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3 years ago

What is the unit for the volt? W V = 9 What is the unit for the volt

Physics

1 answer:

Likurg_2 [28]3 years ago

4 0

the unit for volt is v

The volt (symbol: V) is the derived unit for electric potential, electric potential difference (voltage), and electromotive force.

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What is the net work done on the 20kg block while it moves the 4 meters?

Vinil7 [7]

Answer:

The answer to your question is 784.8 J. None of your answer, did you forget some information?

Explanation:

Data

mass = 20 kg

distance = 4 m

work = ?

Formula

Work = force x distance

Force = mass x gravity

Process

1.- Calculate the weight of the block

Weight = 20 x 9.81

Weight = 196.2 N

2.- Calculate the work done

Work = 196.2 x 4

Work = 784.8 J

5 0

4 years ago

Reaction time is the lapse of time between muscular movement and external stimuli. True or false

riadik2000 [5.3K]

False. reaction time would be lapse of time between external stimuli and muscular movement

3 0

3 years ago

Read 2 more answers

Jimmy held the end of a metal bar over a fire while holding on to the opposite end. After a few minutes, the end he was holding

Pavel [41]

Answer:

A. Conduction

Explanation:

Conduction is the transfer of heat within the material itself, and he's holding the metal bar over the fire.

6 0

4 years ago

In a physics lab, light with a wavelength of 530 nm travels in air from a laser to a photocell in a time of 16.7 ns . When a sla

Harlamova29_29 [7]

Answer:

$\lambda'=78.086\ nm$

Explanation:

Given:

wavelength of light in the air, $\lambda=530\times 10^{-9}\ m$

time taken to travel from the source to the photocell via air, $t=16.7\ s$

time taken to reach the photocell via air and glass slab, $t'=21.3\times 10^{-9}\ s$

thickness of the glass slab, $x=0.87\ m$

<u>Now we have the relation for time:</u>

$\rm time=\frac{distance}{speed}$

hence,

$t=\frac{d}{c}$

c= speed of light in air

$16.7\times 10^{-9}=\frac{d}{3\times 10^8}$

$d=16.7\times 10^{-9}\times 3\times 10^8$

$d=5.01\ m$

For the case when glass slab is inserted between the path of light:

$\frac{(d-x)}{c} +\frac{x}{v} =t'$ (since light travel with the speed c only in the air)

here:

v = speed of light in the glass

$\frac{(5.01-0.87)}{3\times 10^8} +\frac{0.87}{v} =21.3\times 10^{-9}$

$v=4.42\times 10^7\ m.s^{-1}$

Using Snell's law:

$\frac{\lambda}{\lambda'} =\frac{c}{v}$

$\frac{530}{\lambda'} =\frac{3\times 10^8}{4.42\times 10^7}$

$\lambda'=78.086\ nm$

5 0

3 years ago

A 7600 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.35 m/s2 and feels no appreci

ollegr [7]

Answer:

a) The rocket reaches a maximum height of 737.577 meters.

b) The rocket will come crashing down approximately 17.655 seconds after engine failure.

Explanation:

a) Let suppose that rocket accelerates uniformly in the two stages. First, rocket is accelerates due to engine and second, it is decelerated by gravity.

1st Stage - Engine

Given that initial velocity, acceleration and travelled distance are known, we determine final velocity ( $v$ ), measured in meters per second, by using this kinematic equation:

$v = \sqrt{v_{o}^{2} +2\cdot a\cdot \Delta s}$ (1)

Where:

$a$ - Acceleration, measured in meters per square second.

$\Delta s$ - Travelled distance, measured in meters.

$v_{o}$ - Initial velocity, measured in meters per second.

If we know that $v_{o} = 0\,\frac{m}{s}$ , $a = 2.35\,\frac{m}{s^{2}}$ and $\Delta s = 595\,m$ , the final velocity of the rocket is:

$v = \sqrt{\left(0\,\frac{m}{s} \right)^{2}+2\cdot \left(2.35\,\frac{m}{s^{2}} \right)\cdot (595\,m)}$

$v\approx 52.882\,\frac{m}{s}$

The time associated with this launch ( $t$ ), measured in seconds, is:

$t = \frac{v-v_{o}}{a}$

$t = \frac{52.882\,\frac{m}{s}-0\,\frac{m}{s}}{2.35\,\frac{m}{s} }$

$t = 22.503\,s$

2nd Stage - Gravity

The rocket reaches its maximum height when final velocity is zero:

$v^{2} = v_{o}^{2} + 2\cdot a\cdot (s-s_{o})$ (2)

Where:

$v_{o}$ - Initial speed, measured in meters per second.

$v$ - Final speed, measured in meters per second.

$a$ - Gravitational acceleration, measured in meters per square second.

$s_{o}$ - Initial height, measured in meters.

$s$ - Final height, measured in meters.

If we know that $v_{o} = 52.882\,\frac{m}{s}$ , $v = 0\,\frac{m}{s}$ , $a = -9.807\,\frac{m}{s^{2}}$ and $s_{o} = 595\,m$ , then the maximum height reached by the rocket is:

$v^{2} -v_{o}^{2} = 2\cdot a\cdot (s-s_{o})$

$s-s_{o} = \frac{v^{2}-v_{o}^{2}}{2\cdot a}$

$s = s_{o} + \frac{v^{2}-v_{o}^{2}}{2\cdot a}$

$s = 595\,m + \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(52.882\,\frac{m}{s} \right)^{2}}{2\cdot \left(-9.807\,\frac{m}{s^{2}} \right)}$

$s = 737.577\,m$

The rocket reaches a maximum height of 737.577 meters.

b) The time needed for the rocket to crash down to the launch pad is determined by the following kinematic equation:

$s = s_{o} + v_{o}\cdot t +\frac{1}{2}\cdot a \cdot t^{2}$ (2)

Where:

$s_{o}$ - Initial height, measured in meters.

$s$ - Final height, measured in meters.

$v_{o}$ - Initial speed, measured in meters per second.

$a$ - Gravitational acceleration, measured in meters per square second.

$t$ - Time, measured in seconds.

If we know that $s_{o} = 595\,m$ , $v_{o} = 52.882\,\frac{m}{s}$ , $s = 0\,m$ and $a = -9.807\,\frac{m}{s^{2}}$ , then the time needed by the rocket is:

$0\,m = 595\,m + \left(52.882\,\frac{m}{s} \right)\cdot t + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right)\cdot t^{2}$

$-4.904\cdot t^{2}+52.882\cdot t +595 = 0$

Then, we solve this polynomial by Quadratic Formula:

$t_{1}\approx 17.655\,s$ , $t_{2} \approx -6.872\,s$

Only the first root is solution that is physically reasonable. Hence, the rocket will come crashing down approximately 17.655 seconds after engine failure.

7 0

3 years ago