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IceJOKER [234]
3 years ago
6

Es frecuente que en las instalaciones eléctricas domésticas se utilice alambre de cobre de 2.05 mm de diámetro. Determine la res

istencia de un alambre de ese tipo con longitud de 24.0 m
Physics
1 answer:
Tasya [4]3 years ago
7 0

Answer:

The resistance is 0.124 ohm.

Explanation:

It is common for domestic electrical installations to use copper wire with a diameter of 2.05 mm. Determine the resistance of such a wire with a length of 24.0 m.

diameter, d = 2.05 mm

radius, r = 1.025 mm

Length, L = 24 m

resistivity of copper = 1.7 x 10^-8 ohm m

Let the resistance is R.

R =\rho \frac{L}{A}\\\\R = \frac {1.7\times10^{-8}\times 24}{3.14\times1.025\times1.025\times 10^{-6}}\\\\R = 0.124 ohm

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Once​ Kate's kite reaches a height of 52 ft ​(above her​ hands), it rises no higher but drifts due east in a wind blowing 6 ft d
Andru [333]

Answer:

5.213ft

Explanation:

Z² = x² + y²

x = √(z² - y²)

y = 52ft, dx = 6ft, z = 105ft, dz = ?

d(z² = x² + y²)

2zdz = 2xdx

dz = xdx/z

But x = √(z² - y²)

dz = √(z² - y²)/z * dx

dz = [√(105² - 52²)/105] * 6

dz = √(8521)/ 17.5

dz = 5.213ft

7 0
3 years ago
A plane on the ground will weigh:
kenny6666 [7]

On ground weight of plane will be measured as its actual weight which will be given as

W = mg

now when plane is in air its weight is measured as combined effect of earth gravitational force and buoyancy force both

so weight in air will be given as

W_{air} = mg - F_b

here since net effect due to air is opposite to the direction of weight so in air the plane weight will be measure less than its weight on ground.

so answer will be

A)more than a plane in the air

3 0
3 years ago
Read 2 more answers
A cat walks along a plank with mass M= 6.00 kg. The plank is supported by two sawhorses. The center of mass of the plank is a di
dimulka [17.4K]

Answer:

d₂ = 1.466 m

Explanation:

In this case we must use the rotational equilibrium equations

        Στ = 0

        τ = F r

we must set a reference system, we use with origin at the easel B and an axis parallel to the plank , we will use that the counterclockwise ratio is positive

      + W d₁ - w_cat d₂ = 0

      d₂ = W / w d₁

      d₂ = M /m d₁

      d₂ = 5.00 /2.9    0.850

      d₂ = 1.466 m

6 0
4 years ago
If the period of the pendulum is tripled, what is the length of the string increased by?
Nitella [24]
The period of the pendulum doesn't determine the length of the string. 
It's the other way around.

The period of the pendulum is proportional to the square root of its length.
So if you want to triple the period, you have to make the string nine times
as long as it is now.
7 0
3 years ago
If a certain brand of solar panels is rated at a value of 1.50 KW/m2 , and a person needed to generate 2.50 MJ in an hour, what
alisha [4.7K]

Answer:

A=0.462\ m^2

Explanation:

Power rating of a solar panel is 1.50 KW/m²

It generates 2.50 MJ in an hour.

We need to find the area of this type of solar panel would be needed. The power pertaining to generate this energy is given by :

P=\dfrac{2.5\times 10^6}{1\ h}\\\\P=\dfrac{2.5\times 10^6}{3600\ s}\\\\P=694.44\ W

Let A be the area of the solar panel. It is calculated as follows :

\dfrac{P}{A}=1.5\times 10^3\\\\A=\dfrac{694.44}{1.5\times 10^3}\\A=0.462\ m^2

So, the required area of the solar panel is 0.462\ m^2.

4 0
3 years ago
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