The primary purpose of a switch in a circuit is to ___________.

A)either open or close a conductive path
B)change a circuit from parallel to series
C)change a circuit from series to parallel
D)store a charge for later use

Answer:

kinetic and static

Explanation:

hope it helps! ^w^

Answer:

A current of 1ma flows through a copper wire, how many electron will pass a given point in one second? 1 Coulomb = 6.24 x 10^18 electrons (or protons)/1Sec which is also equal to 1 Amp/1 Sec. 1mA is 1/1000th of 1A so only 1/1000th of 6.24 x 10^18 electrons will pass a given point in 1 Sec.

Answer:

Explanation:

F = ma so filling in:

60.0 = 4.50a and

a = 13.3 m/s/s

Answer:

13.3 pounds.

Explanation:

For a spring of constant K, with an attached object of mass M, the period can be written as:

T = 2*π*√(M/K)

Where π = 3.14

First, we know that the period is 4 seconds, then we have:

4s = (2*π)*√(M/K)

We know that if the mass is reduced by 10lb, the period becomes 2s.

Then the new mass of the object will be: (M - 10lb)

Then the period equation becomes:

2s = (2*π)*√((M-10lb)/K)

So we have two equations:

4s = (2*π)*√(M/K)

2s = (2*π)*√((M-10lb)/K)

We want to solve this for M.

First, we need to isolate K in one of the equations.

Let's isolate K in the first one:

4s = (2*π)*√(M/K)

(4s/2*π) = √(M/K)

(2s/π)^2 = M/K

K = M/(2s/π)^2 = M*(π/2s)^2

Now we can replace it in the other equation.

2s = (2*π)*√((M-10lb)/K)

First, let's simplify the equation:

2s/(2*π) = √((M-10lb)/K)

1s/π =  √((M-10lb)/K)

(1s/π)^2 =  ((M-10lb)/K

K*(1s/π)^2 = M - 10lb

Now we can use the equation: K =  M*(π/2s)^2

then we get:

K*(1s/π)^2 = M - 10lb

(M*(π/2s)^2)*(1s/π)^2 = M - 10lb

M/4 = M - 10lb

10lb = M - M/4

10lb = (3/4)*M

10lb*(4/3) = M

13.3 lb = M

Answer:

ωf = 8.8 rad/s

v = 2.2 m/s

Explanation:

We will use the third equation of motion to find the maximum angular velocity of the wheel:

[tex]2\alpha \theta = \omega_f^2 -\omega_I^2[/tex]

where,

α = angular acceleration = 6 rad/s²

θ = angular displacemnt = 1 rev = 2π rad

ωf = max. final angular velocity = ?

ωi = initial angular velocity = 1.5 rad/s

Therefore,

[tex]2(6\ rad/s^2)(2\pi\ rad)=\omega_f^2-(1.5\ rad/s)^2\\\omega_f^2=75.4\ rad/s^2+2.25\ rad/s^2\\\omega_f = \sqrt{77.65\ rad/s^2}[/tex]

ωf = 8.8 rad/s

Now, for linear velocity:

v = rω = (0.25 m)(8.8 rad/s)

v = 2.2 m/s

I guess that we want to find how much money you get each week.

We know that the job pays $8.60 per hour.

We know that you work 20 hours per week.

Then the gross pay (the total money that you earn) in a week is 20 times $8.60, or:

20*$8.60 = $172.

Now we know that your employer witholds:

10% + 7.65% + 5% = 22.65%

Then your employer withholds 22.65% of your gross pay.

if the 100% of your gross pay is $172

Then the 22.65% will be:

(22.65%/100%)*$172 = 0.2265*$172 = $38.96

This means that your employer withholds $38.96 of your weekly gross pay.

Then each week you get:

$172 - $38.96 = $133.04

If you want to learn more, you can read:

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(C)

Explanation:

Doubling the distance will make the force only 1/4 of the original value.

