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3 years ago

How would you decide what numeric scale to use for each axis?

Physics

1 answer:

faltersainse [42]3 years ago

7 0

Look at at highest number for each x and y value given

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A ball of 0.5kg slows down from 5m/s to 3m/s. Calculate the work done inthe process.

dlinn [17]

Answer:

9.8 Joules (rounded to 2 significant figures)

Explanation:

Work done (J)= Force(N) x distance changed (m)

Force= 9.80665 x 0.5kg
Force= 4.90332 Newtons

Distance changed= 5-3
distance changed= 2m/s

--> work done= 4.90332 x 2

work done= 9.8 Joules

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2 years ago

Light waves

irga5000 [103]

Answer:

Explanation:

Because I searched it up

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3 years ago

Read 2 more answers

How many hydrogen and carbon atoms in a diamond

Wewaii [24]

Answer:

Explanation:

Thus, total 4+4=8 C atoms are present per unit cell of diamond. Carbon has an electronic arrangement of 2,4. In diamond, each carbon shares electrons with four other carbon atoms - forming four single bonds.

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3 years ago

Kevin goes bowling. Whenever he bowls the ball, he transfers energy from his hand to the bowling ball. The amount of energy befo

g100num [7]

Answer:the answer is C

Explanation:

3 0

3 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