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riadik2000 [5.3K]

2 years ago

13

A wave hits an object as shown. A vertical line with Medium 1 to the left and Medium 2 to the right. An arrow from the left reac

hes the line. An arrow starts from the same point on the line and goes right at a different angle. Which kind of wave interaction is shown?

2 answers:

nika2105 [10]2 years ago

7 0

Answer:

refraction

Explanation:

Anvisha [2.4K]2 years ago

3 0

Answer:refraction

Explanation:

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A 50 Q resistor in a circuit has a current flowing through it of 2.0 A. What is

Alex17521 [72]

Hello!

We can use the following equation for calculating power dissipated by a resistor:
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7 0

1 year ago

Two particles, one with charge -5.45 × 10^-6 C and one with charge 4.39 × 10^-6 C, are 0.0209 meters apart. What is the magnitud

Levart [38]

Explanation:

Given that,

Charge 1, $q_1=-5.45\times 10^{-6}\ C$

Charge 2, $q_2=4.39\times 10^{-6}\ C$

Distance between charges, r = 0.0209 m

1. The electric force is given by :

$F=k\dfrac{q_1q_2}{r^2}$

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F = -492.95 N

2. Distance between two identical charges, $r=0.0209\ m$

Electric force is given by :

$F=\dfrac{kq_3^2}{r^2}$

$q_3=\sqrt{\dfrac{Fr^2}{k}}$

$q_3=\sqrt{\dfrac{492.95\times (0.0209)^2}{9\times 10^9}}$

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3 0

2 years ago

A vertical cylindrical tank 10 ft in diameter, has an inflow line of 0.3 ft inside diameter and an outflow line of 0.4 ft inside

neonofarm [45]

Answer:

$\frac{dh}{dt} = 1.3 \times 10^{-3} \frac{ft}{s}$ , level is rising.

Explanation:

Since liquid water is a incompresible fluid, density can be eliminated of the equation of Mass Conservation, which is simplified as follows:

$\dot V_{in} - \dot V_{out} = \frac{dV_{tank}}{dt}$

$\frac{\pi}{4}\cdot D_{in}^2 \cdot v_{in}-\frac{\pi}{4}\cdot D_{out}^2 \cdot v_{out}= \frac{\pi}{4}\cdot D_{tank}^{2} \cdot \frac{dh}{dt} \\D_{in}^2 \cdot v_{in} - D_{out}^2 \cdot v_{out} = D_{tank}^{2} \cdot \frac{dh}{dt} \\\frac{dh}{dt} = \frac{D_{in}^2 \cdot v_{in} - D_{out}^2 \cdot v_{out}}{D_{tank}^{2}}$

By replacing all known variables:

$\frac{dh}{dt} = \frac{(0.3 ft)^{2}\cdot (5 \frac{ft}{s} ) - (0.4 ft)^{2} \cdot (2 \frac{ft}{s} )}{(10 ft)^{2}}\\\frac{dh}{dt} = 1.3 \times 10^{-3} \frac{ft}{s}$

The positive sign of the rate of change of the tank level indicates a rising behaviour.

6 0

3 years ago