Answer:

x = 4 t^2 - 15 t + 25        displacement of particle

dx / dt = 8 t - 15        velocity of particle

d^2x / dt^2 = 8       acceleration of particle

If 8 t -15 = o     then t = 8 / 15

Since acceleration is a constant 8 then motion has uniform acceleratkon

Answer:

a) fem = 5.709 V,  b)  t = 0.196 s,  c)  t = 0.589 s, d)   T = 0.785 s

Explanation:

This is an exercise in Faraday's law

          fem= - N [tex]\frac{d \Phi _B}{dt}[/tex]

          fem = - N [tex]\frac{d \ (B A cos \theta)}{dt}[/tex]

The magnetic field and the area are constant

          fem = - N B A [tex]\frac{d \ cos \ \theta}{dt}[/tex]

          fem = - N B A (-sin θ)  [tex]\frac{d \theta}{dt}[/tex]

          fem = N B (π d² / 4) sin θ   w

          fem= [tex]\frac{\pi }{4}[/tex]  N B d² w sin θ

with this expression we can correspond the questions

a) the peak of the electromotive force

this hen the sine of the angle is 1

         sin θ = 1

         fem = [tex]\frac{\pi }{4}[/tex]   77  1.18  0.10² 8.0

         fem = 5.709 V

b) as the system has a constant angular velocity, we can use the angular kinematics relations

          θ = w₀ t

          t = θ/w₀

Recall that the angles are in radians, so the angle for the maximum of the sine is

           θ= π/2

           t = [tex]\frac{\pi }{2} \ \frac{1}{8}[/tex]

           t = 0.196 s

c) for the electromotive force to be negative, the sine function of being

            sin θ= -1

whereby

          θ = 3π/ 2

          t = [tex]\frac{3\pi }{2} \ \frac{1}{8}[/tex]  

          t = 0.589 s

d) This electromotive force has values ​​that change sinusoidally with an angular velocity of

          w = 8 rad / s

angular velocity and period are related

          w = 2π / T

          T = 2π / w

          T = 2π / 8

          T = 0.785 s

Answer:

V = k Q / R       potential at distance R for a charge Q

R = k Q / V

R = 9 * 10E9 * 15 * 10E-9 / 30 = 9 * 15 / 30 = 4.5 m

Note: Our equation says that if R if infinite then V must be zero.

Answer:

The correct answer is "4.26 m".

Explanation:

Given:

Wavelength,

[tex]\lambda = 436.1 \ nm[/tex]

or,

  [tex]=436.1\times 10^{-9} \ m[/tex]

Distance,

[tex]d = 0.31 \ mm[/tex]

or,

  [tex]=0.31\times 10^{-3} \ m[/tex]

Distance between the 1st and 2nd dark fringes,

[tex](y_2-y_1) = 6\times 10^{-3} \ m[/tex]

As we know,

⇒ [tex]\frac{d}{L} (y_2-y_1) = \lambda[/tex]

or,

⇒ [tex]L=\frac{d(y_2-y_1)}{\lambda}[/tex]

By substituting the values, we get

       [tex]=\frac{0.31\times 6\times 10^{-6}}{436.1\times 10^{-9}}[/tex]

       [tex]=\frac{0.31\times 6\times 10^3}{436.1}[/tex]

       [tex]=\frac{1860}{436.1}[/tex]

       [tex]=4.26 \ m[/tex]

Answer:

Explanation:

When a falcon is hovering, the force of up thrust is balanced by the weight.

When it begins to fall towards the ground, the weight acts downwards, kinetic friction is upwards, drag is upwards, normal force is upwards, thrust is upwards.

Answer:

a) The energy efficiency of the out-of-condition professor is 9.082 %.

b) The food calories needed by the well-conditioned athlete is 181.644 kilocalories.

Explanation:

a) The energy efficiency of the food metabolization ([tex]\eta[/tex]), no unit, is defined by following formula:

[tex]\eta = \frac{W}{E}\times 100\,\%[/tex] (1)

Where:

[tex]W[/tex] - Useful work, in joules.

[tex]E[/tex] - Food energy, in joules.

