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3 years ago

In a position-time graph what does it mean when you return to 0? ASAP HELP PLEASE!

Physics

1 answer:

Novosadov [1.4K]3 years ago

6 0

It means that you have returned to the designated starting position.

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A horizontal net force F is exerted on an object at rest. The object starts at x=0 m and has a speed of 4.0 m/s after moving 4.0

Alex_Xolod [135]

Explanation:

The solution is be found in the attachment.

8 0

3 years ago

The capacitor can withstand a peak voltage of 590 volts. If the voltage source operates at the resonance frequency, what maximum

kirill115 [55]

Answer:

The maximum voltage is 41.92 V.

Explanation:

Given that,

Peak voltage = 590 volts

Suppose in an L-R-C series circuit, the resistance is 400 ohms, the inductance is 0.380 Henry, and the capacitance is $1.20×10^{-2}\ \mu F$ .

We need to calculate the resonance frequency

Using formula of frequency

$f=\dfrac{1}{2\pi\sqrt{LC}}$

Put the value into the formula

$f=\dfrac{1}{2\pi\sqrt{0.380\times1.20\times10^{-8}}}$

$f=2356.88\ Hz$

We need to calculate the maximum current

Using formula of current

$I=\dfrac{V_{c}}{X_{c}}$

$I=2\pi\times f\times C\times V_{c}$

$I=2\pi\times2356.88\times1.20\times10^{-8}\times590$

$I=0.1048\ A$

Impedance of the circuit is

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

At resonance frequency $X_{L}=X_{C}$

$Z=R$

We need to calculate the maximum voltage

Using ohm's law

$V=I\times R$

$V=0.1048\times400$

$V=41.92\ V$

Hence, The maximum voltage is 41.92 V.

4 0

3 years ago

We know the frequency range of certain sounds are: 400-560 Hz, what are the ranges of wavelength in meters when the signal trans

Ksivusya [100]

Answer:

Range of wavelength will be $5.35\times 10^5m$ to $7.5\times 10^5m$

Explanation:

We have given range of frequency is 400-560 Hz

Speed of the light $c=3\times 10^8m/sec$

We have to find the range of the wavelength of signal transmitted

Ween know that velocity is given by $v=\lambda f$ , here $\lambda$ is wavelength and f is frequency

So for 400 Hz frequency wavelength will be $\lambda =\frac{c}{f}=\frac{3\times 10^8}{400}=7.5\times 10^5m$

And wavelength for frequency 560 Hz $\lambda =\frac{c}{f}=\frac{3\times 10^8}{560}=5.35\times 10^5m$

So range of wavelength will be $5.35\times 10^5m$ to $7.5\times 10^5m$

8 0

3 years ago

wo fixed charges, A and B are located at x axis. A is at x = 0 m, B is at x = 4 m. QA = +4.0 μC and QB = -5.0 μC. Calculate the

lys-0071 [83]

Answer:

10250 N/C leftwards

Explanation:

QA = 4 micro Coulomb

QB = - 5 micro Coulomb

AP = 6 m

BP = 2 m

A is origin, B is at 4 m and P is at 6 m .

The electric field due to charge QA at P is EA rightwards

$E_{A}=\frac{KQ_{A}}{AP^{2}}=\frac{9\times10^{9}\times4\times10^{-6}}{6^{2}}=1000 N/C (rightwards)$

The electric field due to charge QB at P is EB leftwards

$E_{B}=\frac{KQ_{B}}{BP^{2}}=\frac{9\times10^{9}\times5\times10^{-6}}{2^{2}}=11250 N/C (leftwards)$

The resultant electric field at P due the charges is given by

E = EB - EA

E = 11250 - 1000 = 10250 N/C leftwards

5 0

3 years ago

A 500-kilogram sports car accelerates uni-

timama [110]

Answer:

90 meters

Explanation:

Given:

x₀ = 0 m

v₀ = 0 m/s

v = 30 m/s

t = 6 s

Find:

x = x₀ + ½ (v + v₀)t

x = 0 + ½ (30 + 0)(6)

x = 90

The car travels 90 meters.

3 0

3 years ago