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Bingel [31]
3 years ago
13

A bumblebee is flying to the right when a breeze causes the bee to slow down with a constant leftward acceleration of magnitude

0.50\,\dfrac{\text m}{\text s^2}0.50 s 2 m ​ 0, point, 50, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction. After 2.0\,\text{s}2.0s2, point, 0, start text, s, end text, the bee is moving to the right with a speed of 2.75\,\dfrac{\text m}{\text s}2.75 s m ​ 2, point, 75, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction. What was the velocity of the bumblebee right before the breeze?
Physics
1 answer:
lesya692 [45]3 years ago
7 0

Answer:

3.75 m/s

Explanation:

Given:

v = 2.75 m/s

a = -0.50 m/s²

t = 2.0 s

Find: v₀

v = at + v₀

2.75 m/s = (-0.50 m/s²) (2.0 s) + v₀

v₀ = 3.75 m/s

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A 100g block lies on an inclined plane that makes an angle of 15 degrees with the horizontal. The coefficient of kinetic frictio
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Answer:

Mass that one should put in the container so that the 100 g block slides down the inclined plane at constant speed = 34.16 g

Explanation:

The vertical forces (with respect to the inclined plane) acting on the 100 g block include the component of the weight of the block in the direction vertical to the inclined plane and the normal reaction of the plane on the block.

And sum of upward forces = sum of downward forces.

N = mg cos θ

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N = (0.1×9.8×cos 15°) = 0.946582 N

The horizontal forces (With respect to the inclined plane) include the frictional force (acting upwards for the inclined plane, opposite to the intended direction of motion), the Tension in the rope (acting downwards, away from the 100 g block) and the horizontal component (with respect to the inclined plane) of the weight of the block, F, (also acting downards).

For the body to slide down the inclined plane at constant speed, the downward sloping forces must balance the frictional force, that is, there will be no acceleration.

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Frictional force = Fr = (0.60 × 0.9466) = 0.56796 N = 0.568 N

The horizontal component (with respect to the inclined plane) of the weight of the block (also acting downards) = mg sin θ

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Fr = F + T

T = Fr - F = 0.568 - 0.253624 = 0.314376 N = 0.3144 N

But the balance on the rope now has the total weight on the container (weight of container + weight on the container) to be equal to 2T.

2T = mg

2 × 0.3144 = 9.8m

m = 0.06416 kg = 64.16 g.

Mass of the container = 30 g

So, mass that one should put in the container so that the 100 g block slides down the inclined plane at constant speed = 64.16 - 30 = 34.16 g

Hope this Helps!!!

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