If we know that [tex]W = 1.90\times 10^{5}\,J[/tex] and [tex]E = 2.092\times 10^{6}\,J[/tex], the energy efficiency of the food metabolization is:

[tex]\eta = \frac{1.90\times 10^{5}\,J}{2.092\times 10^{6}\,J} \times 100\,\%[/tex]

[tex]\eta = 9.082\,\%[/tex]

The energy efficiency of the out-of-condition professor is 9.082 %.

b) If we know that [tex]W = 1.90\times 10^{5}\,J[/tex] and [tex]\eta = 25\,\%[/tex], then the quantity of food energy is:

[tex]E = \frac{W}{\eta}\times 100\,\%[/tex]

[tex]E = 1.90\times 10^{5}\,J\times \frac{100\,\%}{25\,\%}[/tex]

[tex]E = 7.60\times 10^{5}\,J[/tex]

[tex]E = 181.644\,kcal[/tex]

The food calories needed by the well-conditioned athlete is 181.644 kilocalories.

Answer:

19m

Explanation:

we have proper length L = 100m

the speed of the train v = 0.95

the speed of light is given as = 3x10⁸

length of the tunnel is given as = 50 meters

we can solve for the lenght contraction as

LX√1-v²/c²

= 100 * √1-(0.95*3x10⁸)²/(3x10⁸ )

= 31.22 metres

the train would be well seen at

50 - 31.22

= 18.78

= this is approximately 19 metres

we conclude tht the trains ends clears the ends of the tunnel by 19 meters.

thank you!

Answer:

T = 712.9 N

Explanation:

First, we will find the speed of the wave:

v = fλ

where,

v = speed of the wave = ?

f = frequency = 890 Hz

λ = wavelength = 0.1 m

Therefore,

v = (890 Hz)(0.1 m)

v = 89 m/s

Now, we will find the linear mass density of the wire:

[tex]\mu = \frac{m}{L}[/tex]

where,

μ = linear mass density of wie = ?

m = mass of wire = 90 g = 0.09 kg

L = length of wire = 1 m

Therefore,

[tex]\mu = \frac{0.09\ kg}{1\ m}[/tex]

μ = 0.09 kg/m

Now, the tension in wire (T) will be:

T = μv² = (0.09 kg/m)(89 m/s)²

T = 712.9 N

Answer:

log T = 3.72 to 3.77

Explanation:

Temperature range is

T = 5200 to 5900

Take the log

So,

log T = log 5200 to log 5900

log T = 3.72 to 3.77

Answer:

Total distance traveled by the sound wave = 2d . Thus we can conclude that the cliff should be at a distance of 877.15 metres from the source of sound to meet the required conditions

Explanation:

this is what i found on the web

Answer:

hgff

Explanation:

Answer:

The charge moves to equilibrium.

E.e = B.e.V

E is electric field force.

e is the charge.

B is magnetic field force.

V is acceleration voltage.

The velocity in the horizontal direction will not change. Then the horizontal speed will be 8.133 meters per second.

An item or particle that is propelled in a gravitational influence, such as from the crust of the Ground, and moves along a curved route while solely being affected by gravity is said to be in projectile motion.

A ball thrown horizontally from a point 24 m above the ground, strikes the ground after traveling horizontally a distance of 18 m.

The initial velocity is zero. Then the time taken to reach the ground is given as,

h = ut + 1/2at²

- 24 = 0×t + 1/2 (-9.8)t²

24 = 4.9t²

t² = 4.8979

t = 2.213 seconds

Then the horizontal speed is given as,

v = 18 / 2.213

v = 8.133 meters per second

The velocity in the horizontal direction will not change. Then the horizontal speed will be 8.133 meters per second.

Learn more about projectile motion:

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#SPJ6

Answer:

b. 1.9 Ω

Explanation:

Here is the complete question

A series LR circuit contains an emf source of 14 V having no internal resistance, a resistor, a 34 H inductor having no appreciable resistance, and a switch. If the emf across the inductor is 80% of its maximum value 4.0 s after the switch is closed, what is the resistance of the resistor? a. 1.5 ? b. 1.9 ? c. 5.0 ? d. 14 ?

Solution

The voltage across the inductor V is

[tex]V = V_{0}e^{-\frac{Rt}{L} }[/tex] where V₀ = emf of source = 14 V, R = resistance, L = inductance = 34 H and t = time

Given that V = 80% of its maximum value after 4.0 s, this implies that V = 80 % of V₀ = 0.8V₀ and t = 4.0 s

Since [tex]V = V_{0}e^{-\frac{Rt}{L} }[/tex] and V = 0.8V₀.

