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2 years ago

1.

Physics

2 answers:

Jet001 [13]2 years ago

4 0

Answer:

1. high frequency and high energy.

2. short wavelengths and high frequencies.

3. less energy and long wavelengths.

4. high frequencies and high energy

5. X-rays

6. infrared waves

7. radio waves.

8. Gamma Rays

9. Microwaves

10. infrared waves

sweet [91]2 years ago

3 0

Answer: 1: High frequency and high energy. 2: Short wavelengths and high frequencies. 3: Less energy and long wavelengths. 4: High frequencies and high energy. 5: X-rays. 6: Infrared waves. 7: Radio waves. 8: Gamma Rays. 9: Microwaves. 10: Infrared waves.

Explanation: 1: High frequency and high energy. 2: Short wavelengths and high frequencies. 3: Less energy and long wavelengths. 4: High frequencies and high energy. 5: X-rays. 6: Infrared waves. 7: Radio waves. 8: Gamma Rays. 9: Microwaves. 10: Infrared waves.

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A large uniform chain is hanging from the ceiling, supporting a block of mass 46 kg. The mass of the chain itself is 19 kg, and

docker41 [41]

Answer:

T = 451.26 N

Explanation:

It is given that,

The mass of block, m = 46 kg

Mass of the chain, m' = 19 kg

Length of the chain, l = 1.9 m

Let T is the the tension in the chain at the point where the chain is supporting the block. It is clearly equal to the product of mass and acceleration.

$T=mg$

$T=46\ kg\times 9.81\ m/s^2$

T = 451.26 N

So, the tension in the chain at the point where the chain is supporting the block is 451.26 N. Hence, this is the required solution.

4 0

2 years ago

Four long wires are each carrying 6.0 A. The wires are located

Firdavs [7]

Answer:

$B_T=2.0*10^-5[-\hat{i}+\hat{j}]T$

Explanation:

To find the magnitude of the magnetic field, you use the following formula for the calculation of the magnetic field generated by a current in a wire:

$B=\frac{\mu_oI}{2\pi r}$

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

I: current = 6.0 A

r: distance to the wire in which magnetic field is measured

In this case, you have four wires at corners of a square of length 9.0cm = 0.09m

You calculate the magnetic field in one corner. Then, you have to sum the contribution of all magnetic field generated by the other three wires, in the other corners. Furthermore, you have to take into account the direction of such magnetic fields. The direction of the magnetic field is given by the right-hand side rule.

If you assume that the magnetic field is measured in the up-right corner of the square, the wire to the left generates a magnetic field (in the corner in which you measure B) with direction upward (+ j), the wire down (down-right) generates a magnetic field with direction to the left (- i) and the third wire generates a magnetic field with a direction that is 45° over the horizontal in the left direction (you can notice that in the image attached below). The total magnetic field will be:

$B_T=B_1+B_2+B_3\\\\B_{T}=\frac{\mu_o I_1}{2\pi r_1}\hat{j}-\frac{\mu_o I_2}{2\pi r_2}\hat{i}+\frac{\mu_o I_3}{2\pi r_3}[-cos45\hat{i}+sin45\hat{j}]$

I1 = I2 = I3 = 6.0A

r1 = 0.09m

r2 = 0.09m

$r_3=\sqrt{(0.09)^2+(0.09)^2}m=0.127m$

Then you have:

$B_T=\frac{\mu_o I}{2\pi}[(-\frac{1}{r_2}-\frac{cos45}{r_3})\hat{i}+(\frac{1}{r_1}+\frac{sin45}{r_3})\hat{j}}]\\\\B_T=\frac{(4\pi*10^{-7}T/A)(6.0A)}{2\pi}[(-\frac{1}{0.09m}-\frac{cos45}{0.127m})\hat{i}+(\frac{1}{0.09m}+\frac{sin45}{0.127m})]\\\\B_T=\frac{(4\pi*10^{-7}T/A)(6.0A)}{2\pi}[-16.67\hat{i}+16.67\hat{j}]\\\\B_T=2.0*10^-5[-\hat{i}+\hat{j}]T$