Ask question

Ask question

All categories

3 years ago

11

What is the equation that relate electric potential (voltage) to electric field?

1 answer:

erma4kov [3.2K]3 years ago

7 0

V= I x R
I= V / R
r= V / I

You might be interested in

A cellist tunes the C string of her instrument to a fundamental frequency of 65.4 Hz. The vibrating portion of the string is 0.5

Dahasolnce [82]

Answer:

a

$\lambda = 1.18 \ m$

b

$v = 77.172 \ m/s$

c

$T = 151.41 \ N$

Explanation:

From the question we are told that

The frequency is $f = 65.4 \ Hz$

The length of the vibrating string is $L = 0.590 \ m$

The mass is $m = 15.0 \ g = 0.015 \ kg$

Generally the wavelength is mathematically represented as

$\lambda = 2 * L$

=> $\lambda = 2 * 0.590$

=> $\lambda = 1.18 \ m$

Generally the wave speed is

$v = \lambda * f$

=> $v = 1.18 * 65.4$

=> $v = 77.172 \ m/s$

Generally the tension on the wire is mathematically represented as

$T = v^2 * \frac{ m }{L }$

=> $T = 77.172 ^2 * \frac{ 0.015 }{0.590}$

=> $T = 151.41 \ N$

7 0

3 years ago

A test pilot flies with an acceleration of of 5 g . what is the acceleration in meter per second

JulijaS [17]

I mean if he flies 5g that means that's his average speed too

5 0

3 years ago

How long must a simple pendulum be if it is to make exactly one swing per four seconds? (That is, one complete oscillation takes

cricket20 [7]

Answer:

Length of the pendulum will be 3.987 m

Explanation:

We have given time period of the pendulum T = 8 sec

Acceleration due to gravity $g=9.81m/sec^2$

We have to find the length of the simple pendulum

We know that time period of the simple pendulum is given by

$T=2\pi \sqrt{\frac{l}{g}}$

$8=2\times 3.14 \sqrt{\frac{l}{9.81}}$

$l=3.987m$

So length of the pendulum will be 3.987 m

3 0

3 years ago

An airplane is moving at 350 km/hr. If a bomb is

bogdanovich [222]

Answers:

a) -171.402 m/s

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

$y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2}$ (1)

$x=V_{ox}t$ (2)

$V_{f}=V_{oy}-gt$ (3)

Where:

$y=0 m$ is the bomb's final height

$y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m$ is the bomb's initial height

$V_{oy}=0 m/s$ is the bomb's initial vertical velocity, since the airplane was moving horizontally

$t$ is the time

$g=9.8 m/s^{2}$ is the acceleration due gravity

$x$ is the bomb's range

$V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s$ is the bomb's initial horizontal velocity

$V_{f}$ is the bomb's final velocity

Knowing this, let's begin with the answers:

<h3>b) Time </h3>

With the conditions given above, equation (1) is now written as:

$y_{o}=\frac{1}{2}gt^{2}$ (4)

Isolating $t$ :

$t=\sqrt{\frac{2 y_{o}}{g}}$ (5)

$t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}}$ (6)

$t=17.49 s$ (7)

<h3>a) Final velocity </h3>

Since $V_{oy}=0 m/s$ , equation (3) is written as:

$V_{f}=-gt$ (8)

$V_{f}=-(97.22)(17.49 s)$ (9)

$V_{f}=-171.402 m/s$ (10) The negative sign only indicates the direction is downwards

<h3>c) Range </h3>

Substituting (7) in (2):

$x=(97.22 m/s)(17.49 s)$ (11)

$x=1700.99 m$ (12)

5 0

3 years ago

1.) True or False.

eimsori [14]

Answer:

Explanation:

According to Newton's third law of motion, forces always act in equal but opposite pairs. Another way of saying this is for every action, there is an equal but opposite reaction. This means that when you push on a wall, the wall pushes back on you with a force equal in strength to the force you exerted. 1.True 2.falues 3.true 4. not really sure on this one

6 0

2 years ago