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3 years ago

What is the impulse that a car with a mass of 6kg experiences when its velocity changes from 23m/s to 6m/s?

1 answer:

8 0

A.
Impulse is calculated through the equation,

I = mΔv

where I is impulse, m is mass and Δv is change is velocity. Substituting the known values,

I = (6 kg)(23 m/s - 6 m/s)
I = 102 kg m/s

Answer: 102 kg m/s

B.
Impulse in this item is calculated by multiplying the force by the time.

I = Force / time = (29 kg m/s²)(7 s) = 203 kg m/s

Answer: 203 kg m/s

C.
To answer this item, we have to multiply the mass and the velocity to get the impulse and divide the answer by the time to get the force.

F = m(v)/t

Substituting,

F = (0.1 kg)(-5.69 m/s)/0.64 s
F = -0.89 N

Answer: -0.89 N

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For the circuit shown, R = 75.0 ohms, L= 55.0 mg, and C = 25.0 μC. The power source has 12.0 V arms and a frequency of 60.0 Hz.

Natasha_Volkova [10]

Explanation:

Given that,

Resistance R = 75.0 ohms

Inductance L = 55.0 mH

Capacitance $C = 25.0\ \mu C$

Voltage V = 12.0 V

Frequency f = 60.0 Hz

We need to calculate the angular frequency

Using formula of angular frequency

$\omega = 2\pi f$

Put the value into the formula

$\omega =2\times3.14\times60.0$

$\omega=376.8\ rad/s$

(a). We need to calculate the value of $X_{L}$

Using formula of $X_{L}$

$X_{L}=\omega\times L$

Put the value into the formula

$X_{L}=376.8\times55.0\times10^{-3}$

$X_{L}=20.724\ \Omega$

(b). We need to calculate the value of $X_{L}$

Using formula of $X_{C}$

$X_{C}=\dfrac{1}{\omega C}$

$X_{C}=\dfrac{1}{376.8\times25.0\times10^{-6}}$

$X_{C}=106.16\ \Omega$

(c). We need to calculate the value of Z

Using formula of impedance

$Z=\sqrt{R^2+(X_{L}-X_{C})^2}$

Put the value into the formula

$Z=\sqrt{75.0^2+(20.724-106.16)^2}$

$Z=113.68\ \Omega$

(d). We need to calculate the rms current

Firstly we need to calculate the current

Using formula of current

$I=\dfrac{V}{R}$

Put the value into the formula

$I=\dfrac{12.0}{75.0}$

$I=0.16\ A$

Using formula of rms current

$I_{rms}=\dfrac{I_{0}}{\sqrt{2}}$

$I_{rms}=\dfrac{0.16}{\sqrt{2}}$

$I_{rms}=0.113\ A$

(e). We need to calculate the rms voltage across the resistor

Using formula of rms voltage

$V_{rms}=I_{rms}\times R$

$V_{rms}=0.113\times75.0$

$V_{rms}=8.475\ V$

(f). We need to calculate the rms voltage across the inductor

Using formula of rms voltage

$V_{rms}=I_{rms}\times X_{L}$

$V_{rms}=0.113\times20.724$

$V_{rms}=2.342\ V$

(g). We need to calculate the rms voltage across the capacitor

Using formula of rms voltage

$V_{rms}=I_{rms}\times X_{C}$

$V_{rms}=0.113\times106.16$

$V_{rms}=11.99\ V$

(h). We need to calculate the dissipated power by the circuit

Using formula of dissipated power

$P=RI^2$

Put the value into the formula

$P=75.0\times0.113^2$

$P=0.958\ W$

Hence, This is the required solution.

3 0

3 years ago