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2 years ago

Choose a system consisting only of ball A. What is the momentum change of the system during the collision?

Physics

1 answer:

shusha [124]2 years ago

3 0

The momentum change of the system during the collision is always equals to one another.

<h3>Momentum change during collision</h3>

In a collision, the momentum change of first object is equal to and opposite of the momentum change of second object with which the first one collided. Or in other words, the momentum lost by first object is equal to the momentum gained by second object.

The total momentum of the system means the collection of two objects is conserved so we can conclude that the momentum change of the system during the collision is always equals and opposite to one another.

Learn more about collisions here: brainly.com/question/7538238

You might be interested in

This is the number of complete movements of a wave per second.

Alona [7]

A unit for measuring frequency, equal to one cycle per second. If a sound wave has a frequency of 20,000 Hz, this means that 20,000 waves are passing through a given point during the interval of one second.

5 0

3 years ago

A uniform disk has a mass of 3.7 kg and a radius of 0.40 m. The disk is mounted on frictionless bearings and is used as a turnta

nevsk [136]

Answer:

1.25 kgm²/sec

Explanation:

Disk inertia, Jd =

Jd = 1/2 * 3.7 * 0.40² = 0.2960 kgm²

Disk angular speed =

ωd = 0.1047 * 30 = 3.1416 rad/sec

Hollow cylinder inertia =

Jc = 3.7 * 0.40² = 0.592 kgm²

Initial Kinetic Energy of the disk

Ekd = 1/2 * Jd * ωd²

Ekd = 0.148 * 9.87

Ekd = 1.4607 joule

Ekd = (Jc + 1/2*Jd) * ω²

Final angular speed =

ω² = Ekd/(Jc+1/2*Jd)

ω² = 1.4607/(0.592+0.148)

ω² = 1.4607/0.74

ω² = 1.974

ω = √1.974

ω = 1.405 rad/sec

Final angular momentum =

L = (Jd+Jc) * ω

L = 0.888 * 1.405

L = 1.25 kgm²/sec

4 0

3 years ago

Who invented the transistor

erma4kov [3.2K]

A transistor is a semiconductor device used to amplify or switch electronic signals and electrical power. It is composed of semiconductor material usually with at least three terminals for connection to an external circuit. A voltage or current applied to one pair of the transistor's terminals changes the current through another pair of terminals. Because the controlled (output) power can be higher than the controlling (input) power, a transistor can amplify a signal. Today, some transistors are packaged individually, but many more are found embedded in integrated circuits.

Some of the earliest work on semiconductor amplifiers emerged from Eastern Europe. In 1922-23 Russian engineer Oleg Losev of the Nizhegorod Radio Laboratory, Leningrad, found that a special mode of operation in a point-contact zincite (ZnO) crystal diode supported signal amplification up to 5 MHz. Although Losev experimented with the material in radio circuits for years, he died in the 1942 Siege of Leningrad and was unable to advocate for his place in history. His work is largely unknown.

Austro-Hungarian physicist, Julius E. Lilienfeld, moved to the US and in 1926 filed a patent for a “Method and Apparatus for Controlling Electric Currents” in which he described a three-electrode amplifying device using copper-sulfide semiconductor material. Lilienfeld is credited with inventing the electrolytic capacitor but there is no evidence that he built a working amplifier. His patent, however, had sufficient resemblance to the later field effect transistor to deny future patent applications for that structure.

<span>German scientists also contributed to this early research. While working at Cambridge University, England in 1934, German electrical engineer and inventor Oskar Heil filed a patent on controlling current flow in a semiconductor via capacitive coupling at an electrode – essentially a field-effect transistor. And in 1938, Robert Pohl and Rudolf Hilsch experimented on potassium-bromide crystals with three electrodes at Gottingen University. They reported amplification of low-frequency (about 1 Hz) signals. None of this research led to any applications but Heil is remembered in audiophile circles today for his air motion transformer used in high fidelity speakers.</span>

4 0

3 years ago

Read 2 more answers

Solving elastic collisions problem the hard way

vova2212 [387]

Answer:

<h2>Solving elastic collisions problem the hard way</h2><h3 />

Explanation:

perfect drawing

4 0

3 years ago

An uncharged series RC circuit is to be connected across a battery. For each of the following changes, determine whether the tim

slavikrds [6]

a) Increase

b) Unchanged

c) Increase

Explanation:

The charge on a capacitor charging in a RC circuit connected to a battery follows the exponential equation:

$Q(t)=Q_0 (1-e^{-\frac{t}{RC}})$

where

$Q_0 = CV$ is the final charge stored in the capacitor, where C is the capacitance and V is the voltage of the battery

t is the time

R is the resistance of the circuit

The capacitor reaches 90% of its final charge when

$Q(t)=0.90Q_0$

Substituting and re-arranging the equation, we find:

$0.90Q_0 = Q_0(1-e^{-\frac{t}{RC}})\\0.90=1-e^{-\frac{t}{RC}}\\e^{-\frac{t}{RC}}=0.10\\-\frac{t}{RC}=ln(0.10)\\t=-RCln(0.10)=2.30RC$

We see that if we double the RC constant, then $(RC)'=2(RC)$

So the time taken will double as well:

$t'=2.30(RC)'=2.30(2RC)=2(2.30RC)=2t$

So, the answer is "increase"

In this second part, the battery voltage is doubled.

According to the equation written in part a),

$Q_0 =CV$

this means also that the final charge stored on the capacitor will also double.

However, the equation that gives us the time needed for the capacitor to reach 90% of its full charge is

$t=2.30 RC$

We see that this equation does not depend at all on the voltage of the battery.

Therefore, if the battery voltage is doubled, the final charge on the capacitor will double as well, but the time needed for the capacitor to reach 90% of its charge will not change.

So the correct answer is

"unchanged"

In this case, a second resistor is added in series with the original resistor of the circuit.

We know that for two resistors in series, the total resistance of the circuit is given by the sum of the individual resistances:

$R=R_1+R_2$

Since each resistance is a positive value, this means that as we add new resistors, the total resistance of the circuit increases.

Therefore in this problem, if we add a resistor in series to the original circuit, this means that the total resistance of the circuit will increase.

The time taken for the capacitor to reach 90% of its final charge is still

$t=2.30 RC$

As we can see, this time is directly proportional to the resistance of the circuit, R: therefore, if we add a resistor in series, the resistance of the circuit will increase, and therefore this time will increase as well.

So the correct answer is

"increase"

8 0

3 years ago