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Tasya [4]
3 years ago
13

A wire of length L and cross-sectional area A has resistance R.

Physics
1 answer:
Leona [35]3 years ago
6 0

Answer:

The resistance R stretched of the wire if it is stretched to twice its original length is [Δ(2L) / A ]* 2

Explanation:

Original Resistance R = ΔL / A

Given length is doubled

New length  L' = 2L

New Volume is constant, resistivity and density of material is also assumed to be constant

AL = A'L'

AL = A'(2L)

A' = AL / 2L

A' = A / 2

∴ new Resistance R' = ΔL' / A'

                                 =  Δ(2L) / A/2

                                  = [Δ(2L) / A ]* 2

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Consider the following arrangement with a frictionless/massless pulley. Determine the force F required to move block A if the co
Delicious77 [7]

Answer: The free - body diagrams for blocks A and B. frictionless surface by a constant horizontal force F = 100 N. Find the tension in the cord between the 5 kg and 10 kg blocks. The string that attaches it to the block of mass M2 passes over a frictionless pulley of negligible mass. The coefficient of kinetic friction Hk between M.

Explanation: Hope this helped :)

6 0
3 years ago
A 9-μC positive point charge is located at the origin and a 6 μC positive point charge is located at x = 0.00 m, y = 1.0 m. Find
sukhopar [10]

Answer:

The coordinates of the point is (0,0.55).

Explanation:

Given that,

First charge q_{1}=9\times10^{-6}\ C at origin

Second charge q_{2}=6\times10^{-6}\ C

Second charge at point P = (0,1)

We assume that,

The net electric field between the charges is zero at mid point.

Using formula of electric field

E=\dfrac{kq}{r^2}

0=\dfrac{k\times9\times10^{-6}}{d^2}+\dfrac{k\times6\times10^{-6}}{(1-d)^2}

\dfrac{(1-d)}{d}=\sqrt{\dfrac{6}{9}}

\dfrac{1}{d}=\dfrac{\sqrt{6}}{3}+1

\dfrac{1}{d}=1.82

d=\dfrac{1}{1.82}

d=0.55\ m

Hence, The coordinates of the point is (0,0.55).

3 0
3 years ago
Sean climbs a tower that is 71.3 m high to make a jump with a parachute. The mass of Sean plus the parachute is 81.4 kg. If U =
myrzilka [38]

Answer:

U = 56877.4 J

Explanation:

The potential energy of a body is that which it possesses because it is located at a certain height above the surface of the earth and can be calculated using the following formula:

U = mgh Formula (1)

Where:

U is the potential energy in Joules (J)

m is the mass of the body in kilograms (kg)

g is the acceleration due to gravity (m/s²)

h is the height at which the body is found from the surface of the earth in meters (m)

Data

m= 81.4 kg

g= 9.8 m/s²

h = 71.3 m

Potential energy of Sean and the parachute at the top of the tower

We replace data in the formula (1)

U = m*g*h

U = (81.4 kg)*(9.8 m/s²)*(71.3 m)

U = 56877.4 N*m

U = 56877.4 J

3 0
4 years ago
A plece of titanium has a mass of 67.5g and a volume of 15cm<br> What is the density?
sveta [45]

Answer:

4.5g/cm^3

Explanation:

Here, Mass(m)=67.5g

         Volume(v)=15cm^3

Now, According to formula,

        Density(p)=m/v

                         =67.5/15

                         =4.5g/cm^3

8 0
3 years ago
Someone please help its a simple power problem.
SOVA2 [1]
Well 200 doubled or (x2)=400 if that’s what it means
7 0
3 years ago
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