Since we need to find R, we make R subject of the formula, we have

[tex]V = V_{0}e^{-\frac{Rt}{L} }[/tex]

[tex]V/V_{0}= e^{-\frac{Rt}{L} }[/tex]

taking natural logarithm of both sides, we have

㏑(V/V₀) = -Rt/L

R = -L㏑(V/V₀)/t

Substituting the values of the variables into the equation, we have

R = -34㏑(0.8V₀/V₀)/4.0 s

R = -34㏑(0.8)/4.0 s

R = -34 × -0.2231/4.0 s

R = 7.587/4

R = 1.896 Ω

R ≅ 1.9 Ω

So, B is the answer

Answer:

First remember that the distance between two points (a, b) and (c, d) is given by the equation:

[tex]d = \sqrt{(a - c)^2 + (b - d)^2}[/tex]

Now let's define the position of the radar as:

(0mi, 0mi)

Then we can write the position of the plane as:

(480mi/h*t, 1mi)

where t is time in hours.

Then we can write the distance equation as:

[tex]d(t) = \sqrt{(480\frac{mi}{h}*t - 0mi)^2 + (1mi -0mi)^2 } \\\\d(t) = \sqrt{(480\frac{mi}{h}*t )^2 + (1mi)^2 }[/tex]

Now we want to get:

the rate at which the distance from the plane to the station is increasing when it is 3 mi away from the station.

So first we want to find the value of t such that:

d(3) = 3mi

We will look at the positive value of t, because at this point the plane is increasing its distance to the station.

[tex]3mi = \sqrt{(480\frac{mi}{h}*t )^2 + (1mi)^2 }\\\\(3mi)^2 = (480\frac{mi}{h}*t )^2 + (1mi)^2\\\\9mi^2 - 1mi^2 = (480\frac{mi}{h}*t )^2\\\\8mi^2 = (230,400 mi^2/h^2)*t^2\\\\\\\sqrt{\frac{8mi^2}{230,400 mi^2/h^2} } = t = 0.0059 h[/tex]

The rate of change when the plane is 3 mi away from the station is:

d'(0.0059h)

remember that:

d'(t) = dd(t)/dt

We can write:

d(t) = h( g(t) )

such that:

h(x) = √x

g(t) = (480mi/h*t)^2 + (1mi)^2

then:

d'(t) = h'(g(t))*g'(t)

This is:

[tex]d'(t) = \frac{dd(t)}{dt} = \frac{1}{2}*\frac{2*t*480mi/h}{\sqrt{(480mi/h*t)^2 + (1mi)^2} }[/tex]

The rate of change at t = 0.0059h is then:

[tex]d'(0.0059h) = \frac{1}{2}*\frac{2*0.0059h*(480mi/h)^2}{\sqrt{(480mi/h*0.0059h)^2 + (1mi)^2} } =452.6 mi/h^2[/tex]

A particle, mass 0.25 kg is at a position (-7i + 7j + 5k) m, has a velocity (6i - j + 4k) m/s, and is subject to a force (-5i + 0j - k) N. What is the magnitude of the torque on the particle about the origin?

47.94Nm

The torque (τ) on a particle subject to a force (represented as force vector F) at a position (represented as position vector r) about the origin is given by the cross product of the position vector r for the point of application of a force and the force F. i.e

τ = r x F

Given:

r = (-7i + 7j + 5k) m

F = (-5i + 0j - k) N

                    |   i             j              k    |

r x F  =         |   -7            7              5  |

                    |  -5           0              -1   |

r x F  =       i(-7 - 0) - j(7+25) + k(0+35)

r x F  =       i(-7) - j(32) + k(35)

r x F  =       -7i - 32j + 35k

Therefore the torque τ = -7i - 32j + 35k

The magnitude of the torque is therefore;

|τ| = [tex]\sqrt{(-7)^2 + (-32)^2 + (35)^2}[/tex]

|τ| = [tex]\sqrt{49 + 1024 + 1225}[/tex]

|τ| = [tex]\sqrt{2298}[/tex]

|τ| = 47.94Nm

The magnitude of the torque on the particle about the origin is 47.94Nm

Answer:

49 N

Explanation:

Given,

Mass ( m ) = 50 kg

To find : Weight ( W ) = ?

