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3 years ago

5

A variable is something that _____.

1 answer:

babymother [125]3 years ago

5 0

Answer:

<h2>Changes</h2>

Explanation:

<h3>Variable is something that varies and doesn't remain constant.</h3>

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Two falling inflated balls of different masses<br> land at the same time.

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3 years ago

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3 years ago

Read 2 more answers

A 3.0-kg block starts at rest at the top of a 37° incline, which is 5.0 m long. Its speed when it reaches the bottom is 2.0 m/s.

Mama L [17]

Answer: $f_{r}$ = 16.49N

Explanation: The object is placed on an inclined plane at an angle of 37° thus making it weight have two component,

$W_{x}$ = horizontal component of the weight = mgsinФ

$W_{y}$ = vertical component of weight = mgcosФ

Due to the way the object is positioned, the horizontal component of force will accelerate the object thus acting as an applied force.

by using newton's law of motion, we have that

mgsinФ - $f_{r}$ = ma

where m = mass of object=5kg

a = acceleration= unknown

Ф = angle of inclination = 37°

g = acceleration due to gravity = 9.8 $m/s^{2}$

$f_{r}$ = frictional force = unknown

we need to first get the acceleration before the frictional force which is gotten by using the equation below

$v^{2} = u^{2} + 2aS$

where v = final velocity = 2m/s

u = initial velocity = 0m/s (because the object started from rest)

a= unknown

S= distance covered = length of plane = 5m

$2^{2} = 0^{2} + 2*a*5\\\\4= 10 *a\\\\a = \frac{4}{10} \\a = 0.4m/s^{2}$

we slot in a into the equation below to get frictional force

mgsinФ - $f_{r}$ = ma

3 * 9.8 * sin 37 - $f_{r}$ = 3* 0.4

17.9633 - $f_{r}$ = 1.2

$f_{r}$ = 17.9633 - 1.2

$f_{r}$ = 16.49N

4 0

3 years ago

A source at rest emits light of wavelength 500 nm. When it is moving at 0.90c toward an observer, the observer detects light of

Vesna [10]

Answer:

The observer detects light of wavelength is 115 nm.

(b) is correct option

Explanation:

Given that,

Wavelength of source = 500 nm

Velocity = 0.90 c

We need to calculate the wavelength of observer

Using Doppler effect

$\lambda_{o}=\sqrt{\dfrac{1-\beta}{1+\beta}}\lambda_{s}$

Where, $\beta=\dfrac{c}{v}$

$\lambda_{o}=\sqrt{\dfrac{c-0.90c}{c+0.90c}}\times500\times10^{-9}$

$\lambda_{o}=115\ nm$

Hence, The observer detects light of wavelength is 115 nm.

8 0

3 years ago