What is the angular momentum of a 0.25 kg mass rotating on the end of a piece of

Consider the classic problem with holiday lights, one little bulb goes out and the whole string goes out. First consider a strin

Flura [38]

Answer:

a. 0.33 Amp

b. 2.4 Volt

c. 0

d. 2.45 Amp

e. Infinite

f. Series is safer

Explanation:

Series Connection of Resistors

When two or more resistors are connected in series, the current through each one of them is the same, and the voltage divides depending on the particular value of each resistance. If all the resistances are equal, then the voltage is equally divided.

a. The string of 50 bulbs is connected to a 120 VAC outlet and consumes 40 W. The power of a circuit is given by

$P=V.I$

Solving for I

$\displaystyle I=\frac{P}{V}=\frac{40}{120}=0.33\ Amp$

Since all the bulbs are connected in series the current is the same for all of them.

b. The voltage is equally divided, so each bulb has 120/50= 2.4 V

c. If one of the bulbs burns out and its resistance becomes infinite, then the series circuit is open and no current flows through it, neither through the rest of the bulbs. The typical case of the whole string going out.

d. If one of the bulbs short circuits, the resistance of that bulb is zero and the voltage is distributed by the 49 remaining bulbs. Thus the new current is

$\displaystyle I=\frac{V}{R}=\frac{120}{49}=2.45\ A$

e. If the bulbs were connected in parallel, all of them would have the same voltage, and the total current will be equally divided among them. In that case, a short circuit in one of the bulbs will cause a parallel short, theoretically producing an infinite current and making the short circuit protection blow up.

f. The condition described above makes the strings be made of series-connected bulbs which is safer than the parallel circuit. If a single bulb shorts, the entire string goes out in a series connection, but the breaker would trigger disconnection of the house circuit if it's a parallel connection. That is why we must deal with unusable strings instead of burning cables.

6 0

3 years ago

What is the frequency of a wave that has a wavelength of 20 cm and a speed of 10 m/s

jarptica [38.1K]

You would probably have a low frequency due to how much the wavelength is spread out.

8 0

3 years ago

Does magnets exert a force. explain

Ratling [72]

Answer:

Magnets exert forces and torques on each other due to the rules of electromagnetism. The forces of attraction field of magnets are due to microscopic currents of electrically charged electrons orbiting nuclei and the intrinsic magnetism of fundamental particles (such as electrons) that make up the material. Hope this helps you! :)

7 0

4 years ago

2. What is the water pressure at a depth of 24 m in a lake? [Density of water, p = 1 000 kg m- and gravitational acceleration, g

Andrei [34K]

Answer:

Pressure = ρgh

pressure (p) is measured in pascals (Pa)

density (ρ) is measured in kilograms per metre cubed (kg/m3)

The fore of gravitational field strength (g) is measured in N/kg or m/s 2

height of column (h) is measured in metres (m)

Answer = 235,200 Pa

Explanation:

Pressure = ρgh

Pressure = 1,000 x 9.8 x 24

Pressure = 235,200 Pa

4 0

3 years ago

A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0-m-high frictionless chute to a waiti

LuckyWell [14K]

Answer:

The speed of the package of mass m right before the collision $= 7.668\ ms^-1$

Their common speed after the collision $= 2.56\ ms^-1$

Height achieved by the package of mass m when it rebounds $= 0.33\ m$

Explanation:

Have a look to the diagrams attached below.

a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

where $K$ is Kinetic energy and $U$ is Potential energy.

$K= \frac{mv^2}{2}$ and $U= mgh$

Considering the fact $K_{initial} = 0\ and U_{final} =0$ we will plug out he values of the given terms.

So $V_{1}{(initial)} =\sqrt{2gh} = \sqrt{2\times9.8\times3} = 7.668\ ms^-1$

Keypoints:

Sum of energies and momentum are conserved in all collisions.
Sum of KE and PE is also known as Mechanical energy.
Only KE is conserved for elastic collision.
for elastic collison we have $e=1$ that is co-efficient of restitution.

KE = Kinetic Energy and PE = Potential Energy

b.Now when the package stick together there momentum is conserved.

Using law of conservation of momentum.

$m_1V_1(i) = (m_1+m_2)V_f$ where $V_1{i} =7.668\ ms^-1$ .

Plugging the values we have

$m\times 7.668 = (3m)\times V_{f}$

Cancelling m from both sides and dividing 3 on both sides.

$V_f = 2.56\ ms^-1$

Law of conservation of energy will be followed over here.

c.Now the collision is perfectly elastic $e=1$

We have to find the value of $V_{f}$ for m mass.

As here $V_{f}=-2.56\ ms^-1$ we can use that if both are moving in right ward with $2.56$ then there is a $-2.56$ velocity when they have to move leftward.

The best option is to use the formulas given in third slide to calculate final velocity of object $1$ .

So

$V_{1f} = \frac{m_1-m_2}{m_1+m_2} \times V_{1i}= \frac{m-2m}{3m} \times7.668=\frac{-7.668}{3} = -2.56\ ms^-1$

Now using law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

$\frac{m\times V(f1)^2}{2} + 0 = 0 +mgh$

$\frac{v(f1)^2}{2g} = h$

$h= \frac{(-2.56)^2}{9.8\times 3} =0.33\ m$

The linear momentum is conserved before and after this perfectly elastic collision.

So for part a we have the speed $=7.668\ ms^-1$ for part b we have their common speed $=2.56\ ms^-1$ and for part c we have the rebound height $=0.33\ m$ .

3 0

3 years ago