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3 years ago

15

If a light ray is aimed so that it bounces off a metal drain in a pool, the light ray will reflect off the drain according to __

_______.
A. Snell's law
B. the law of reflection
C. the law of refraction
D. the law of incidence

1 answer:

professor190 [17]3 years ago

8 0

B.the law of reflection

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A system is a group of objects that’s analyzed as one unit. Consider a car moving along a road that has a flat section and a hil

vladimir1956 [14]

Answer:

a) Em= K +U, b) Em= K

Explanation:

The system in this case is formed by the mobilizes and the hill.

Let's write the expressions correctly and completely.

a) When the car moves in the path, the mechanical energy is the siua of the kinetic energy of the car and the potential energy of the car when going up the hill.

Em = K + U

be) when the car moves in the flat part all the mechanical energy is formed by its kinetic energy that is calculated with the mass and speed of the car

Em = K

c) When the car goes up the hill the energy the mechanical energy is conserved, but part of the kinetic energy is transformed into potential energy.

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3 years ago

The volume of water in the Pacific Ocean is about 7.0 × 10 8 km 3 . The density of seawater is about 1030 kg/m3. (a) Determine t

Novay_Z [31]

To solve the problem it is necessary to consider the concepts related to Potential Energy and Kinetic Energy.

Potential Energy because of a planet would be given by the equation,

$PE=\frac{GMm}{r}$

Where,

G = Gravitational Universal Constant

M = Mass of Ocean

M = Mass of Moon

r = Radius

From the data given we can calculate the mass of the ocean water through the relationship of density and volume, then,

$m = \rho V$

$m = (1030Kg/m^3)(7*10^8m^3)$

$m = 7.210*10^{11}Kg$

It is necessary to define the two radii, when the ocean is far from the moon and when it is facing.

When it is far away, it will be the total diameter from the center of the earth to the center of the moon.

$r_1 = 3.84*10^8 + 6.4*10^6 = 3.904*10^8m$

When it's near, it will be the distance from the center of the earth to the center of the moon minus the radius,

$r_2 = 3.84*10^8-6.4*10^6 - 3.776*10^8m$

PART A) Potential energy when the ocean is at its furthest point to the moon,

$PE_1 = \frac{GMm}{r_1}$

$PE_1 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.904*10^8}$

$PE_1 = 9.05*10^{15}J$

PART B) Potential energy when the ocean is at its closest point to the moon

$PE_2 = \frac{GMm}{r_2}$

$PE_2 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.776*10^8}$

$PE_2 = 9.361*10^{15}J$

PART C) The maximum speed. This can be calculated through the conservation of energy, where,

$\Delta KE = \Delta PE$

$\frac{1}{2}mv^2 = PE_2-PE_1$

$v=\sqrt{2(PE_2-PE_1)/m}$

$v = \sqrt{\frac{2*(9.361*10^{15}-9.05*10^{15})}{7.210*10^{11}}}$

$v = 29.4m/s$

8 0

4 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:

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

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4 years ago

What percentage of the nations demand for electricity could be produce in the future by wind energy?

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