We often refer to the electricity at a typical household outlet as being 120 V. In fact, the voltage of this AC source varies; the 120 V is . We often refer to the electricity at a typical household outlet as being 120 V. In fact, the voltage of this AC source varies; the 120 V is . the minimum value of the voltage the peak value of the voltage the average value of the voltage the rms value of the voltage

Answers

Answer 1

Answer:

Explanation:

We often refer to the electricity at a typical household outlet as being 120 V. In fact, the voltage of this AC source varies; the 120 V is "the rms value of the voltage".

The rms value of voltage is given by :

$V_(rms)=(V_(pk))/(√(2))$

Where

$v_(pk)$ is the peak value of voltage

So, the correct option is (d). " rms value of voltage".

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In this example we will use pendulum motion to actually measure the acceleration of gravity on a different planet. An astronaut on the surface of Mars measures the frequency of oscillation of a simple pendulum consisting of a ball on the end of a string. He finds that the pendulum oscillates with a period of 1.5 s. But the acceleration due to gravity on Mars is less than that on earth, gMars=0.38gearth. Later, during a journey to another planet, the astronaut finds that his simple pendulum oscillates with a period of 0.92 s. What planet is he now on?SOLUTIONSET UP Each planet has a different value of the gravitational acceleration g near its surface. The astronaut can measure g at his location, and from this he can determine what planet he's on. First we use the information about Mars to find the length L of the string that the astronaut is swinging. Then we use that length to find the acceleration due to gravity on the unknown planet.

Answers

Answer:

Explanation:

Let length of the pendulum be l . The expression for time period of pendulum is as follows

T = 2π $\sqrt{(l)/(g) }$

For Mars planet ,

1.5 = $2\pi\sqrt{(l)/(.38*9.8) }$

For other planet

.92 = $2\pi\sqrt{(l)/(g_1) }$

Squiring and dividing the two equations

$(1.5^2)/(.92^2) = (g_1)/(3.8*9.8)$

$g_1 = 9.9$

The second planet appears to be earth.

Suppose a boat moves at 16.4 m/s relative to the water. If the boat is in a river with the current directed east at 2.70 m/s, what is the boat's speed relative to the ground when it is heading east, with the current, and west, against the current? (Enter your answers in m/s.)

Answers

Answer:

with the current: 19.1 m/s eastern

against the current: 13.7 m/s western

Explanation:

The boat speed relative to ground is the sum of the boat speed relative to water + the water speed relative to ground.

Suppose eastern is positive, water speed due east is 2.7m/s.

When the boat is heading east, its speed relative to water is 16.4m/s, its speed relative to ground is 16.4 + 2.7 = 19.1 m/s eastern

When the boat is heading west, its speed relative to water is -16.4m/s, its speed relative to ground is -16.4 + 2.7 = -13.7 m/s western

Determine how many wavelengths will fit into the glass container when it is a vacuum. Since the light passes through the container twice, you need to determine how many wavelengths will fit into a glass container that has a length of 2L.

Answers

Answer:

# = 2L /λ

Explanation:

For this exercise we can use a direct proportion rule. If there is 1 wave in 1 wavelength in 2L wave how many lengths are there

# = 2L /λ 1 wave

let's calculate

# = 2L /λ

we see that the longer the wavelength the fewer waves fit in the container

Assuming a vertical trajectory with no drag, derive the applicable form of the rocket equation for this application

Answers

Answer:

The vertical trajectory is governed by Ordinary Differential Equation.

Time derivatives of each state variables.

d(d)/dt = v, d(m)/dt = -d(m-fuel)/dt, d(v)/dt = F/m.

Where V is velocity positive upwards, t is time, m is mass, m-fuel is fuel mass, F is Total force, positive upwards.

Therefore,

F = -mg - D + T, If V is positive and

F = -mg + D - T, If T is negative.

D is drag and the questions gave it as zero.

Explanation:

The two sign cases in derivative equations above are required because F is defined positive up, so the drag D and thrust T can subtract or add to F depending in the sign of V . In contrast, the gravity force contribution mg is always negative. In general, F will be some function of time, and may also depend on the characteristics of the particular rocket. For example, the T component of F will become zero after all the fuel is expended, after which point the rocket will be ballistic, with only the gravity force and the aerodynamic drag force being p

In part (a), suppose that the box weighs 128 pounds, that the angle of inclination of the plane is θ = 30°, that the coefficient of sliding friction is μ = 3 /4, and that the acceleration due to air resistance is numerically equal to k m = 1 3 . Solve the differential equation in each of the three cases, assuming that the box starts from rest from the highest point 50 ft above ground. (Assume g = 32 ft/s2 and that the downward velocity is positive.)

Answers

Answer:

v(t) = 21.3t

v(t) = 5.3t

$v(t) = 48 -48 e ^{ (t)/(9)}$

Explanation:

When no sliding friction and no air resistance occurs:

$m(dv)/(dt) = mgsin \theta$

where;

$(dv)/(dt) = gsin \theta , 0 < \theta < ( \pi)/(2)$

Taking m = 3 ; the differential equation is:

$3 (dv)/(dt)= 128*(1)/(2)$

$3 (dv)/(dt)= 64$

$(dv)/(dt)= 21.3$

By Integration;

$v(t) = 21.3 t + C$

since v(0) = 0 ; Then C = 0

v(t) = 21.3t

ii)

When there is sliding friction but no air resistance ;

$m (dv)/(dt)= mg sin \theta - \mu mg cos \theta$

Taking m =3 ; the differential equation is;

$3 (dv)/(dt)=128*(1)/(2) -(√(3) )/(4)*128*(√(3) )/(4)$

$(dv)/(dt)= 5.3$

By integration; we have ;

v(t) = 5.3t

iii)

To find the differential equation for the velocity (t) of the box at time (t) with sliding friction and air resistance :

$m (dv)/(dt)= mg sin \theta - \mu mg cos \theta - kv$

The differential equation is :

= $3 (dv)/(dt)=128*(1)/(2) - ( √( 3))/(4)*128 *( √( 3))/(2)-(1)/(3)v$

= $3 (dv)/(dt)=16 -(1)/(3)v$

By integration

$v(t) = 48 + Ce ^{(t)/(9)$

Since; V(0) = 0 ; Then C = -48

$v(t) = 48 -48 e ^{ (t)/(9)}$

Instantaneous speed is...a) A speed of 1000 km/h
b) The speed attained at a particular instant in time.
c) The speed that can be reached in a particular amount of time.

PLEASE HURRY

Answers

Answer:

The speed attained at a particular instant in time.

Explanation:

Instantaneous speed is the speed attained at a particular instant in time.

It is given by :

$v=(dx)/(dt)$

It is equal to the rate of change of speed.

It can be also defined as when the speed of an object is constantly changing, the instantaneous speed is the speed of an object at a particular moment (instant) in time.

Hence, the correct option is (b).

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