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3 years ago

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A ball is rolled uphill a distance of 12 meters before it slows, stops, and begins to roll back. The ball rolls downhill 20 mete

rs before coming to rest against a tree. What is the magnitude of the ball’s displacement________ m

1 answer:

Sedaia [141]3 years ago

5 0

The ball rolled a distance of
d = 12m + 20m.
But the change of position is
x = + 12m - 20m

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Δt = 0.02 sec

In given question, wave function

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the explanation is attached

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2 years ago

If the car is traveling at a velocity of 15 m/s, what is the approximate centripetal acceleration of the car?.

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The vehicle's centripetal acceleration is equal to 22.5m/s²

Radius, r = 10 meter

Speed, V = 15 m/s

To ascertain the car's centripetal acceleration

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A(c) = 225/10

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<h3>What is Centripetal acceleration ?</h3>

When an item moves in a circular route, one of its motion characteristics is centripetal acceleration. Any motion in a circle with an acceleration vector pointing in the direction of the circle's centre is referred to as centripetal acceleration.

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1 year ago

What causes the electric charges to flow from one end of the battery to the other? a balance in electric potential a balance in

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3 years ago

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The Professor's centripetal acceleration is 0.044 m/s²

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It is given by:

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Given that the radius (r) = 4 m, velocity (v) = 0.419 m/s, hence:

a = v²/r = 0.419²/4 = 0.044 m/s²

The Professor's centripetal acceleration is 0.044 m/s²

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3 years ago

A 0.150 kg stone rests on a frictionless, horizontal surface. A bullet of mass 9.50 g, traveling horizontally at 380 m/s, strike

Anvisha [2.4K]

Answer:

(a)Magnitude=28.81 m/s

Direction=33.3 degree below the horizontal

(b) No, it is not perfectly elastic collision

Explanation:

We are given that

Mass of stone, M=0.150 kg

Mass of bullet, m=9.50 g= $9.50\times 10^{3} kg$

Initial speed of bullet, u=380 m/s

Initial speed of stone, U=0

Final speed of bullet, v=250m/s

a. We have to find the magnitude and direction of the velocity of the stone after it is struck.

Using conservation of momentum

$mu+ MU=mv+ MV$

Substitute the values

$9.5\times 10^{-3}\times 380 i+0.150(0)=9.5\times 10^{-3} (250)j+0.150V$

$3.61i=2.375j+0.150V$

$3.61 i-2.375j=0.150V$

$V=\frac{1}{0.150}(3.61 i-2.375j)$

$V=24.07i-15.83j$

Magnitude of velocity of stone

= $\sqrt{(24.07)^2+(-15.83)^2}$

|V|=28.81 m/s

Hence, the magnitude and direction of the velocity of the stone after it is struck, |V|=28.81 m/s

Direction

$\theta=tan^{-1}(\frac{y}{x})$

= $tan^{-1}(\frac{-15.83}{24.07})$

$\theta=tan^{-1}(-0.657)$

=33.3 degree below the horizontal

(b)

Initial kinetic energy

$K_i=\frac{1}{2}mu^2+0=\frac{1}{2}(9.5\times 10^{-3})(380)^2$

$K_i=685.9 J$

Final kinetic energy

$K_f=\frac{1}{2}mv^2+\frac{1}{2}MV^2$

= $\frac{1}{2}(9.5\times 10^{-3})(250)^2+\frac{1}{2}(0.150)(28.81)^2$

$K_f=359.12 J$

Initial kinetic energy is not equal to final kinetic energy. Hence, the collision is not perfectly elastic collision.

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3 years ago