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4 years ago

In right ΔPQR, sin P = 3 over 5 , and cos P = 4 over 5. What is tan P?

1 answer:

5 0

Not sure if you learned S=O/H,C= A/H, T=O/A, but Sin is equal to opposite over hypotenuse, so draw a triangle, and label those angles (opposite= 3 and hypotenuse=5). Also, Cos is equal to adjacent over hypotenuse, so label those as well (you already have hypotenuse from Sin, and the Adjacent angle = 4). Therefore, Tan which equals opposite over adjacent is 3/4

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It may seem strange that the selected velocity does not depend on either the mass or the charge of the particle. (For example, w

Charra [1.4K]

Answer:

b) q large and m small

Explanation:

q is large and m is small

We'll express it as :

q > m

As we know the formula:

F = Eq

And we also know that :

F = Bqv

F = $\frac{mv^{2} }{r}$

Bqv = $\frac{mv^{2} }{r}$

or Eq = $\frac{mv^{2} }{r}$

Assume that you want a velocity selector that will allow particles of velocity v⃗ to pass straight through without deflection while also providing the best possible velocity resolution. You set the electric and magnetic fields to select the velocity v⃗ . To obtain the best possible velocity resolution (the narrowest distribution of velocities of the transmitted particles) you would want to use particles with q large and m small.

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3 years ago

A square loop of side 7 cm is placed with the nearest side 2 cm from a long wire carrying a current that varies with time at a c

ioda

Answer:

Explanation:

side of the square loop, a = 7 cm

distance of the nearest side from long wire, r = 2 cm = 0.02 m

di/dt = 9 A/s

Integrate on both the sides

$\int _{0}^{i}di =9\int _{0}^{t}dt$

i = 9t

(a) The magnetic field due to the current carrying wire at a distance r is given by

$B = \frac{\mu_{0}i}{2\pi r}$

$B = \frac{\mu_{0}\times 9t}{2\pi r}$

(b)

Magnetic flux,

$\phi=\int B\times a dr$

$\phi=\int \frac{\mu_{0}\times 9t}{2\pi r}\times a dr$

$\phi=\frac{\mu_{0}\times 9t\times a}{2\pi}\times ln\left ( \frac{2 + 7}{2} \right )$

$\phi=\frac{\mu_{0}\times 9t\times 0.07}{2\pi}\times ln(4.5)$

$\phi = 1.89 \times 10^{-7}t$

(c)

R = 3 ohm

$e = -\frac{d\phi}{dt}$

magnitude of voltage is

e = 1.89 x 10^-7 V

induced current, i = e / R = (1.89 x 10^-7) / 3

i = 6.3 x 10^-8 A

8 0

3 years ago

LVULAN

3241004551 [841]

For this case we have that by definition, the kinetic energy is given by the following formula:

$k= \frac {1} {2} * m * v ^ 2$

Where:

m: It is the mass

v: It is the velocity

According to the data we have to:

$m = 100 \ kg\\v = 9 \frac {m} {s}$

Substituting the values we have:

$k = \frac {1} {2} * (100) * (9) ^ 2\\k = \frac {1} {2} * (100) * 81\\k = 50 * 81\\k = 4050$

finally, the kinetic energy is $4050 \ J$

Answer:

Option A

7 0

4 years ago

HAaaaaaaaaaaaLPPP thank you.

lara31 [8.8K]

15 guesting i think it not right

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3 years ago

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3 years ago

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