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fenix001 [56]
3 years ago
14

a 1.2x10^3 kilogram car is accelerated uniformly from 10. meters per second to 20 meters per second in 5.0 seconds. what is the

magnitude of the net force acting on the car during this 5.0 second interval
Physics
1 answer:
irinina [24]3 years ago
7 0
Force , F = ma

F =  m(v - u)/t               

Where m = mass in kg, v= final velocity in m/s, u = initial velocity in m/s
t = time, Force is in Newton.

m= 1.2*10³ kg,  u = 10 m/s,  v = 20 m/s, t = 5s

F =  1.2*10³(20 - 10)/5

F = 2.4*10³ N = 2400 N


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An AC generator with a variable frequency and an RMS voltage output of 120 Volts is connected in series to a 3.1 μF Capacitor an
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Answer:

The RMS voltage across the resistor = 28 V

Explanation:

Capacitor: A capacitor is an electrical device that has the ability to store electrical charges in an electrical circuit. It is expressed in Farad (F)

Resistor: A resistor is an electrical device that oppose the flow of electric current in a circuit. It is expressed in ohms (Ω)

RMS Voltage : RMS voltage  value of an alternating voltage is defined as that value of steady voltage which would dissipate heat at the same rate in a given resistance

Since the it is a series circuit, the total voltage is divided across the resistance and the capacitor.

Vt = V₁ + V₂...........................Equation 1

Where Vt = total Rms voltage = 120 V ,  V₁ = Rms voltage across the Capacitor = 92 V, V₂ = Rms voltage across the resistor.

Making V₂ the subject of the equation in equation 1 above,

V₂ = Vt - V₁  = 120 - 92

V₂ = 28 V.

The RMS voltage across the resistor = 28 V

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An amplifier.

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Are most plants and animals single-cell organisms
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Read 2 more answers
Two stationary positive point charges, charge 1 of magnitude 3.90 nC and charge 2 of magnitude 1.80 nC, are separated by a dista
soldi70 [24.7K]

Answer:

v = 7793150 m/s

Explanation:

First, we are going to calculate the electrical potential in the point middle between the two charges

Remember that the electrical potential can be calculated as:

v = \frac{kQ}{r}

                 Where     k = 8.9874 x 10^{9} \frac{Nm^{2} }{C^{2} }

and it is satisfy the superposition principle, thus

v = \frac{8.9874x10^{9}(3.90x10^{-9} ) }{0.23} +  \frac{8.9874x10^{9}(1.80x10^{-9} ) }{0.23}

v = 222.73v

The electrical potential at 10 cm from charge 1 is:

v = \frac{8.9874x10^{9}(3.90x10^{-9} ) }{0.1} +  \frac{8.9874x10^{9}(1.80x10^{-9} ) }{0.36}

v = 395.44 v

Since the work - energy theorem, we have:

q\Delta v = \frac{mv^{2} }{2}

                     where q is the electron's charge and m is the electron's mass

Therefore:

v = \sqrt{\frac{2q\Delta v}{m} }

v = 7793150 m/s

6 0
3 years ago
A piano string having a mass per unit length equal to 4.70 10-3 kg/m is under a tension of 1 400 N. Find the speed with which a
sweet [91]

Answer:

The speed of the sound wave on the string is 545.78 m/s.

Explanation:

Given;

mass per unit length of the string, μ = 4.7 x 10⁻³ kg/m

tension of the string, T = 1400 N

The speed of the sound wave on the string is given by;

v = \sqrt{\frac{T}{\mu} }

where;

v is the speed of the sound wave on the string

Substitute the given values and solve for speed,v,

v = \sqrt{\frac{T}{\mu} }\\\\v = \sqrt{\frac{1400}{4.7*10^{-3}} }\\\\v = \sqrt{297872.34}\\\\v = 545.78 \ m/s

Therefore, the speed of the sound wave on the string is 545.78 m/s.

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2 years ago
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