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3 years ago

How is velocity different from speed? Based on distance and/or direction

Physics

1 answer:

Mnenie [13.5K]3 years ago

6 0

Answer:

Speed is solved with time and distance but has no direction

Average velocity is solved with Δx/Δt and has a direction

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1. You place an object 63 cm in front of a converging lens, with a 40 cm focal length.

NikAS [45]

Answer:

109.6 cm , - 1.74 , real

1.5

Explanation:

d₀ = object distance = 63 cm

f = focal length of the lens = 40 cm

d = image distance = ?

using the lens equation

$\frac{1}{f} = \frac{1}{d_{o}} + \frac{1}{d}$

$\frac{1}{40} = \frac{1}{63} + \frac{1}{d}$

d = 109.6 cm

magnification is given as

$m = \frac{-d}{d_{o}}$

$m = \frac{-109.6}{63}$

m = - 1.74

The image is real

d₀ = object distance = a

d = image distance = - (a + 5)

f = focal length of lens = 30 cm

using the lens equation

$\frac{1}{f} = \frac{1}{d_{o}} + \frac{1}{d}$

$\frac{1}{30} = \frac{1}{a} + \frac{1}{- (a + 5)}$

a = 10

magnification is given as

$m = \frac{-d}{d_{o}}$

$m = \frac{- (- (a +5))}{a}$

$m = \frac{(5 + 10)}{10}$

m = 1.5

6 0

3 years ago

Atoms can join together to form ______ substance

ELEN [110]

Answer:

Molecules

Explanation:

7 0

3 years ago

A wave has a wavelength of 20 mm and a frequency of 5 Hz what is the speed?

yKpoI14uk [10]

The answer is 100mm/s. I hope this helps :)

7 0

3 years ago

What type of mirror can produce both<br> converging and diverging rays?

Ivahew [28]

Answer:

A convex mirror is a diverging mirror (f is negative) and forms only one type of image. It is a case 3 image—one that is upright and smaller than the object, just as for diverging lenses.

Explanation:

hope this helps have a good night :)

4 0

3 years ago

Read 2 more answers

Answer True or Flase1-Electric potential due to a uniform E field doesn’t change with location.2-The equipotential surfaces asso

TEA [102]

Answer:

1. False

2. True

3. True

Explanation:

1- False —> The relation between electric potential and electric field is given such that

$-\int\limits^a_b \vec{E}d\vec{l} = V_{ab}$

Therefore, for a uniform E field, electric potential is linearly proportional to the distance.

2- True —> The electric field lines always cross the equipotential lines perpendicularly.

3- True —> In order to be a potential difference, one source of electric field is enough. The electric potential will decrease radially according to the following formula:

$V = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}$

There is no test charge in the formula, only the source charge. Even when there is no test charge, the potential difference between points in space can exist.

3 0

3 years ago