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3 years ago

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An object with a weight of 20 kg will move _______ an object with a weight of 20 g if the same amount of force is placed on the

two objects.
A.
Neither object will move at all.
B.
a smaller distance than
C.
a greater distance than
D.
the same distance as

1 answer:

elena55 [62]3 years ago

7 0

The answer should be D

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The net force acting on a 50kg crate is 0 N. what is the crates accelratiin

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Answer:

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Explanation:

From the question we are told that:

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The graph represents velocity over time. What is the acceleration? –0.4 m/s2 –0.2 m/s2 0.2 m/s2 0.4 m/s2

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Missing graph. I attach it in the answer.

In a uniformly accelerated motion, the velocity at time t is given by:
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Given the previous equation, if we plot v(t) versus t, we find a straight line; moreover, a (the acceleration) represents the slope of the curve.

Looking at the graph, we see that when the time goes from 10 s to 20 s, the velocity increases from 4 m/s to 6 m/s. Therefore the slope of the curve is
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3 years ago

Read 2 more answers

A 30.0-Î¼F capacitor is connected to a 49.0-Î© resistor and a generator whose rms output is 30.0 V at 60.0 Hz. (a) Find the rms

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Explanation:

Given that,

Capacitor = 30μC

Resistor = 49.0Ω

Voltage = 30.0 V

Frequency = 60.0 Hz

We need to calculate the impedance

Using formula of impedance

$Z=\sqrt{R^2+X_{c}^2}$ .....(I)

We need to calculate the value of $X_{c}$

Using formula of $X_{c}$

$X_{c}=\dfrac{1}{2\pi f c}$

$X_{c}=\dfrac{1}{2\times\pi\times60.0\times30\times10^{-6}}$

$X_{c}=88.42\ \Omega$

Put the value of $X_{c}$ into the formula of impedance

$Z=\sqrt{(49.0)^2+(88.42)^2}$

$Z=101.08\ \Omega$

(a). We need to calculate the rms current in the circuit

Using formula of rms current

$I_{rms}=\dfrac{V}{Z}$

$I_{rms}=\dfrac{30.0}{101.08}$

$I_{rms}=0.30\ A$

The rms current in the circuit is 0.30 A.

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

Using formula of rms voltage

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

Put the value into the formula

$V_{rms}=0.30\times49.0$

$V_{rms}=14.7\ V$

The rms voltage drop across the resistor is 14.7 V

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

Using formula of rms voltage

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

$V_{rms}=0.30\times88.42$

$V_{rms}=26.53\ V$

The rms voltage drop across the capacitor is 26.53 V.

Hence, This is the required solution.

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