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3 years ago

An electron moves 5 m in the direction of an electric field of strength 300 N/C. The change in electrical potential energy is:

1 answer:

6 0

First we need to find the voltage difference between the initial and final location of the electron.
Since the electron travelled for a distance d=5 m in an uniform electric field of intensity E=300 N/C, the voltage difference between the final and initial location is
$\Delta V= Ed=(300 N/C)(5 m)=1500 V$

And then, we can calculate the change in potential energy of the electron, which is the product between the charge of the electron and the voltage difference:
$\Delta U = q\Delta V=(-1.6 \cdot 10^{-19}C)(1500 V)=-2.4 \cdot 10^{-16}J$
And the negative sign is due to the fact that we assumed the electron traveled in the natural direction of the electric field, so traveling from a point at lower voltage to a point at higher voltage (the sign of $\Delta V$ is positive), so since it's a negative charge the electron is losing potential energy.

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If you put a object in front of the mirror why does it appear the opposite way

UNO [17]

It is because that is how mirrors work, they reflect light, and since we see objects because we are seeing the light these objects reflect, what is reflected back by the mirror is what we see.

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2 years ago

A coil of 40 turns is wrapped around a long solenoid of cross-sectional area 7.5×10−3m2. The solenoid is 0.50 m long and has 500

defon

To solve this problem it is necessary to apply the concepts related to mutual inductance in a solenoid.

This definition is described in the following equation as,

$M = \frac{\mu_0 N_1 N_2A_1}{l_1}$

Where,

$\mu =$ permeability of free space

$N_1 =$ Number of turns in solenoid 1

$N_2 =$ Number of turns in solenoid 2

$A_1=$ Cross sectional area of solenoid

l = Length of the solenoid

Part A )

Our values are given as,

$\mu_0 = 4\pi *10^{-7}H/m$

$N_1 = 500$

$N_2 = 40$

$A = 7.5*10^{-4}m^2$

$l = 0.5m$

Substituting,

$M = \frac{\mu_0 N_1 N_2A_2}{l_1}$

$M = \frac{(4\pi *10^{-7})(500)(40)(7.5*10^{-4})}{0.5}$

$M = 3.77*10^{-4}H$

PART B) Considering that many of the variables remain unchanged in the second solenoid, such as the increase in the radius or magnetic field, we can conclude that mutual inducantia will appear the same.

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3 years ago

A flywheel is a solid disk that rotates about an axis that is perpendicular to the disk at its center. Rotating flywheels provid

Sergeeva-Olga [200]

Answer:

$8.37*10^5$ rpm

Explanation:

Given that rotational kinetic energy = $4.66*10^9J$

Mass of the fly wheel (m) = 19.7 kg

Radius of the fly wheel (r) = 0.351 m

Moment of inertia (I) = $\frac{1}{2} mr ^2$

Rotational K.E is illustrated as $(K.E)_{rt} = \frac{1}{2} I \omega^2$

$\omega = \sqrt{\frac{2(K.E)_{rt}}{I} }$

$\omega = \sqrt{\frac{2(KE)_{rt}}{1/2 mr^2} }$

$\omega = \sqrt{\frac{4(K.E)_{rt}}{mr^2} }$

$\omega = \sqrt{\frac{4*4.66*10^9J}{19.7kg*(0.351)^2} }$

$\omega = 87636.04$

$\omega = 8.76*10^4 rad/s$

Since 1 rpm = $\frac{2 \pi}{60} rad/s$

$\omega = 8.76*10^4(\frac{60}{2 \pi})$

$\omega = 836518.38$

$\omega = 8.37 *10^5 rpm$

3 0

2 years ago

When the core of a star like the sun uses up its supply of hydrogen for fusion, the core begins to ________?

pav-90 [236]

When the core of a star like the sun uses up its supply of hydrogen for fusion, the core begins to contract in size. The gravitational forces becomes strong and cause the star to reduce in size. as the star contracts, the temperature of the star also rises since the heat is distributed inside smaller Volume now.

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3 years ago

Read 2 more answers

What is the force that the gas exerts on each of the six sides of the box when the gas temperature is 20.0∘C?

Taya2010 [7]

Incomplete question as number of moles and length is missing.So I have assumed 3 moles and length of 0.300 m.So the complete question is here:

Three moles of an ideal gas are in a rigid cubical box with sides of length 0.300 m.What is the force that the gas exerts on each of the six sides of the box when the gas temperature is 20.0∘C?

Answer:

The Force act on each side is 2.43×10⁴N

Explanation:

Given data

n=3 mol

L=0.3 m

Temperature=20.0°C=293 K

To find

Force F

Solution

To get force act on each side it would employ by

F=P.A

Where P is pressure

A is Area

First we need to find pressure by applying ideal gas law

$P.V=nRT\\P=\frac{nRT}{V}\\ P=\frac{(3mol)(8.315J/mol)(293K)}{(0.3m*0.3m*0.3m)}\\P=27.069*10^{4}Pa$

So The Force is given as:

$F=P.A\\F=(27.069*10^{4} )(0.3m*0.3m)\\F=2.43*10^{4}N$

The Force act on each side is 2.43×10⁴N

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2 years ago