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PIT_PIT [208]
1 year ago
15

The capacitors in each circuit are fully charged before the switch is closed. Rank, from longest to shortest, the length of time

the bulbs (resistors) stay lit in each circuit.
Physics
1 answer:
8_murik_8 [283]1 year ago
3 0

The concept used in this is circuit analysis using the simplification of resistors and capacitors.

Explanation:

The time constant for each of the circuits in figure A, B, C, D and E. Therefore, rank the length of time the bulbs stay lit from longest to shortest by using the value of time constant for each circuit. The rank of the time constant of the circuit is C > A = E > B > DC  > A = E > B > D.

Capacitance is the central concept in electrostatics and constructed devices called capacitors are essential elements of electronic circuits.

If more charge is placed on the conductor the voltage increases proportionately. The ratio of the charge to the voltage is called the capacitance C of the conductor C= q/v.

The resistance increases if you add resistors in series and decreases if you add them in parallel. on the other hand the capacitors increases if it is added parallel and decreases if added in series. hence the circuit longest time constant takes longest time to discharge.

You might be interested in
Suppose Earth's mass increased but Earth's diame-
navik [9.2K]

Answer: It would increase.

Explanation:

The equation for determining the force of the gravitational pull between any two objects is:

F = G \frac{m1m2}{r^2}

Where G is the universal gravitational constant, m1 is the mass of one body, m2 is the mass of the other body, and r^2 is the distance between the two objects' centers squared.

Assuming the Earth's mass but not its diameter increased, in the equation above m1 (the term usually indicative of the object of larger mass) would increase, while the r^2 would not.

Thus, it goes without saying that, with some simple reasoning about fractions, an increasing numerator over a constant denominator would result in a larger number to multiply by G, thus also meaning a larger gravitational strength between Earth and whatever other object is of interest.

7 0
2 years ago
A skateboarder with mass ms = 54 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then
Elan Coil [88]

Answer:

A) W_{ff} =-744.12J

B) F_f=-W_{ff}*sin\theta /hy = 112.75N

C) F_{f2}=207.58N

Explanation:

This question is incomplete. The full question was:

<em>A skateboarder with mass ms = 54 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then jumps on his skateboard and descends down the ramp. His speed at the bottom of the ramp is vf = 6.2 m/s.  </em>

<em>Part (a) Write an expression for the work, Wf, done by the friction force between the ramp and the skateboarder in terms of the variables given in the problem statement.  </em>

<em>Part (b) The ramp makes an angle θ with the ground, where θ = 30°. Write an expression for the magnitude of the friction force, fr, between the ramp and the skateboarder.  </em>

<em>Part (c) When the skateboarder reaches the bottom of the ramp, he continues moving with the speed vf onto a flat surface covered with grass. The friction between the grass and the skateboarder brings him to a complete stop after 5.00 m. Calculate the magnitude of the friction force, Fgrass in newtons, between the skateboarder and the grass.</em>

For part A), we make a balance of energy to calculate the work done by the friction force:

W_{ff}=\Delta E

W_{ff}=1/2*m*vf^2-m*g*hy

W_{ff}=-744.12J

For part B), we use our previous value for the work:

W_{ff}=-F_f*(hy/sin\theta)   Solving for friction force:

F_f=-W_{ff}*sin\theta /hy

F_f=112.75N

For part C), we first calculate the acceleration by kinematics and then calculate the module of friction force by dynamics:

Vf^2=Vo^2+2*a*d

Solving for a:

a=-3.844m/s^2

Now, by dynamics:

|F_f|=|m*a|

|F_f|=207.58N

8 0
1 year ago
jesse is swinging miguel in a circle at a tangential speed of 3.50 m/s. if the radius of the circle is 0.600 m and miguel has a
Morgarella [4.7K]
Centripetal acceleration = (speed)² / (radius) .

Force = (mass) · (acceleration)

Centripetal force = (mass) · (speed)² / (radius) .

                             = (11 kg) · (3.5 m/s)² / (0.6 m)

                             = (11 kg) · (12.25 m²/s²) / (0.6 m)

                             =  (11 · 12.25) / 0.6  kg-m/s²

                             =      224.58 newtons.    (about 50.5 pounds)

That's the tension in Miguel's arm or leg or whatever part of his body
Jesse is swinging him by.  It's the centripetal force that's needed in
order to swing 11 kg in a circle with a radius of 0.6 meter, at 3.5
meters/second.  If the force is less than that, then the mass has to
either swing slower or else move out to follow a bigger circle.
6 0
2 years ago
Read 2 more answers
A stunt car driver testing the use of air bags drives a car at a constant speed of25 m/s for a total of 100m. He applies his bra
PIT_PIT [208]

Answer:

The graphs are attached

Explanation:

We are told that he starts with a constant speed of 25 m/s for a distance of 100 m.

At constant velocity, v = distance/time

time(t) = distance(d)/velocity(v)

t1 = 100/25

t1 = 4 s

Now, we are told that he applies his brakes and accelerates uniformly to a stop just as he reaches a wall 50m away.

It means, he decelerate and final velocity is zero.

Thus;

v² = u² + 2as

0² = 25² + 2a(50)

25² = - 100a

625 = - 100a

a = - 625/100

a = - 6.25 m/s²

v = u + at

0 = 25 + (-6.25t)

25 = 6.25t

t = 25/6.25

t = 4 s

With the values gotten, kindly find attached the distance-time and velocity-time graphs.

4 0
2 years ago
Two thin lenses with a focal length of magnitude 12.0cm, the first diverging and the second converging, are located 9.00cm apart
attashe74 [19]

Answer:

Explanation:

b ) First is concave lens with focal length f₁ = - 12 cm .

object distance u = - 20 cm .

Lens formula

1 / v - 1 / u = 1 / f

1 / v + 1 / 20 = -1 / 12

1 / v =  - 1 / 20  -1 / 12

= - .05 - .08333

= - .13333

v = - 1 / .13333

= - 7.5 cm

first image is formed before the first lens on the side of object.

This will become object for second lens

distance from second lens = 7.5 + 9 = 16.5 cm

c )

For second lens

object distance u = - 16.5 cm

focal length f₂ = + 12 cm ( lens is convex )

image distance = v

lens formula ,

1 / v - 1 / u = 1 / f₂

1 / v + 1 / 16.5 = 1 / 12

1 / v =   1 / 12 -  1 / 16.5

= .08333- .0606

= .02273

v = 1 /  .02273

= 44 cm ( approx )

It will be formed on the other side of convex lens

distance from first lens

= 44 + 9 = 53 cm .

magnification by first lens = v / u

= -7.5 / -20 = .375 .

magnification by second lens = v / u

= 44 / - 16.5

= - 2.67

d )

total magnification

= .375 x - 2.67

= - 1.00125

height of final image

= 2.50 mm x 1.00125

= 2.503mm

e )

The final image will be inverted with respect to object  because total magnification is negative .

6 0
1 year ago
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