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3 years ago

In transverse waves, the medium moves perpendicular to the direction of energy transport? True or false?

Physics

1 answer:

saul85 [17]3 years ago

8 0

Answer:

This statement is true

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By what means can the internal energy of a closed system increase?.

kirill115 [55]

Answer:

when work is done on the system or heat comes into the system

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2 years ago

Read 2 more answers

In an RLC series circuit that includes a source of alternating current operating at fixed frequency and voltage, the resistance

maw [93]

Answer:

Capacitive Reactance is 4 times of resistance

Solution:

As per the question:

R = $X_{L} = j\omega L = 2\pi fL$

where

R = resistance

$X_{L} = Inductive Reactance$

f = fixed frequency

Now,

For a parallel plate capacitor, capacitance, C:

$C = \frac{\epsilon_{o}A}{x}$

where

x = separation between the parallel plates

Thus

C ∝ $\frac{1}{x}$

Now, if the distance reduces to one-third:

Capacitance becomes 3 times of the initial capacitace, i.e., x' = 3x, then C' = 3C and hence Current, I becomes 3I.

Also,

$Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}$

Also,

Z ∝ I

Therefore,

$\frac{Z}{I} = \frac{Z'}{I'}$

$\frac{\sqrt{R^{2} + (R - X_{C})^{2}}}{3I} = \frac{\sqrt{R^{2} + (R - \frac{X_{C}}{3})^{2}}}{I}$

${R^{2} + (R - X_{C})^{2}} = 9({R^{2} + (R - \frac{X_{C}}{3})^{2}})$

${R^{2} + R^{2} + X_{C}^{2} - 2RX_{C} = 9({R^{2} + R^{2} + \frac{X_{C}^{2}}{9} - 2RX_{C})$

Solving the above eqn:

$X_{C} = 4R$

6 0

2 years ago

Where does most star formation occur in the milky way galaxy?

k0ka [10]

Answer:

Star formation occurs most rapidly in the spiral arms, where the density of interstellar matter is highest.

4 0

1 year ago

Cardiorespiratory fitness improves the efficiency of the cardiovascular and the respiratory systems in delivering oxygen to the

algol [13]

Your answer should be T

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3 years ago

A 873-kg (1930-lb) dragster, starting from rest completes a 401.4-m (0.2509-mile) run in 4.945 s. If the car had a constant acce

Delvig [45]

To solve this problem it is necessary to apply the kinematic equations of motion.

By definition we know that the position of a body is given by

$x=x_0+v_0t+at^2$

Where

$x_0 =$ Initial position

$v_0 =$ Initial velocity

a = Acceleration

t= time

And the velocity can be expressed as,

$v_f = v_0 + at$

Where,

$v_f = Final velocity$

For our case we have that there is neither initial position nor initial velocity, then

$x= at^2$

With our values we have $x = 401.4m, t=4.945s$ , rearranging to find a,

$a=\frac{x}{t^2}$

$a = \frac{ 401.4}{4.945^2}$

$a = 16.41m/s^2$

Therefore the final velocity would be

$v_f = v_0 + at$

$v_f = 0 + (16.41)(4.945)$

$v_f = 81.14m/s$

Therefore the final velocity is 81.14m/s

8 0

2 years ago