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3 years ago

Which situation would deplete freshwater?

Physics

1 answer:

sasho [114]3 years ago

5 0

Answer:

If freshwater consumption was greater than freshwater renewal.

Explanation:

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Two vectors are being added, one at an angle of 20.0 , and the other at 80.0. The only thing you know about the magnitudes is th

postnew [5]

Answer:a

Explanation: they are all positive

4 0

3 years ago

Julianne went to a restaurant to have a taste of her favorite fried chicken and spaghetti. She drove 2 km, east and then 8.5 km,

GREYUIT [131]

Julianne’s displacement from her origin is equal to 10.015 kilometers.

Given the following data:

Distance A = 2 km, East.

Distance B = 8.5 km, Northeast.

To calculate Julianne’s displacement from her origin:

<h3>How to calculate displacement.</h3>

We would denote the two (2) unit vectors along the East and Northeast directions by i and j respectively.

Note: Northeast is at angle of 45° with the East.

In terms of vectors, the distances becomes:

Distance A = 2i

$Distance\;B=8.5 [(cos 45i + sin 45j)]\\\\Distance\;B=(\frac{8.5}{\sqrt{2} } i \;+\;\frac{8.5}{\sqrt{2} } j)$

For the resultant displacement:

$D^2 = [(2+\frac{8.5}{\sqrt{2} } )^2+ (\frac{8.5}{\sqrt{2} } )^2\\\\D =\sqrt{[(2+\frac{8.5}{\sqrt{2} } )^2+ (\frac{8.5}{\sqrt{2} } )^2} \\\\D=2+\frac{8.5}{\sqrt{2} } + \frac{8.5}{\sqrt{2} }$

D = 10.015 kilometers.

Read more on displacement here: brainly.com/question/13416288

5 0

2 years ago

To test the performance of its tires, a car

velikii [3]

Answer:

The coefficient of static friction between the tires and the road is 1.987

Explanation:

Given:

Radius of the track, r = 516 m

Tangential Acceleration $a_r$ = 3.89 m/s^2

Speed,v = 32.8 m/s

To Find:

The coefficient of static friction between the tires and the road = ?

Solution:

The radial Acceleration is given by,

$a_{R = \frac{v^2}{r}$

$a_{R = \frac{(32.8)^2}{516}$

$a_{R = \frac{(1075.84)}{516}$

$a_{R = 2.085 m/s^2$

Now the total acceleration is

$\text{ total acceleration} = \sqrt{\text{(tangential acceleration)}^2 +{\text{(Radial acceleration)}^2$

=> $= \sqrt{ (a_r)^2+(a_R)^2}$

=> $\sqrt{ (3.89 )^2+( 2.085)^2}$

=> $\sqrt{ (15.1321)+(4.347)^2}$

=> $19.4791 m/s^2$

The frictional force on the car will be f = ma------------(1)

And the force due to gravity is W = mg--------------------(2)

Now the coefficient of static friction is

$\mu =\frac{f}{W}$

From (1) and (2)

$\mu =\frac{ma}{mg}$

$\mu =\frac{a}{g}$

Substituting the values, we get

$\mu =\frac{19.4791}{9.8}$

$\mu =1.987$

8 0

3 years ago

Find the resultant of an easterly force of 100 N and a southeast force of 80 N acting at 65 degrees to the 100 N force

saveliy_v [14]

Answer:

Resultant is 152 N at 28.5 degrees south to the 100 N force

Explanation:

7 0

3 years ago

أسقط عامل حجرة من سطح بناية سقوطأ حرة. أوجد ما يلي :

Ilia_Sergeevich [38]

Answer:

??? i don't no what you just said

Explanation:

7 0

2 years ago