Temperature is a measure of

Two charges, Q1 and Q2, are separated by 6·cm. The repulsive force between them is 25·N. In each case below, find the force betw

Misha Larkins [42]

Answer:

a) 5 N b) 225 N c) 5 N

Explanation:

a) Per Coulomb's Law the repulsive force between 2 equal sign charges, is directly proportional to the product of the charges, and inversely proportional to the square of the distance between them, acting along the line that joins the charges, as follows:

F₁₂ = K Q₁ Q₂ / r₁₂²

So, if we make Q1 = Q1/5, the net effect will be to reduce the force in the same factor, i.e. F₁₂ = 25 N / 5 = 5 N

b) If we reduce the distance, from r, to r/3, as the factor is squared, the net effect will be to increase the force in a factor equal to 3² = 9.

So, we will have F₁₂ = 9. 25 N = 225 N

c) If we make Q2 = 5Q2, the force would be increased 5 times, but if at the same , we increase the distance 5 times, as the factor is squared, the net factor will be 5/25 = 1/5, so we will have:

F₁₂ = 25 N .1/5 = 5 N

3 0

3 years ago

A body which has surface area 5cm² and temperature of 727°C radiates 300J of energy in one minute. Calculate it's emissivity giv

cestrela7 [59]

<h2>Answer: 0.17</h2>

Explanation:

The Stefan-Boltzmann law establishes that a black body (an ideal body that absorbs or emits all the radiation that incides on it) "emits thermal radiation with a total hemispheric emissive power proportional to the fourth power of its temperature":

$P=\sigma A T^{4}$ (1)

Where:

$P=300J/min=5J/s=5W$ is the energy radiated by a blackbody radiator per second, per unit area (in Watts). Knowing $1W=\frac{1Joule}{second}=1\frac{J}{s}$

$\sigma=5.6703(10)^{-8}\frac{W}{m^{2} K^{4}}$ is the Stefan-Boltzmann's constant.

$A=5cm^{2}=0.0005m^{2}$ is the Surface area of the body

$T=727\°C=1000.15K$ is the effective temperature of the body (its surface absolute temperature) in Kelvin.

However, there is no ideal black body (ideal radiator) although the radiation of stars like our Sun is quite close. So, in the case of this body, we will use the Stefan-Boltzmann law for real radiator bodies:

$P=\sigma A \epsilon T^{4}$ (2)

Where $\epsilon$ is the body's emissivity

(the value we want to find)

Isolating $\epsilon$ from (2):

$\epsilon=\frac{P}{\sigma A T^{4}}$ (3)

Solving:

$\epsilon=\frac{5W}{(5.6703(10)^{-8}\frac{W}{m^{2} K^{4}})(0.0005m^{2})(1000.15K)^{4}}$ (4)

Finally:

$\epsilon=0.17$ (5) This is the body's emissivity

3 0

2 years ago

Two parallel metal plates are at a distance of 8.00 m apart.The electric field between the plates is uniform directed towards th

yuradex [85]

Answer:

C

Explanation:

Formula E=F/C also E=V/d

In this case use the second formula; E=V/d

Data given; E=4N/C d=8m

So v=E X d

V=4x8=32V

k.e=eV= 2X32=64eV

3 0

3 years ago

Six keplerian element table

Vikentia [17]

That’s the answer hope you enjoy

7 0

2 years ago

An elevator and its load have a combined mass of 1650 kg. Find the tension in the supporting cable when the elevator, originally

gizmo_the_mogwai [7]

Answer:

Tension in the supporting cable is = 4,866 N ≅4.9 KN

Explanation:

First of all, we need to understand that tension is a force, so the motion law

F = Ma applies perfectly.

From Newtons third law of motion, action and reaction are equal and opposite. This means that the force experienced by the elevator, is equal to the tension experienced by the spring.

Parameters given:

Mass of load = 1650 kg

Acceleration of load = ?

The acceleration of the load can be obtained by diving the change in velocity by the time taken. But we need to know the time taken for the motion to 41 m.

Time taken = distance covered / velocity

= $\frac{41m}{11m/s}$ = 3.73 seconds

∴Acceleration = ( initial velocity - final velocity )/ time taken

Note: Final velocity is = 0 since the body came to a rest.

Acceleration = $\frac{11 - 0 m/s}{3.73s}$ = 2.95m/ $s^{2}$

Force acting on the cable = mass of elevator × acceleration of elevator

= 1650 × 2.95 = 4869.5 kg ≅ 4.9 KN

6 0

2 years ago