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3 years ago

The net force acting on an object equals the applied force plus the force of friction.

Physics

1 answer:

Georgia [21]3 years ago

3 0

Answer:

False

Explanation:

The net force is equal to the applied force minus the force of friction. It is possible for friction to act in the same direction as an applied force, but that would mean there would have to be more than two forces acting on the object.

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A flat circular plate of copper has a radius of 0.262 m and a mass of 61.5 kg.

Evgesh-ka [11]

Answer:

0.031 m

Explanation:

Density of copper = ρ = 8960 kg/m³

r = Radius = 0.262 m

m = Mass of plate = 61.5 kg

v = Volume of plate = Volume of cylinder = πr²h

$\rho=\frac{m}{v}\\\Rightarrow 8940=\frac{61.5}{\pi r^2h}\\\Rightarrow h=\frac{61.5}{\pi 0.262^2\times 8940}\\\Rightarrow h=0.031\ m$

So, thickness of plate is 0.031 m

5 0

4 years ago

A 5kg book rests on a table. How much force is it exerting on the table?

valentinak56 [21]

Answer:

5 Kg of Force

Explanation:

3 0

3 years ago

Read 2 more answers

When a 9.00 V voltage is applied to a resistor, it drives 0.325 A of current through the resistor. How much power does the resis

solong [7]

Answer: 2.925

Explanation: :)

8 0

3 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Gelneren [198K]

Work = force x distance (displacement) or the change in Kinetic energy

Carrying a box down a corridor has no external force acting on the box and therefore no work is being done.

Pushing against a locked door doesn't displace the door, so no work is being done.

Suspending a heavy weight with a strong chain requires, no force and actually prevents a force from acting on the chain.

Pulling a trailer up a hill on the other hand requires a person to apply a force to the trailer and displace it up the hill and therefore it requires work.

6 0

3 years ago

An AC source of maximum voltage V0 = 30 V is connected to a resistor R = 50 Ω, an inductor L = 0.6 H, and a capacitor C = 20 µF.

ivolga24 [154]

Hello!

We can begin by solving for the resonance ANGULAR frequency of the circuit.

For an RCL circuit, the resonance angular frequency is given as:
$\omega_0^2 = \frac{1}{LC}\\\\w_0 = \sqrt{\frac{1}{LC}}$

ω₀ = resonance angular frequency (rad/s)

L = Inductance (0.6 H)
C = Capacitance (20 μF)

Plug in the values and solve.

$\omega_0 = \sqrt{\frac{1}{(0.6)(0.00002)} } = 288.675 \frac{rad}{s}$

For an AC power source, the output is usually expressed as:

$V(t) = V_{max}sin(\omega_0 t})$

So, using the appropriate values and setting the source angular frequency equivalent to the circuit's resonance angular frequency:

$V(t) = 30sin(288.675t)$

To find the maximum charge on the capacitor when the frequency of the source is equivalent to the resonance frequency of the circuit (or the angular frequencies are equal), we can begin by finding the maximum voltage across the capacitor.

To find this, however, we must solve for the maximum current across the circuit by finding the total impedance of the circuit. When the circuit is at resonance, the impedance is equivalent to the resistance of the RESISTOR.

So, solve for the maximum current in the circuit using Ohm's Law:

$i = \frac{V}{R}$

In this instance AT RESONANCE:

$I_{Max} = \frac{V_Max}{R}\\\\I_{Max} = \frac{30}{50} = 0.6 A$

Now, we must solve for the capacitive reactance in order to find the maximum voltage across the capacitor. Using the following equation for capacitive reactance:
$X_c = \frac{1}{\omega C}\\\\X_c = \frac{1}{(288.675)(0.00002)} = 173.205 \Omega$

Now that we found the maximum current and capacitive reactance, we can now solve for the maximum voltage across the capacitor:
$V_{C, max} = X_C I_{Max}\\\\V_{C, max} = 173.205 * 0.6 = 103.923 V$

Finally, we can easily solve for the maximum charge on the capacitor using the relationship:
$C = \frac{Q}{V}\\\\Q = CV$

Plug in the values solved for above.

$Q = (0.00002)(103.923) = 0.00208 C = \boxed{2.078 mC}$

6 0

2 years ago