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2 years ago

After striking both mirrors, at what angle relative to the incoming ray does the outgoing ray emerge?

1 answer:

4 0

The appropriate response is Zero degrees. The beam will leave the two mirrors along a way parallel to the one it came in on. This is the guideline of the corner reflector, which is frequently utilized as a radar target. Take note of that the corner reflector utilizes three reflecting surfaces (that are set up at 90o from each other) rather than the two like are being utilized here. Wikipedia has a truly awesome drawing that shows this two-dimentional issue pleasantly. A moment connection is given to the article on the corner reflector and the 3-D angles.

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Anyone knows how to do this? I think it’s 0 because it’s not moving————

dedylja [7]

Answer:

1 m/s

Explanation:

2 to 3 second interval would be zero.

3 to 4 is 2 m/s

so average is 1 m/s

8 0

2 years ago

Which of the following controls the loudness of a sound?

GarryVolchara [31]

The answer is amplitude hope you get it right

8 0

3 years ago

A ship travels north at 15 km/h for 5 h. What is the ship's displacement in km?

aliya0001 [1]

So you would do 15*5 to get 75

3 0

3 years ago

Read 2 more answers

A point charge of +3 C is located at the origin of a coordinate system and a second point charge of -6 C is at x = 1.0 m. At w

Butoxors [25]

Answer:

The point at which the electrical potential is zero is x = +0.33 m.

Explanation:

By definition the electrical potential is:

$V_{E} = \frac{K*q}{r}$

Where:

K: is Coulomb's constant = 9x10⁹ N*m²/C²

q: is the charge

r: is the distance

The point at which the electrical potential is zero can be calculated as follows:

$V_{1} + V_{2} = 0$

$K(\frac{q_{1}}{r_{1}} + \frac{q_{2}}{r_{2}}) = 0$ (1)

q₁ is the first charge = +3 mC

r₁ is the distance from the point to the first charge

q₂ is the first charge = -6 mC

r₂ is the distance from the point to the second charge

By replacing r₁ = 1 - r₂ into equation (1) we have:

$K(\frac{q_{1}}{1 - r_{2}} + \frac{q_{2}}{r_{2}}) = 0$ (2)

By solving equation (2) for r₂:

$r_{2} = \frac{q_{1}}{q_{1} - q_{2}} = \frac{3 mC}{3 mC - (-6 mC)} = +0.33 m$

Therefore, the point at which the electrical potential is zero is x = +0.33 m.

I hope it helps you!

8 0

3 years ago

I need help with this question I’m not sure where to even begin

Strike441 [17]

We will have the following:

a. We determine the tension force of T2 as follows:

We know that the system must be at equilibrium on the horizontal axis:

$\sum F_x=T_1cos(42.5)+T_2cos(36.5)=0$

So:

$\begin{gathered} T_2cos(36.5)=-(1235N)cos(42.5)\Rightarrow T_2=-\frac{(1235N)cos(42.5)}{cos(36.5)} \\ \\ \Rightarrow T_2=1132.711003...N\Rightarrow T_2\approx1132.7N \end{gathered}$

So, the value of T2 is approximately 1132.7 N.

b. We will determine the torques created by T1 and T2 as follows:

T1:

$\tau_{T1}=(10m)(1235N)sin(42.5)\Rightarrow\tau_{T1}\approx8343.5N\ast m$

T2:

$\tau_{T2}=(10m)(1132.7N)sin(36.5)\Rightarrow\tau_{T2}\approx6737.6N\ast m$

So the torques of T1 and T2 on the base are approximately 8343.5 N*m and 6737.6 N*m respectively.

c. The torques around that axis generated by the normal force and the weight are both 0 N*, since they are parallel to the axis.

d. We will determine the angular acceleration as follows:

$\begin{gathered} \alpha=\frac{\tau_{T1}}{I}\Rightarrow\alpha=\frac{\tau_{T1}}{(1/3mL^2)} \\ \\ \Rightarrow\alpha=\frac{(8343.5N\ast m)}{(1/3(200kg)(10m)^2)}\Rightarrow\alpha=1.251525rad/s^2 \\ \\ \Rightarrow\alpha\approx1.25rad/s^2 \end{gathered}$

So, the angular acceleration is approximately 1.25 radians/ s^2.

6 0

1 year ago