I need help for a review sheet

When you take pictures with a camera, the distance between the lens and the film or chip has to be adjusted, depending on the di

Shalnov [3]

Answer:

Explanation:

When a camera shifts focus from a faraway object to a nearby object, the lens-to-film distance must increase. Likewise, when it shifts focus from a nearby object to a distant object, there must be an increase in the lens to film distance (that is, the image distance).

Therefore, if the picture of an object that is far away, the lens must move towards the film.

The focal length cannot be changed because it is fixed for a lens. Nevertheless, in order to focus on an object, the image distance can be changed.

8 0

4 years ago

A horizontal disk with a radius of 23 m rotates about a vertical axis through its center. The disk starts from rest and has a co

lys-0071 [83]

Answer:

time is 0.42 sec

Explanation:

Given data

radius = 23 m

angular acceleration = 5.7 rad/s²

to find out

time

solution

we know that radius is constant so that

tangential acceleration At = angular acceleration × radius ............. 1

tangential acceleration = 5.7 × 23 = 131.1 m/s²

and

radial acceleration Ar = (angular velocity)² × radius ........................2

we consider angular velocity = ω

this is acting toward center

so

compare 1 and 2

At = Ar

5.7 r =ω³ r

ω = √5.7 = 2.38746 rad/s

so

ω = 5.7 t

2.387 = 5.7 t

t = 2.387 / 5.7

t = 0.4187

time is 0.42 sec

8 0

3 years ago

Two blocks A and B have a weight of 11 lb and 5 lb , respectively. They are resting on the incline for which the coefficients of

Alchen [17]

Answer:

$\theta=10.20^{\circ}$

$\Delta l=0.10 ft$

Explanation:

First of all, we analyze the system of blocks before starting to move.

$\Sum F_{x}=P_{A}sin(\theta)+P_{B}sin(\theta)-F_{fA}-F_{fB}=0$

$\Sum F_{x}=11sin(\theta)+5sin(\theta)-0.16N_{A}-0.23N_{B}=0$

$11sin(\theta)+5sin(\theta)-0.16P_{A}cos(\theta)-0.23P_{B}cos(\theta)=0$

$11sin(\theta)+5sin(\theta)-0.16*11cos(\theta)-0.23*5cos(\theta)=0$

$16sin(\theta)-2.91cos(\theta)=0$

$tan(\theta)=0.18$

$\theta=arctan(0.18)$

$\theta=10.20^{\circ}$

Hence, the incline angle θ for which both blocks begin to slide is 10.20°.

Now, if we do a free body diagram of block A we have that after the block moves, the spring force must be taken into account.

$P_{A}sin(\theta)-F_{fA}-F_{spring}=0$

Where:

$F_{spring} = k\Delta l=2.1\Delta l$

$P_{A}sin(\theta)-0.16*11cos(\theta)-2.1\Delta l=0$

$\Delta l=\frac{11sin(\theta)-0.16*11cos(\theta)}{2.1}$

$\Delta l=0.10 ft$

Therefore, the required stretch or compression in the connecting spring is 0.10 ft.

I hope it helps you!

4 0

4 years ago

Which is a unit rate

avanturin [10]

D. 18 miles driven in a week

6 0

3 years ago

Read 2 more answers

What is the average acceleration of the object represented in the graph above over the 10 seconds? I know the answer isn’t C

polet [3.4K]

It's always a good idea to start with the definition of the thing you're trying to find.

This problem is just trying to find out whether you KNOW the definition of acceleration. You may know it, but you haven't used it yet.

Average acceleration = (change in velocity) divided by (total time).

Change in velocity = (end value) minus (start value)

Change in velocity = (20m/s) - 0

Change in velocity = 20 m/s

Time = 10 s

Average acceleration = (20m/s)/(10s)

Average acceleration = 2 m/s^2

3 0

3 years ago

Read 2 more answers