What is a non example of linear relationship

Physics

1 answer:

xxTIMURxx [149]3 years ago

4 0

Neither the speed nor the distance of a falling object is linearly related to time.

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A 0.68 kg squirrel is resting on a branch 8 meters above the ground. What is the gravitational potential energy of a squirrel? A

ANEK [815]

Answer:

The gravitational potential energy of a squirrel is 53.312 J.

Explanation:

We have,

Mass of a squirrel is 0.68 kg

It is placed at a height of 8 m above the ground.

It is required to find the gravitational potential energy of a squirrel. It is possessed by an object due to its position. Its formula is given by :

$E=mgh\\\\E=0.68\times 9.8\times 8\\\\E=53.312\ J$

So, the gravitational potential energy of a squirrel is 53.312 J.

7 0

3 years ago

Where could convection currents form? Check all that apply.

Sladkaya [172]

In a fresh water lake
atmosphere
earths mantle

8 0

4 years ago

Read 2 more answers

A 10-cm-long archerfish begins such a leap with its body vertical in the water, using its fins and undulating its body to propel

siniylev [52]

Answer:

Explanation:

Velocity will be equal to the area under the curve of the acceleration-time plot.

v = ½(45)(0.075) = 1.6875

v = 1.7 m/s

h = (0² - 1.6875²) / (2(-9.8)) = 0.145288...

h = 15 cm

3 0

3 years ago

The radius of an atom of krypton (kr) is about 1.9 å. (a) express this distance in nanometers (nm). nm express this distance in

I am Lyosha [343]

Note:
1 A (armstrong) = 10⁻¹⁰ m
1 nm (nanometer) = 10⁻⁹ m

Given:
Radius of a krypton atom = 1.9 A = 1.9 x 10⁻¹⁰ m

Part (a)
$1.9\,A = (1.9\,A)*(10^{-10}\, \frac{m}{A})*( \frac{1}{10^{-8}} \frac{nm}{m}) } =0.019\,nm$
Answer: 0.019 nm

Part (b)
The diameter of a krypton atom = 2*1.9A = 3.8 A = 3.8 x 10⁻¹⁰ m.
The number of krypton atoms within a length of 1.0 mm is
$\frac{1.0\, mm}{3.8 \time 10^{-10}\, m} = \frac{10^{-3}\, m}{3.8 \times 10^{-10} \,m} =2.632 \times 10^{6}$

Answer: About 2.632 x 10⁶ atoms

Part (c)
The radius of a krypton atom is
1.9 A = (1.9 x 10⁻¹⁰ m)*(10² cm/m) = 1.9 x 10⁻⁸ cm
The volume of a krypton atm is
$\frac{4 \pi }{3} (1.9 \times 10^{-8} \, cm)^{3} = 2.873 \times 10^{-23} \, cm^{3}$

Answer: 2.873 x 10⁻²³ cm³

8 0

3 years ago

A cannon shoots an artillery shell with a initial velocity of 400 meters/second at an indirect fire angle of 60 on level ground

notsponge [240]

Answer: 14139.19 m

Explanation:

This situation is related to parabolic motion and can be solved using the following equations:

$x=V_{o}cos \theta t$ (1)