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chubhunter [2.5K]
3 years ago
7

Someone help please !!?

Physics
1 answer:
Tanzania [10]3 years ago
6 0
Mmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm
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A block is sitting on a platform 20 m high. It weighs 50 kg. How much potential energy does it contain?
kifflom [539]

Answer:

9806.65 Joules

Explanation:

5 0
3 years ago
A wire with a current of 3.40 A is to be formed into a circular loop of one turn. If the required value of the magnetic field at
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Answer:

0.107 m

Explanation:

The magnetic field at the center of a current-carrying loop is given by

B=\frac{\mu_0 I}{2r}

where

\mu_0 is the vacuum permeability

I is the current

r is the radius of the loop

In this problem we have

I = 3.40 A is the current in the loop

B=20 \mu T=20\cdot 10^{-6}T is the magnetic field at the centre of the loop

So, solving the formula for r we find

r=\frac{\mu_0 I}{2B}=\frac{(12.56\cdot 10^{-7} H/m)(3.40 A)}{2(20\cdot 10^{-6} T)}=0.107 m

7 0
3 years ago
A) explain how the data matches each line on the graph (give name and line letter for each)
stepladder [879]
The answer you have is right good job 12
3 0
2 years ago
Monochromatic light of wavelength λ=136.8μ m is shone at normal incidence through a thin film of thickness t resting atop a full
inn [45]

Answer:

1.8 × 10⁻⁸ Hm

Explanation:

Given that:

The refractive index of the film = 19

The wavelength of the light = 136.8 μ m

The thickness can be calculated by using the formula shown below as:

Thickness=\frac {\lambda}{4\times n}

Where, n is the refractive index of the film

{\lambda} is the wavelength

So, thickness is:

Thickness=\frac {136.8\ \mu\ m}{4\times 19}

Thickness = 1.8 μ m

Since,

1 μ m = 10⁻⁸ Hm

So,

Thickness = 1.8 × 10⁻⁸ Hm

7 0
3 years ago
A uniform 2.50m ladder of mass 7.30kg is leaning against a vertical wall while making an angle of 51.0degree with the floor. A w
Zigmanuir [339]

Answer:

19.95 J

Explanation:

The center of mass of the ladder is initially at a height of:

h_1=\frac{L}{2}sin\theta

The center of mass of the ladder ends at a height of:

h_2= \frac{L}{2}sin90 =L/2

So, the work done is equal to the change in potential energy which is:

W = PE = mg(h_2-h_1)

now h_2-h_1= 1-sin\theta

therefore

W = [mgL/2]×[1 - sin(theta)]

W = [(7.30)(9.81)(2.50)/2]×[1-sin(51°)]

solving this we get

W = 19.95 J

8 0
3 years ago
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