Take the value of acceleration due to gravity as 9.8 m/s^2

Formula : -

W = mg

W = 50 x 9.8

W = 49 N

Answer:

Accuracy refers to how close a measurement is to the true or accepted value. ... Precision is independent of accuracy. That means it is possible to be very precise but not very accurate, and it is also possible to be accurate without being precise. The best quality scientific observations are both accurate and precise.

Answer:

True

Explanation:

This is because of the point where the forces are applied by our muscles and

the angle they have about the bones. Take for example the  diagram I uploaded.

If we do a free body diagram and a sum of torques, we would get that:

[tex]F_{muscle}sin \theta r1 - mg r2 = 0[/tex]

In this case, mg is the same in magnitude as the force made by the hand to hold the ball, so:

[tex]F_{muscle}sin \theta r_{1} - F_{hand} r_{2} = 0[/tex]

If we solve the equation for the force of the muscle we would get that:

[tex]F_{muscle}=\frac{F_{hand}r_{2}}{r_{1}sin \theta}[/tex]

Since r2 is greater than r1 and the sin function can only return values that are less than 1, this means that the force of the muscle is much greater than the force used by the hand to hold the weight.

Let's use some standard values to prove this, let's say that r1=10cm, r2=35cm and theta=60 degrees. When inputing the values into the equation we get:

[tex]F_{muscle}=\frac{F_{hand}(35cm)}{(10cm)sin (60^{o})}[/tex]

which yields:

[tex]F_{muscle}=4.04 F_{hand}[/tex]

so in this example, the force made by the muscle is 4 times as big as the force exerted by the hand.

Answer:

Mass, M = 1000 kg

Speed, v = 90 km/h = 25 m/s

time, t = 6 sec.

Distance:

[tex]{ \tt{distance = speed \times time }} \\ { \tt{distance = 25 \times 6}} \\ { \tt{distance = 150 \: m}}[/tex]

Force:

[tex]{ \tt{force = mass \times acceleration}} \\ { \bf{but \: for \: acceleration : }} \\ from \: second \: equation \: of \: motion : \\ { \bf{s = ut + \frac{1}{2} {at}^{2} }} \\ \\ { \tt{150 = (0 \times 6) + ( \frac{1}{2} \times a \times {6}^{2} ) }} \\ \\ { \tt{acceleration = 8.33 \: {ms}^{ - 2} }} \\ \\ { \tt{force = 1000 \times 8.33}} \\ { \tt{force = 8333.3 \: newtons}}[/tex]

Answer:

(a) 1504 J

(b) - 884.35 J

(c) 0 J

(d) 0 J

Explanation:

Mass, m = 47 kg

displacement, s = 8 m

Force, F = 188 N

Coefficient of friction = 0.24

(a) Work done by applied force

W = F s = 188 x 8 = 1504 J

(b) Work done by the friction force

W' = - 0.24 x 47 x 9.8 x 8 = - 884.35 J

(c) Work done by the gravitational force

W''= m g s cos 90 = 0 J

(d) Work done by the normal force

W''' = m g scos 90 = 0 J

Explanation:

The energy stored in a capacitor is given by

[tex]U = \dfrac{1}{2}QV = \dfrac{Q^2}{2C}[/tex]

In the case of a parallel plate capacitor, the capacitance C is given by

[tex]C = \dfrac{\epsilon_0A}{d}[/tex]

so we can rewrite the expression for the energy as

[tex]U = \dfrac{Q^2d}{2\epsilon_0 A}[/tex]

Increasing the charge from [tex]3\:\mu\text{C}\:\text{to}\:9\:\mu\text{C}[/tex] means that you're tripling the charge. The same thing is true when you increase the distance from 1.8 mm to 5.4 mm, i.e., you triple the separation distance. So the new energy U' is given by

[tex]U' = \dfrac{Q'^2d'}{2\epsilon_0 A}[/tex]

[tex]\:\:\:\:\:\:= \dfrac{(3Q)^2(3d)}{2\epsilon_0 A}[/tex]

[tex]\:\:\:\:\:\:= 27\left(\dfrac{Q^2d}{2\epsilon_0 A}\right)[/tex]

[tex]\:\:\:\:\:\:= 27U[/tex]

As we can see, tripling both the charge and the separation distance result in the 27-fold increase in its stored energy U.