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3 years ago

How to calculate velocity with acceleration and time

Physics

1 answer:

Pani-rosa [81]3 years ago

8 0

Answer:

Multiply the acceleration by time to obtain the velocity change:

Example:

Velocity change = 6.95 * 4 = 27.8 m/s . Since the initial velocity was zero, the final velocity is equal to the change of speed. You can convert units to km/h by multiplying the result by 3.6: 27.8 * 3.6 ≈ 100 km/h

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a block with a mass of 3 kg is swung on a cord in a horizontal circle with a radius of 2m. the speed of the block is 6m/s^. the

Ivenika [448]

Given Information:

Mass = m = 3 kg

Speed = v = 6 m/s

Radius = r = 2 m

Required Information:

Magnitude of the acceleration = a = ?

Answer:

Magnitude of the acceleration = 18 m/s²

Explanation:

The acceleration of the block traveling along a circular path with some velocity is given by

a = v²/r

a = 6²/2

a = 36/2

a = 18 m/s²

Therefore, the magnitude of the acceleration of the block is most nearly equal to 18 m/s².

Bonus:

The corresponding force acting on the block can be found using

F = ma (a = v²/r)

F = mv²/r

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3 years ago

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As the mass and distance between two objects decrease, so does the gravitational force exerted between them. true or false

marta [7]

I believe it's true,..hope this helps!

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3 years ago

Which statements about acceleration are true?

larisa86 [58]

C is the correct answer, hope it helps

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4 years ago

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A civil engineer wishes to redesign the curved roadway in the example What is the Maximum Speed of the Car? in such a way that a

vlabodo [156]

Answer:

24.3 degrees

Explanation:

A car traveling in circular motion at linear speed v = 12.8 m/s around a circle of radius r = 37 m is subjected to a centripetal acceleration:

$a_c = \frac{v^2}{r} = \frac{12.8^2}{37} = 4.43 m/s^2$

Let α be the banked angle, as α > 0, the outward centripetal acceleration vector is split into 2 components, 1 parallel and the other perpendicular to the road. The one that is parallel has a magnitude of 4.43cosα and is the one that would make the car slip.

Similarly, gravitational acceleration g is split into 2 component, one parallel and the other perpendicular to the road surface. The one that is parallel has a magnitude of gsinα and is the one that keeps the car from slipping outward.

So $gsin\alpha = 4.43cos\alpha$