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3 years ago

Calculate the magnitude of the focal length and power of a spherical mirror that has a radius of 5.00 cm.

Physics

1 answer:

yarga [219]3 years ago

5 0

Answer:

2.5 cm

40 D

Explanation:

When the radius of curvature of a lens is divided by 2 we get the focal length of the lens.

Focal length is given by

$f=\dfrac{R}{2}\\\Rightarrow f=\dfrac{5}{2}\\\Rightarrow f=2.5\ cm$

The focal length of the lens is 2.5 cm

When we divide 1 by the focal length in the unit of meters we get the power of a lens

Power of a lens is given by

$P=\dfrac{1}{f}\\\Rightarrow P=\dfrac{1}{0.025}\\\Rightarrow P=40\ D$

The power of the lens is 40 D

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An electron and a proton are each placed at rest in an electric field of 500 N/C. Calculate the speed (and indicate the directio

Bumek [7]

Answer:

For proton: 2592 m/s In the same direction of electric field.

For electron: 4752000 m/s In the opposite direction of electric field.

Explanation:

E = 500 N/C, t = 54 ns = 54 x 10^-9 s,

Acceleration = Force /mass

Acceleration of proton, ap = q E / mp

ap = (1.6 x 10^-19 x 500) / (1.67 x 10^-27) = 4.8 x 10^10 m/s^2

Acceleration of electron, ae = q E / me

ae = (1.6 x 10^-19 x 500) / (9.1 x 10^-31) = 8.8 x 10^13 m/s^2

For proton:

u = 0, ap = 4.8 x 10^10 m/s^2, t = 54 x 10^-9 s

use first equation of motion

v = u + at

vp = 0 + 4.8 x 10^10 x 54 x 10^-9 = 2592 m/s In the same direction of electric field.

For electron:

u = 0, ae = 8.8 x 10^13 m/s^2, t = 54 x 10^-9 s

use first equation of motion

v = u + at

vp = 0 + 8.8 x 10^13 x 54 x 10^-9 = 4752000 m/s In the opposite direction of electric field.

6 0

3 years ago

While entering a freeway, a car accelerates from rest at a rate of 2.40 m/s2 for 12.0 s. (a) Draw a sketch of the situation. (b)

ArbitrLikvidat [17]

Answer:

a) See attached picture, b) We know the initial velocity = 0, initial position=0, time=12.0s, acceleration= $2.40m/s^{2}$ , c) the car travels 172.8m in those 12 seconds, d) The car's final velocity is 28.8m/s

Explanation:

a) In order to draw a sketch of the situation, I must include the data I know, the data I would like to know and a drawing of the car including the direction of the movement and its acceleration, just like in the attached picture.

b) From the information given by the problem I know:

initial velocity =0

acceleration = $2.40m/s^{2}$

time = 12.0 s

initial position = 0

unknown:

displacement.

in order to choose the appropriate equation, I must take the knowns and the unknown and look for a formula I can use to solve for the unknown. I know the initial velocity, initial position, time, acceleration and I want to find out the displacement. The formula that contains all this data is the following:

$x=x_{0}+V_{x0}t+\frac{1}{2}a_{x}t^{2}$

Once I got the equation I need to find the displacement, I can plug the known values in, like this:

$x=0+0(12s)+\frac{1}{2}(2.40\frac{m}{s^{2}} )(12s)^{2}$

after cancelling the pertinent units, I get that my answer will be given in meters. So I get:

$x=\frac{1}{2} (2.40\frac{m}{s^{2}} )(12s)^{2}$

which solves to:

$x=172.8m$

So the displacement of the car in 12 seconds is 172.8m, which makes sense taking into account that it will be accelerating for 12 seconds and each second its velocity will increase by 2.4m/s.

d) So, like the previous part of the problem, I know the initial position of the car, the time it travels, the initial velocity and its acceleration. Now I also know what its final position is, so we have more than enough information to find this answer out.

I need to find the final velocity, so I need to use an equation that will use some or all of the known data and the unknown. In order to solve this problem, I can use the following equation:

$a=\frac{V_{f}-V_{0} }{t}$

Next, since I need to find the final velocity, I can solve the equation just for that, I can start by multiplying both sides by t so I get:

$at=V_{f}-V_{0}$

and finally I can add $V_{0}$ to both sides so I get:

$V_{f}=at+V_{0}$

and now I can proceed and substitute the known values:

$V_{f}=at+V_{0}$

$V_{f}=(2.40\frac{m}{s^{2}}} (12s)+0$

which solves to:

$V_{f}=28.8m/s$

8 0

2 years ago

Read 2 more answers

Assume light is traveling through an optical medium of n1 and is incident on a boundary with a different optical medium with n2.

Elza [17]

That n2 = 2*n1. That is, the index of refraction is twice as big in medium 2 since v=c/n

5 0

2 years ago

Why do you keep moving forward when you slam on the brakes of your bike?

g100num [7]

Because the act of braking is an example of negative acceleration.
Example: if the rate of braking was say 2 meters per second^2, and the starting velocity was 10 m/s, it would take 5 seconds to come to a stop(during those 5 seconds you would still be moving).

6 0

3 years ago

Read 2 more answers

A stuntman of mass 48 kg is to be launched horizontally out of a spring-

Damm [24]

The velocity of the stuntman, once he has left the cannon is 5 m/s.

The right option is O A. 5 m/s

The Kinetic energy of the stuntman is equal to the elastic potential energy of the spring.

<h3 /><h3>Velocity: </h3>

This is the ratio of displacement to time. The S.I unit of Velocity is m/s. The velocity of the stuntman can be calculated using the formula below.

⇒ Formula:

mv²/2 = ke²/2
mv² = ke².................. Equation 1

⇒ Where:

m = mass of the stuntman
v = velocity of the stuntman
k = force constant of the spring
e = compression of the spring

⇒ Make v the subject of the equation

v = √(ke²/m)................. Equation 2

From the question,

⇒ Given:

m = 48 kg
k = 75 N/m
e = 4 m

⇒ Substitute these values into equation 2

v = √[(75×4²)/48]
v = √25
v = 5 m/s.

Hence, The velocity of the stuntman, once he has left the cannon is 5 m/s.

The right option is O A. 5 m/s

Learn more about velocity here: brainly.com/question/10962624

6 0

2 years ago