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2 years ago

6

An object with a mass of 0.5 kilometre start from rest and achieves a maximum speed of 20 metre per second in 0.01 second, what

average unbalanced force accelerated this object

1 answer:

Katarina [22]2 years ago

3 0

Answer:

Hiii how are you <u>doing?</u><u>?</u><u>I </u><u>don't</u><u> </u><u>understand</u><u> </u><u>that</u>

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A 400.0 kg storage box is held 10 m above ground by a forklift. What is its gravitational potential energy? (PE = mgh)

castortr0y [4]

Answer: D.) 39,200 J

Via the equation of potential energy PE = mgh where m is mass, g is the average gravity on earth and h is the height. In this case m = 400 kg, g = 9.8, h = 10 m thus:

$P.E.=(400kg)(9.8\frac{m}{s^2} )(10m)=39,200 J$

P.E.= 39,200 Joules

7 0

3 years ago

Read 2 more answers

The graph below shows the position of an ant as it crawls over a flat picnic blanket. The total time for the ant to go from the

Mice21 [21]

The average speed of the ant is 0.276 cm/s and the average velocity is 0.136 cm/s.

The correct answer is option D.

In the given graph, we can deduce the following;

the total time of the motion, = 1 mins + 45 s = 60 s + 45 s = 105 s

The average speed of the ant is calculated as;

$average \ speed = \frac{total \ distance }{total \ time }$

The total distance from the graph is calculated as follows;

first horizontal distance from 2 cm to 8 cm = 8 - 2 = 6 cm

first upward distance from 3 cm to 5 cm = 5 - 3 = 2 cm

second horizontal distance from 8 cm to 6 cm = 8 - 6 = 2 cm

second upward distance from 5 cm to 12 cm = 12 - 5 = 7 cm

third horizontal distance from 6 cm to 13 cm = 13 - 6 = 7 cm

fourth downward distance from 12 cm to 9 cm = 3 cm

final horizontal distance from 13 cm to 15 cm = 2cm

The total distance = (6 + 2 + 2 + 7 + 7 + 3 + 2) cm = 29 cm

$average \ speed = \frac{total \ distance }{total \ time } = \frac{29 \ cm}{105 \ s} = 0.276 \ cm/s$

The average velocity is calculated as the change in displacement per change in time.

The displacement is the shortest distance between the start and end positions.

This shortest distance is the straight line connecting the start and end position. Call this line P

From the end position at x = 15 cm, draw a vertical line from y = 9 cm, to y = 3 cm. The displacement = 9 cm - 3 cm = 6 cm

Also, draw a horizontal line from start at x = 2 cm to x = 15 cm. The displacement = 15 cm - 2 cm = 13 cm

Notice, you have a right triangle, now calculate the length of line P.

↓end

↓

↓ 6cm

↓

start -------------13 cm------------

Use Pythagoras theorem to solve for P.

$P^2 = 6^2 + 13^2\\\\P^2 = 36 + 169\\\\P^2 = 205\\\\P= \sqrt{205} \\\\P = 14.318 \ cm$

The average velocity of the ant is calculated as;

$average \ velocity= \frac{\Delta displacemnt }{total\ time }= \frac{14.318 \ cm}{105 \ s} = 0.136 \ cm/s \\\\$

Thus, the average speed of the ant is 0.276 cm/s and the average velocity is 0.136 cm/s.

Learn more here: brainly.com/question/589950

5 0

3 years ago

The stopcock connecting a 2.14 L bulb containing oxygen gas at a pressure of 8.19 atm, and a 9.84 L bulb containing krypton gas

marshall27 [118]

Answer : The final pressure of the system in atm is, 3.64 atm

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

$P\propto \frac{1}{V}$

or,

$P_1V_1+P_2V_2=P_fV_f$

where,

$P_1$ = first pressure = 8.19 atm

$P_2$ = second pressure = 2.65 atm

$V_1$ = first volume = 2.14 L

$V_2$ = second volume = 9.84 L

$P_f$ = final pressure = ?

$V_f$ = final volume = 2.14 L + 9.84 L = 11.98 L

Now put all the given values in the above equation, we get:

$8.19atm\times 2.14L+2.65atm\times 9.84L=P_f\times 11.98L$

$P_f=3.64atm$

Therefore, the final pressure of the system in atm is, 3.64 atm

4 0

3 years ago

A 200kg bucket of cement<br>

ozzi

Answer:

Yes. A 200 kg bucket of cement = About 440.925 pounds of cements.

Explanation:

7 0

3 years ago

A 57 g ice cube can slide without friction up and down a 33 ∘ slope. The ice cube is pressed against a spring at the bottom of t

beks73 [17]

The springs stored energy is transferred to the cube as kinetic energy and then by the slop the KE is converted to height energy.

<span>0.5 . k . x^2 = 0.5 . m . v^2 = m . g . ∆h </span>

<span>0.5 . 50 . (0.1^2) = 0.05 . 9.8 . ∆h </span>

<span>∆h = 0.51 m = 51 cm </span>

<span>This is the height gained </span>
<span>Distance along the slope = ∆h / sin 60 = 0.589 = 59 cm </span>

<span>In the second case, the stored spring energy is converted into height energy AND frictional heat energy. </span>

<span>The height energy is m . g . d sin 60 where d is the distance the cube moves along the slope. </span>

<span>The Frictional energy converted is F . d </span>

<span>F ( the frictional force ) = µ . N </span>

<span>N ( the reaction to the component of the gravity force perpendicular to the surface of the slope ) = m . g . cos60 </span>

<span>Total energy converted </span>

<span>0.5 . k . x^2 = (m . g . dsin60) + (µ . m . g . cos60 . d ) </span>

<span>Solve for d </span>

<span>d = 0.528 = 53 cm</span>

5 0

3 years ago