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3 years ago

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A block is given an initial velocity of 8.0 m/s up a frictionless 28° inclined plane. (a) What is its velocity when it reaches t

he top of the plane? (b) How far horizontally does it land after it leaves the plane?

1 answer:

kramer3 years ago

4 0

Answer:

A.) 8 m/s

B.) 7.0 m

Explanation:

Given that a block is given an initial velocity of 8.0 m/s up a frictionless 28° inclined plane.

(a) What is its velocity when it reaches the top of the plane?

Since the plane is frictionless, the final velocity V will be the same as 8 m/s

The velocity will be 8 m/s as it reaches the top of the plane.

(b) How far horizontally does it land after it leaves the plane?

For frictionless plane,

a = gsinø

Acceleration a = 9.8sin28

Acceleration a = 4.6 m/s^2

Using the third equation of motion

V^2 = U^2 - 2as

Substitute the a and the U into the equation. Where V = 0

0 = 8^2 - 2 × 4.6 × S

9.2S = 64

S = 64/9.2

S = 6.956 m

S = 7.0 m

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4 years ago

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3 years ago

series RC circuit is built with a 15 kΩ resistor and a parallel-plate capacitor with 18-cm-diameter electrodes. A 18 V, 36 kHz s

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<u>Capacitance</u>

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3 years ago

An engineer is designing a runway. She knows that a plane, starting at rest, needs to reach a speed of 180mph at take-off. If th

kvv77 [185]

Answer:

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Explanation:

Convert unit of the the take-off velocity of this plane to $\rm ft\cdot s^{-1}$ :

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Notice that this equation does not require the value of acceleration. Rather, this equation make use of the fact that the distance travelled (under constant acceleration) is equal to duration $t$ times average velocity $(u + v) / 2$ .

The distance that the plane need to cover would be:

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4 0

3 years ago