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qaws [65]
4 years ago
11

A 1200-kg ore cart is rolling at 10.8 m/s across a flat friction-free surface. a crane suddenly drops of ore vertically into the

cart. how fast does the cart move just after being loaded with the ore?
Physics
1 answer:
Andre45 [30]4 years ago
5 0
The value of the final speed depends on the mass of the ore.

Let's call m the mass of the ore. We can solve the exercise by requiring the conservation of momentum, which must be the same before and after the ore is loaded.

Initially, there is only the cart, so the momentum is
p=Mv=(1200 kg)(10.8 m/s)=12960 kg m/s
After the ore is loaded, the new mass will be (1200 kg+m), and the new speed is v_f. The momentum p is conserved, so it is still 12960 kg m/s. Therefore, we have
p=12960 kg m/s =(1200 kg+m)v_f
and so the final speed is
v_f =  \frac{12960 kg m/s}{1200 kg +m}
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The ball's horizontal position in the air is

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It hits the wall when x=17.5\,\mathrm m, which happens at

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Meanwhile, the ball's vertical position is

y=\left(27.5\dfrac{\rm m}{\rm s}\right)\sin38.5^\circ t-\dfrac g2t^2

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At the time the ball hits the wall, its vertical position (relative to its initial position) is

y=\left(27.5\dfrac{\rm m}{\rm s}\right)\sin38.5^\circ(0.813\,\mathrm s)-\dfrac g2(0.813\,\mathrm s)^2\approx\boxed{10.7\,\mathrm m}

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3 years ago
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Explanation:

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A uniform ladder of length 10.8 m is leaning against a vertical frictionless wall. The weight of the ladder is 323 N, and it mak
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0.3625

Explanation:

From the given information:

Consider the equilibrium conditions;

On the ladder, net torque= 0

Thus,

\tau_{net} = 0; and

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However, by rearrangement;

fL \s in \theta =m_L g \dfrac{L}{2} cos \theta + mg (7.46\ m) cos \theta  \\ \\  \mu(m_L + m) gL \ sin \theta = (323  \ N) ( 10.8 \ meters) \ cos 56^0 + (734 \ N) (7.46 \ m) \ cos  \  66.46^0

\mu= \dfrac{ (323  \ N) ( 10.8 \ m) \ cos 56^0 + (734 \ N) (7.46 \ m) \ cos  \  66.46^0}{\Big [(323 \ N)+(734 \ N) \Big] (10.8 \ m)}

\mathbf{\mu= 0.3625 }

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3 years ago
A long straight wire carries a current I. If the magnetic field at a distance d from the wire has a magnitude B, what is the mag
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Answer:

Explanation:

As the field of current wire passing through the loop is same in direction (normally inward) but not uniform in magnitude.

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Magnetic field due to a long straight wire carrying current I is given by B=μ0I2πx, and area, dS=l×dx.

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