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hodyreva [135]
3 years ago
15

A 15 kg cart is pushed on a frictionless surface from rest horizontally by a 30 N force. What is the cart's acceleration?

Physics
1 answer:
Rainbow [258]3 years ago
5 0

Answer:

<em>a. The cart's acceleration is 2 m/s^2</em>

<em>b. The cart will travel 100 m</em>

<em>c. The speed is 20 m/s</em>

Explanation:

a. The acceleration of the cart can be calculated using Newton's second law:

F = m.a

Solving for a:

\displaystyle a=\frac{F}{m}

The cart has a mass of m=15 Kg and is applied a net force of F=30 N, thus:

\displaystyle a=\frac{30}{15}

a=2\ m/s^2

b.

Now we use kinematics to find the distance and speed:

\displaystyle x = v_o.t+\frac{at^2}{2}

The cart starts from rest (vo=0). The distance traveled in t=10 seconds is:

\displaystyle x = 0*10+\frac{2*10^2}{2}

x = 100\ m

The cart will travel 100 m

c.

The final speed is calculated by:

v_f=0+2*10=20\ m/s

The speed is 20 m/s

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Complete Question:

Two small objects each with a net charge of Q (where Q is a positive number) exert a force of magnitude "F" on each other. We

replace one of the objects with another whose net charge is 4Q. The original magnitude of the force on the Q charge was "F"; what is the magnitude of the force on the Q charge now?

Answer:

4 F₀

Explanation:

Assuming that we can treat to both objects as point charges, we can find the force "F" that one charge exerts upon the other applying Coulomb´s law, as follows:

F₀ = K*Q₀² / r₁₂²

If we replace one of the charges by one with a 4Q₀ charge, the new value of F will be as follows:

F₁ = K*Q₀*4Q₀ / r₁₂² =( K*Q₀² / r₁₂²)* 4 = 4* F₀

This value is reasonable, as the electrostatic force is a linear - type one, so it is possible to use the superposition principle (we can get the force exerted by one charge on another without considering the ones due to another charges)

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A parallel-plate, air-gap capacitor has a capacitance of 0.14 mu F. The plates are 0.5 mm apart, What is the area of each plate?
Marysya12 [62]

Answer:

7.9060 m²

8.57 Volts

5.142×10⁻⁶ Joule

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Explanation:

C = Capacitance between plates = 0.14 μF = 0.14×10⁻⁶ F

d = Distance between plates = 0.5 mm = 0.5×10⁻³ m

Q = Charge = 1.2 μC = 1.2×10⁻⁶ C

ε₀ = Permittivity = 8.854×10⁻¹² F/m

Capacitance

C=\frac{\epsilon_{0}A}{d}\\\Rightarrow A=\frac{Cd}{\epsilon_{0}}\\\Rightarrow A=\frac{0.14\times 10^{-6}\times 0.5\times 10^{-3}}{8.854\times 10^{-12}}\\\Rightarrow A=7.9060\ m^2

∴ Area of each plate is 7.9060 m²

Voltage

V=\frac{Q}{C}\\\Rightarrow V=\frac{1.2\times 10^{-6}}{0.14\times 10^{-6}}\\\Rightarrow V=8.57\ Volts

∴ Potential difference between the plates if the capacitor is charged to 1.2 μC  is 8.57 Volts.

Energy stored

E=0.5CV²

⇒E = 0.5×0.14×10⁻⁶×8.57²

⇒E = 5.142×10⁻⁶ Joule

∴ Stored energy is 5.142×10⁻⁶ Joule

Charge

Q = CV

⇒Q = 0.14×10⁻⁶×8.57

⇒Q = 1.2×10⁻⁶ C

∴ Charge the capacitor carries before a spark occurs between the two plates is 1.2×10⁻⁶ Coulomb

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