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2 years ago

Why do moving back and forth, and moving in a circle look the same on a graph of position versus time?

Physics

1 answer:

Tomtit [17]2 years ago

3 0

Answer:

<u>Because in both kind of motions, the position is periodically repeated. Both are simple harmonic motions.</u>

Explanation:

<em>Moving back and forth</em> means that the moving object goes through the same position every time a certain period of time elapses.

The same is true for the <em>circular motion</em>, every time the object completes a turn, the object is in the same position.

They are examples of simple harmonic motions. The time it takes the object to repeat a complete cycle is called period.

Therefore, in a grahp of position versus time, the curve that represents the position repeats itself every so often equal to one period, both for moving back and forth and for moving in a circle.

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2 examples of balanced forces

valina [46]

Answer:

<h2>Here are some examples of situations involving balanced forces. </h2><h2>Hanging objects. The forces on this hanging crate are equal in size but act in opposite directions.</h2><h2>Floating in water. Objects float in water when their weight is balanced by the upthrust from the water.</h2><h2>Standing on the ground.</h2>

I hope this helps

6 0

3 years ago

How much force is needed to keep the bowling ball moving towards the pins once it has

vazorg [7]

Answer:

Explanation:

The amount of force needed needs to be greater than all the forces acting in the opposite direction that the bowling ball was thrown. This includes air resistance, floor friction, gravity, and any other force involved. As long as the force acting on the bowling ball that is causing it to go in the direction of the pins is slightly greater than the opposite acting forces then it will continue in that direction. Since no values are provided we cannot calculate the actual precise value of force needed.

6 0

2 years ago

A. A land speed car can decelerate at 9.8m/s. How long does it take the car to come to a complete stop from a run of 885 km/hr (

Nimfa-mama [501]

Answer:

A. 25.08 s

B. 3082.53 m

C. 3×10⁵ m/s²

Explanation:

A. Determination of the time.

This can be obtained as illustrated below:

Acceleration (a) = –9.8 m/s²

Initial velocity (u) = 245.8 m/s

Final velocity (v) = 0 m/s

Time (t) =.?

v = u + at

0 = 245.8 + (–9.8 × t)

0 = 245.8 – 9.8t

Collect like terms

0 – 245.8 = – 9.8t

– 245.8 = – 9.8t

Divide both side by –9.8

t = –245.8 / –9.8

t = 25.08 s

Therefore, it will take 25.08 s for the car to come to a complete stop.

B. Determination of the distance travelled by the car.

Acceleration (a) = –9.8 m/s²

Initial velocity (u) = 245.8 m/s

Final velocity (v) = 0 m/s

Distance (s) =?

v² = u² + 2as

0² = 245.8² + (2 × –9.8 × s)

0 = 60417.64 – 19.6s

Collect like terms

0 – 60417.64 = – 19.6s

– 60417.64 = – 19.6s

Divide both side by –19.6

s = –60417.64 / –19.6

s = 3082.53 m

Thus, the car travelled a distance of 3082.53 m before stopping completely.

C. Determination of the acceleration of the object.

Initial velocity (u) = 0 m/s

Final velocity (v) = 600 m/s

Distance (s) = 0.6 m

Acceleration (a) =?

v² = u² + 2as

600² = 0² + (2 × a × 0.6)

360000 = 0 + 1.2a

360000 = 1.2a

Divide both side by 1.2

a = 360000 / 1.2

a = 300000 = 3×10⁵ m/s²

7 0

3 years ago

Which object has the larger magnitude of its momentum?

Charra [1.4K]

The object that had the most 1000 ton weight has the most momentum

5 0

3 years ago

Your electric drill rotates initially at 5.21 rad/s. You slide the speed control and cause the drill to undergo constant angular

RUDIKE [14]

Answer:

The drill's angular displacement during that time interval is 24.17 rad.

Explanation:

Given;

initial angular velocity of the electric drill, $\omega _i$ = 5.21 rad/s

angular acceleration of the electric drill, α = 0.311 rad/s²

time of motion of the electric drill, t = 4.13 s

The angular displacement of the electric drill at the given time interval is calculated as;

$\theta = \omega _i t \ + \ \frac{1}{2}\alpha t^2\\\\\theta = (5.21 \ \times \ 4.13) \ + \ \frac{1}{2}(0.311)(4.13)^2\\\\\theta = (21.5173 ) \ + \ (2.6524)\\\\\theta =24.17 \ rad$

Therefore, the drill's angular displacement during that time interval is 24.17 rad.

6 0

3 years ago