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3 years ago

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Planet X has a moon similar to Earth’s moon. Which path would this moon’s orbit take?

1 answer:

victus00 [196]3 years ago

3 0

What are the choices?

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When 15 newtons of force is applied to the 0.5 kg book, the friction keeps the book from sliding down the wall. What is the mini

valkas [14]

Answer:

15 N and 3.061

Explanation:

From the question,

The minimum force of friction to keep the book from sliding = 15 N.

using

F = mgμ................. Equation 1

Where F = Frictional Force, m = mass of the book, g = acceleration due to gravity, μ = coefficient of friction.

make μ the subject of the equation

μ = F/mg............... Equation 2

Given: F = 15 N, m = 0.5 kg, g = 9.8 m/s²

Substitute into equation 2

μ = 15(0.5×9.8)

μ = 15/4.9

μ = 3.061

Hence the coefficient of friction to keep the book from sliding = 3.061

4 0

4 years ago

Read 2 more answers

I really need help with this problem ;)

Nesterboy [21]

They're never the same ray.

4 0

3 years ago

Read 2 more answers

A car moving in a straight line starts at x = 0 at t = 0 . It passes the point x = 30.0 m with a speed of 10.0 m/s at t = 3.00 s

Bogdan [553]

Answer:

a) $v=20.3m/s$

b) $a=2.35m/s^2$

Explanation:

From the exercise we know:

$x_{1}=30m$ ; $v_{1}=10m/s$ ; $t_{1}=3s$

$x_{2}=375m$ ; $v_{2}=50m/s$ ; $t_{2}=20s$

The formula for average velocity is:

$v=\frac{x_{2}-x_{1}}{t_{2}-t_{1}}$

a) $v=\frac{375m-30m}{20s-3s}=20.3m/s$

The formula for average acceleration is:

$a=\frac{v_{2}-v_{1}}{t_{2}-t_{1}}$

b) $a=\frac{(50-10)m/s}{(20-3)s}=2.35m/s^2$

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3 years ago

What are the STEPS to using a triple beam balance

Gwar [14]

A good diet. and exersise is the answer

8 0

4 years ago

The output voltage of a voltage amplifier has been found to decrease by 20% when a load resistance of 1 k is connected. What is

yKpoI14uk [10]

We use the voltage division problem between the load resistance, amplifier output resistance as

$V_{out} = V_{amlifire} \times \frac{R_{load} }{R_{load} + R_{out} }$ .

Here, $V_{out}$ is the output voltage, $V_{amlifire}$ is the amplifier voltage, $R_{load}$ is the load resistance and $R_{out}$ is the amplifier output resistance.

Therefore,

$1-\frac{20}{100} = \frac{1 \ k\Omega }{1 \ k\Omega +R_{out} } \\\\ R_{out} = \frac{1 \ k\Omega }{0.8} -1 \ k\Omega =1250 \Omega -1000 \Omega =250 \Omega$ .

Thus, the amplifier output resistance is $250 \ \Omega$ .

4 0

3 years ago