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3 years ago

Physics help please!

Physics

1 answer:

Oliga [24]3 years ago

6 0

Answer:

student attach a save block to a horizontal spring so that the block spring system will oscillator with the block spring system released from rest horizontal position that is not the systems equilibrium position well this question regards about the energy used the answer may be 0.73 Joel ok you just try it ok verified

Explanation:

apply applied the potential energy value mean the formula MGH write it means what mass into gravitation in to height

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A machine accelerates a 5 kg missile from rest to a speed of 5 km/s. The net force accelerating the missile 500,000 N. How long

Gnoma [55]

Final velocity = 5km/sec = 5000 m/sec
Initial velocity = 0, mass = 5kg and force = 500,000 N
F = m(Vf-Vi)/time
Time = m(Vf-Vi)/force = 5*5000/500000 = 0.05 seconds
So, it takes 0.05 sec to arrive at the speed of 5km/s.

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3 years ago

A capacitor with an initial potential difference of 185 V is discharged through a resistor when a switch between them is closed

GrogVix [38]

Answer:

a. $\tau = 2.1161 s$

b. $V(18.8 \ s) = 0.0256 \ V$

Explanation:

The equation for the voltage V of discharging capacitor in an RC circuit at time t is:

$V(t) = V_0 e^{(- \frac{t}{\tau}) }$

where $V_0$ is the initial voltage, and $\tau$ is the time constant.

For our problem, we know

$V_0 = 185 \ V$

and

$V(10 \ s) = V_0 e^{(- \frac{10 \ s}{\tau}) } = 1.64 \ V$

$185 \ V \ e^{(- \frac{10 \ s}{\tau}) } = 1.64 \ V$

$e^{(- \frac{10 \ s}{\tau}) } = \frac{1.64 \ V}{ 185 \ V }$

$ln (e^{(- \frac{10 \ s}{\tau}) } ) = ln (\frac{1.64 \ V}{ 185 \ V })$

$- \frac{10 \ s}{\tau} = ln (\frac{1.64 \ V}{ 185 \ V })$

$\tau = \frac{-10 \s}{ln (\frac{1.64 \ V}{ 185 \ V }) }$

This gives us

$\tau = 2.1161 s$

and this is the time constant.

At t = 18.8 s we got:

$V(18.8 \ s) = 185 \ V \ e^{(- \frac{18.8 \ s}{2.1161 s}) }$

$V(18.8 \ s) = 0.0256 \ V$

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3 years ago

Two charges that are separated by one meter exert 1-n forces on each other. if the magnitude of each charge is doubled, the forc

Goshia [24]

The electrostatic force between the two charges is
$F=k_E \frac{q_1 q_2}{r^2}$
where q1 and q2 are the magnitudes of the two charges, and r the distance between them.

We can see from the formula that F is proportional to the product between the two charges:
$F \sim q_1 q_2$
so, if the magnitude of each charge is doubled, the new force will get a factor 4:
$F' \sim (2 q_1 )(2 q_2 )=4 q_1 q_2 =4 F$
So, the new force will be 4 times the original force:
$F' = 4 \cdot 1N= 4N$

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3 years ago

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