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3 years ago

What is the formula that describes the magnitude of impulse on an object?

Physics

1 answer:

vladimir1956 [14]3 years ago

8 0

Answer:

Option C.

Impulse = mass × change in velocity

Explanation:

Impulse is defined by the following the following formula:

Impulse = force (F) × time (t)

Impulse = Ft

From Newton's second law of motion,

Force = change in momentum /time

Cross multiply

Force × time = change in momentum

Recall:

Impulse = Force × time

Thus,

Impulse = change in momentum

Recall:

Momentum = mass x velocity

Momentum = mv

Chang in momentum = mass × change in velocity

Change in momentum = mΔv

Thus,

Impulse = change in momentum

Impulse = mass × change in velocity

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1. A 1.30 kg ball strikes a wall with a velocity of -10.5 m/s. The ball bounces off with a velocity of 6.50 m/s. If the ball is

bekas [8.4K]

Let F be the magnitude of the force. The impulse of this force while the ball is in contact with the wall is

Ft = F (0.0210 s)

and this impulse is equal to the change in the ball's momentum,

m ∆v = (1.30 kg) (6.50 m/s - (-10.5 m/s)) = (1.30 kg) (17.0 m/s)

Solve for F :

F (0.0210 s) = (1.30 kg) (17.0 m/s)

F = (1.30 kg) (17.0 m/s) / (0.0210 s)

F ≈ 1050 N

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3 years ago

Two tiny conducting spheres are identical and carry charges of -19.8μC and +40.7μC. They are separated by a distance of 3.59 cm.

romanna [79]

Answer:

(a): $\rm -5.627\times 10^3\ N.$

(b): $\rm 7.626\times 10^2\ N.$

Explanation:

<u>Given:</u>

Charge on one sphere, $\rm q_1 = -19.8\ \mu C = -19.8\times 10^{-6}\ C.$

Charge on second sphere, $\rm q_2 = +40.7\ \mu C = +40.7\times 10^{-6}\ C.$
Separation between the spheres, $\rm r=3.59\ cm = 3.59\times 10^{-2}\ m.$

Part (a):

According to Coulomb's law, the magnitude of the electrostatic force of interaction between two static point charges is given by

$\rm F=k\cdot\dfrac{q_1q_2}{r^2}$

where,

k is called the Coulomb's constant, whose value is $\rm 9\times 10^9\ Nm^2/C^2.$

From Newton's third law of motion, both the spheres experience same force.

Therefore, the magnitude of the force that each sphere experiences is given by

$\rm F=k\cdot\dfrac{q_1q_2}{r^2}\\=9\times 10^9\times \dfrac{(-19.8\times 10^{-6})\times (+40.7\times 10^{-6})}{(3.59\times 10^{-2})^2}\\=-5.627\times 10^3\ N.$

The negative sign shows that the force is attractive in nature.

Part (b):

The spheres are identical in size. When the spheres are brought in contact with each other then the charge on both the spheres redistributes in such a way that the net charge on both the spheres distributed equally on both.

Total charge on both the spheres, $\rm Q=q_1+q_2=-19.8\ \mu C+40.7\ \mu C = 20.9\ \mu C.$

The new charges on both the spheres are equal and given by

$\rm q_1'=q_2'=\dfrac Q2 = \dfrac{20.9}{2}\ \mu C=10.45\ \mu C = 10.45\times 10^{-6}\ C.$

The magnitude of the force that each sphere now experiences is given by

$\rm F'=k\cdot \dfrac{q_1'q_2'}{r^2}'\\=9\times 10^9\times \dfrac{10.45\times 10^{-6}\times 10.45\times 10^{-6}}{(3.59\times 10^{-2})^2}\\=7.626\times 10^2\ N.$

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80 divided by 20 is 2

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3 years ago

An AC source operating at 59 Hz with a maximum voltage of 170 V is connected in series with a resistor (R = 1.2 kΩ) and an induc

Alexxandr [17]

I = V/Z

V = voltage, I = current, Z = impedance

First let's find the total impedance of the circuit.

The impedance of the resistor is:

$Z_{R}$ = R

R = resistance

Given values:

R = 1200Ω

Plug in:

$Z_{R}$ = 1200Ω

The impedance of the inductor is:

$Z_{L}$ = j2πfL

f = source frequency, L = inductance

Given values:

f = 59Hz, L = 2.4H

Plug in:

$Z_{L}$ = j2π(59)(2.4) = j889.7Ω

Add up the individual impedances to get the Z, and convert Z to polar form:

Z = $Z_{R}$ + $Z_{L}$

Z = 1200 + j889.7

Z = 1494∠36.55°Ω

I = V/Z

Given values:

V = 170∠0°V (assume 0 initial phase)

Z = 1494∠36.55°Ω

I = 170∠0°/1494∠36.55°Ω

I = 0.1138∠-36.55°A

Round the magnitude of I to 2 significant figures and now you have your maximum current:

I = 0.11A

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