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3 years ago

What is the current in a 48 kω resistor with 150 v across it?

Physics

1 answer:

AleksandrR [38]3 years ago

8 0

Current = (voltage) / (resistance)

= (150 volts) / (48,000 ohms)

= 3.125 milliamperes

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Lightning flashes one mile (1609 m) away from you. How much time does it take the light to travel that distance?

hichkok12 [17]

Answer:

$t=5.36\times 10^{-6}\ s$

Explanation:

Given that,

Lightning flashes one mile (1609 m) away from you.

We need to find the time it take the light to travel that distance. Let the time be t. We know that,

speed = distance/time

$t=\dfrac{d}{v}\\\\t=\dfrac{1609}{3\cdot10^{8}}\\\\=5.36\times 10^{-6}\ s$

So, the required time is $5.36\times 10^{-6}\ s$

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3 years ago

All but one of the following is an important component of soil.

viktelen [127]

Air is the answer i do believe

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4 years ago

Read 2 more answers

Jan is holding an ice cube. What causes the ice to melt?

zmey [24]

It would be the temperature of her hand and also because it is going through the process of the water cycle it is changing from a solid to a liquid

6 0

4 years ago

The maximum current output of a 60 ω circuit is 11 A. What is the root mean square voltage of the circuit?

goldenfox [79]

Answer:

660V

Explanation:

V=IR

V=?, I=11A,R=60w

V=60 ×11

=660V

8 0

3 years ago

A typical laboratory centrifuge rotates at 4000 rpm. Test tubes have to be placed into a centrifuge very carefully because of th

cestrela7 [59]

Answer:

1. $a_{rad}=17545.2\frac{m}{s^{2}}$

2. $a=4429.45 \frac{m}{s^{2}}$

Explanation:

Radial acceleration is:

$a_{rad}=\frac{v^2}{r}$ (1)

With r the radius respects the axis of rotation and v the tangential velocity that is related with angular velocity (ω) by:

$v= \omega r$ (2)

By (2) on (1):

$a_{rad}= \frac{(\omega r)^2}{r}= (\omega )^2r=(418,88\frac{rad}{s})^2(0.1m)$

$a_{rad}=17545.2\frac{m}{s^{2}}$

To find the acceleration of the tube with the fall, we can use the expression:

$\overrightarrow{J}=\overrightarrow{F}_{avg}(\varDelta t)$ (3)

Due impulse-momentum theorem:

$\overrightarrow{J}=\overrightarrow{p}_{f}-\overrightarrow{p}_{i}$ (4)

with p the momentum and J the impulse. By (4) on (3):

$\overrightarrow{p}_{f}-\overrightarrow{p}_{i}=\overrightarrow{F}_{avg}(\varDelta t)$

And using Newton's second law (F=ma) and that (P=mv):

$mv_f-mv_i=(ma)(\varDelta t)$ (5)

Final velocity is the velocity just after the encounter with hard floor, and initial momentum us just before that moment so the first one is zero and the second one can be found sing conservation of energy:

$\frac{mv_i}{2}=mgh$

$v_i=\sqrt{2gh}=\sqrt{2(9.81)(1.0)}=4.43\frac{m}{s}$

So (5) is:

$-m(4.43)=(ma)(\varDelta t)$

solving for a:

$a=\frac{4.43}{\varDelta t}=\frac{4.43}{1.0\times10^{-3}}=-4429.45$

It’s negative because is opposed to the tube movement.

5 0

4 years ago