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3 years ago

Which of the following is most appropriate to follow motion in one dimension?

1 answer:

5 0

The linear scale is applicable only as it moves in one dimension. From the word "linear" it means it deals with one equation only. Unlike the other options, the dimensions are many because it involves 2 or more variables for its equation.

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Formula for working maximum velocity?

Varvara68 [4.7K]

Answer:

v=Aω.

Explanation:

hope this helps you!!

8 0

3 years ago

Read 2 more answers

1<br> A truck increases its speed from 15 m/s to 60 m/s in 15 s. Its acceleration is

MAVERICK [17]

Acceleration = (Vf - Vi)/t
Since Vf= 60m/s
Vi= 15m/s
T= 15s
=> a= (60m/s - 15m/s)/15s
= 3
So the acceleration is 3m/s^2

3 0

3 years ago

A satellite completes one revolution of a planet in almost exactly one hour. At the end of one hour, the satellite has traveled

topjm [15]

average velocity is vector displacement / time

time is "almost exactly one hour"

disp = -10m

v= -10/1x60x60 = -1/360m/s

5 0

3 years ago

A grandfather clock is "losing" time because its pendulum moves too slowly. Assume that the pendulum is a massive bob at the end

IgorC [24]

Answer:

d) shortening the string

Explanation:

Time period of a pendulum clock is dependent on two factors namely:length and acceleration due to gravity.

When a clock loses time, the time period of the pendulum clock increases.

This however can be corrected by decreasing the length of the pendulum.The time period of the pendulum clock is not dependent on the mass of the bob. The time period of the pendulum clock can be corrected only by changing the length of the pendulum string.

5 0

3 years ago

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A 1.0-cm-tall object is 13 cm in front of a converging lens that has a 40 cm focal length.

kicyunya [14]

A) Image position: -19.3 cm

B) Image height: 1.5 cm, upright

Explanation:

In order to calculate the image position, we can use the lens equation:

$\frac{1}{p}+\frac{1}{q}=\frac{1}{f}$

where

p is the distance of the object from the lens

q is the distance of the image from the lens

f is the focal length

In this problem, we have:

p = 13 cm (object distance)

f = 40 cm (focal length, positive for a converging lens)

So the image distance is

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{40}-\frac{1}{13}=-0.0519\\q=\frac{1}{-0.0519}=-19.3 cm$

The negative sign means that the image is virtual.

In order to calculate the image height, we use the magnification equation:

$\frac{y'}{y}=-\frac{q}{p}$

where

y' is the image height

y is the object height

In this problem, we have:

y = 1.0 cm (object height)

p = 13 cm

q = -19.3 cm

Therefore, the image heigth is

$y'=-\frac{qy}{p}=-\frac{(-19.3)(1.0)}{13}=1.5 cm$

And the positive sign means the image is upright.

6 0

3 years ago