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Softa [21]
3 years ago
8

A low resistance light bulb and a high resistance light bulb are connected in parallel with each other. Which bulb is brighter i

f the parallel combination is in series with a battery?
Physics
1 answer:
sweet [91]3 years ago
6 0
<h2>Answer:</h2>

The bulb with low resistance will be brighter.

<h2>Explanation:</h2>

The brightness of a bulb is a function of both the voltage across the bulb and current flowing through the bulb. The higher the voltage, the higher the current. Hence the brighter the bulb.

Now, according to the question, the bulbs (the high resistance bulb and the low resistance bulb) are connected in parallel with each other. This means that the same voltage passes across them.

Also, we know that according to Ohm's law, the voltage (V) and current (I) through a conductor are related by the following equation;

V =  I x R                -------------------(i)

Where;

R is the resistance of the conductor.

We can re-write equation (i) as follows;

I = V / R               -----------------------(ii)

According to equation (ii), at fixed voltage (V), the current (I) will increase as the resistance (R) decreases.

Now, since the two bulbs have the same voltage, the bulb with the low resistance will allow a larger flow of current than the bulb with high resistance.  Therefore, as said earlier that brightness is dependent on voltage and current, the bulb with the low resistance (and having larger current at some voltage) will be brighter than the bulb with the high resistance (having smaller current at same voltage).

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Consider a uniformly charged thin-walled right circularcylindrical shell having total charge Q, radiusR, and height h. Determine
Vinil7 [7]

Answer:

\frac{k_eQ}{2h}

Explanation:

The question is missing an image. I have added this as an attachment to my answer.

Given;

Q = total charge

R = radius of cylindrical shell

h = height of cylindrical shell

d = distance of point from the right side of the cylinder

Let the thickness of the cylindrical shell be dx , and the charge  \frac{Qdx}{h},

Now, using the formula for finding the electric field due to a ring at a chosen point:

dE = \frac{k_ex}{(x^2 + R^2)^{\frac{3}{2}}} {\frac{Qdx}{h}i}

where x = center of the ring to the point

k_e = electrostatic constant

We integrate on both sides from the limits d to d + h  in order to determine the electric field at the point E

\int\limits dE = \int\limits^{d + h}_d {\frac{k_eQxdx}{h(x^2 + R^2)^{\frac{3}{2}}}i}

E = \int\limits^{d + h}_d {\frac{k_eQdx}{h(x^2 + R^2)^{\frac{3}{2}}}i}  =  \frac{k_eQ}{2h}  

6 0
3 years ago
You often hear people, particularly TV announcers, talk about a "high rate of speed." What do you think they mean? According to
NikAS [45]

TV announcer intend to mean "chnage of speed or velocity" by "High rate of speed" and in Physics domain it would mean "acceleration"

Explanation:

It is common to observe TV announcer saying certain events were occurring “with a high rate of speed”. By saying this they intend to mean that the event was rapid in its occurrence. It can also mean that the change in speed of the happening was very rapid/fast.

However, the same terms connote altogether a different expression in Physics domains. Speed is a scalar quantity with no direction. Hence most of the times speed mean velocity when the direction is also provided. “high rate of speed” would mean a change of velocity per unit time which is acceleration. Hence in Physics domain, the term would stand for acceleration.

3 0
3 years ago
A 4,000-kg car traveling at 20m/s hits a wall with a force of 80,000 N and comes to a stop. What was the impact time?
Nookie1986 [14]
Rate of change of momentum = impact force
(m*v-m*u)/t = F
4000*20/t = 80000 (note: v is zero as it stopped)
<span>soo, t = 1 sec</span>
7 0
3 years ago
The following represents a mass attached to a spring oscillating in simple harmonic motion. X(t) = 4.0 cos(3.0t +0.10) units of
kolbaska11 [484]

Answer:

a) A = 4.0 m , b)   w = 3.0 rad / s , c)  f = 0.477 Hz , d) T = 20.94 s

Explanation:

The equation that describes the oscillatory motion is

          x = A cos (wt + fi)

In the exercise we are told that the expression is

          x = 4.0 cos (3.0 t + 0.10)

let's answer the different questions

a) the amplitude is

         A = 4.0 m

b) the frequency or angular velocity

         w = 3.0 rad / s

c) angular velocity and frequency are related

          w = 2π f

           f = w / 2π

           f = 3 / 2π

           f = 0.477 Hz

d) the period

frequency and period are related

           T = 1 / f

           T = 1 / 0.477

           T = 20.94 s

e) the phase constant

          Ф = 0.10 rad

f) velocity is defined by

          v = dx / dt

         

         v = - A w sin (wt + Ф)

speed is maximum when sine is + -1

         v = A w

          v = 4 3

          v = 12 m / s

g) the angular velocity is

          w² = k / m

          k = m w²

          k = 1.2 3²

          k = 10.8 N / m

h) the total energy of the oscillator is

          Em = ½ k A²

           Em = ½ 10.8 4²

          Em = 43.2 J

i) the potential energy is

           Ke = ½ k x²

for t = 0 x = 4 cos (0 + 0.1)

               x = 3.98 m

j) kinetic energy

           K = ½ m v²

for t = 00.1 ²

    v = A w sin 0.10

    v = 4 3 sin 0.10

    v = 1.98 m / s

3 0
3 years ago
The formula used to find force is F=m*v.<br> true or false
zepelin [54]

It's true IF ' m ' stands for mass and ' v ' stands for acceleration. Otherwise it's false.

4 0
3 years ago
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