 22-mm diameter is subjected to a tension force of 75kN. Knowing that v = 0.30 and E =
VladimirAG [237]

Answer:

(a) Elongation of rod = 0.19732 mm

(b) Change in diameter = 0.00651 mm

Explanation:

Circular area at end of steel rod = pi * diameter^2 / 4

Area = pi * (22 * 10^-^3)/4

Area = 3.801 * 10^(-4)    meter squared

Stress = force / area

Stress = 75000 / (3.801 * 10^-4)

Stress = 197316495 Pa     OR       0.197 GPa

Modulus of elasticity = stress / strain

200 = 0.19732 / Strain

Strain = 0.0009866     (longitudinal)

(a) Strain = change in length / initial length

0.0009866 = Elongation of rod / 200

Elongation of rod = 0.19732 mm

(b) Poisson ratio = lateral strain / longitudinal strain

0.3 = lateral strain / 0.0009866

Lateral strain = 0.000296

Lateral strain = Change in diameter / original diameter

0.000296 = Change in diameter / 22

Change in diameter = 0.00651 mm

4 0
3 years ago
Read 2 more answers
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