Question 3 (5 points)

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 light beam travels at 1.94×108 in quartz. The wavelength of the light in quartz is 355 .Part AWhat is the index of refraction

Alja [10]

A) 1.55

The speed of light in a medium is given by:

$v=\frac{c}{n}$

where

$c=3\cdot 10^8 m/s$ is the speed of light in a vacuum

n is the refractive index of the material

In this problem, the speed of light in quartz is

$v=1.94\cdot 10^8 m/s$

So we can re-arrange the previous formula to find n, the index of refraction of quartz:

$n=\frac{c}{v}=\frac{3\cdot 10^8 m/s}{1.94\cdot 10^8 m/s}=1.55$

B) 550.3 nm

The relationship between the wavelength of the light in air and in quartz is

$\lambda=\frac{\lambda_0}{n}$

where

$\lambda$ is the wavelenght in quartz

$\lambda_0$ is the wavelength in air

n is the refractive index

For the light in this problem, we have

$\lambda=355 nm\\n=1.55$

Therefore, we can re-arrange the equation to find $\lambda_0$ , the wavelength in air:

$\lambda_0 = n\lambda=(1.55)(355 nm)=550.3 nm$

4 0

4 years ago

A car travels 36 km in 30 minutes. The speed of the car is

iogann1982 [59]

Answer:

72km/hr

Explanation:

Speed in Km is usually represented in hours. so if the car is in constant velocity, and if the car travels 36km in 30 min then it travels 72km in 1 hour.

so the speed of the car is 72km/hr

7 0

3 years ago

Automobile traveling at 65 mph constant on the road described below. Find rate at which radar must rotate when theta = 15 deg. A

Finger [1]

Answer:

The rate at which radar must rotate is 0.335 rad/s.

Explanation:

Given that,

Velocity = 65 m/h = 29.0576 m/s

Angle = 15°

Suppose, the radius given by

$r=(100\cos2\theta)\ m$

We need to calculate the rate at which radar must rotate

Using formula of linear velocity

$v=r\omega$

$\omega=\dfrac{v}{r}$

Where, v = velocity

r = radius

Put the value into the formula

$\omega=\dfrac{29.0576}{100\cos30}$

$\omega=0.335\ rad/s$

Hence, The rate at which radar must rotate is 0.335 rad/s.

3 0

4 years ago

Will the body be travelling with a constant speed or a constant acceleration if the distance travelled by the body is directly p

Anna35 [415]

Distance ( say d ) is directly proportional to time ( say t ) .
mathematically ,
d = kt
where k is constant if proportionality .
Now differentiating the above equation with respect to time ,
we get ;
v = k
=> velocity is constant .
=> No acceleration .

6 0

4 years ago