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MrRa [10]
3 years ago
10

A basketball player throws a ball at 60° to the horizontal at a net 4.0m away. The height of the net is 3.3m. The ball leaves hi

s hands at a height 2.0m from the ground. What is the initial speed of the ball?
Th pic above is my ans, but it's quite wierd. So if it's wrong, what is the correct ans, and why my ans is wrong(what mistakes i did). Thanks

Physics
1 answer:
Hoochie [10]3 years ago
4 0
Excellent work!
Your calculations are correct, but near the end, you have forgotten to cancel u.  If you cancel the u, the linear term becomes a constant, and the resultant equation becomes a simple quadratic which is much easier to solve.
I get u=7.468 m/s using g=9.81 (as you did)
You will rework and should get 7.468 m/s as well.

Congrats for the good work!

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Answer:

trying to push a rock that never moves

Explanation:

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3 years ago
Two firefighters are trying to break through a door. One firefighter is heavy, and the other is light. If they run at the same s
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Answer:

The Heavier Firefighter

Explanation:

Generally, more massive objects will have more intertia than less massive objects.  As such it takes more force to halt a more massive object if its moving at the same speed as a smaller object. This can also be thought of in the context of Newton's second law. The more force needed to accelerate an object means the more force the object will have.

6 0
2 years ago
Find the magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a
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Answer:

E=\frac{KQ}{2\sqrt 2a^2}

Explanation:

We are given that

Charge on ring= Q

Radius of ring=a

We have to find the magnitude of electric filed on the axis at distance a from the ring's center.

We know that the electric field at distance x from the center of ring of radius R is given by

E=\frac{kQx}{(R^2+x^2)^{\frac{3}{2}}}

Substitute x=a and R=a

Then, we get

E=\frac{KQa}{(a^2+a^2)^{\frac{3}{2}}}

E=\frac{KQa}{(2a^2)^{\frac{3}{2}}}

E=\frac{KQa}{2\sqrt 2a^3}

E=\frac{KQ}{2\sqrt 2a^2}

Where K=9\times 10^9 Nm^2/C^2

Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center=\frac{KQ}{2\sqrt 2a^2}

4 0
3 years ago
As a wave travels through a medium, it displaces particles in a direction parallel to the motion of the wave. We can conclude th
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We can conclude that it is a longitudinal wave because the wave is traveling through a medium displacing particles<span>
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8 0
3 years ago
Read 2 more answers
A certain quantity of steam has a temperature of 100.0 oC. To convert this steam into ice at 0.0 oC, energy in the form of heat
KonstantinChe [14]

Answer:

2452.79432 m/s

Explanation:

m = Mass of ice

L_s = Latent heat of steam

s_w = Specific heat of water

L_i = Latent heat of ice

v = Velocity of ice

\Delta T = Change in temperature

Amount of heat required for steam

Q_1=mL_s\\\Rightarrow Q_1=m(2.256\times 10^6)

Heat released from water at 100 °C

Q_2=ms_w\Delta T\\\Rightarrow Q_2=m4186\times (100-0)\\\Rightarrow Q_2=m0.4186\times 10^6

Heat released from water at 0 °C

Q_3=mL_i\\\Rightarrow Q_3=m(333.5\times 10^3)\\\Rightarrow Q_3=m(0.3335\times 10^6)

Total heat released is

Q=Q_1+Q_2+Q_3\\\Rightarrow Q=m(2.256\times 10^6)+m0.4186\times 10^6+m(0.3335\times 10^6)\\\Rightarrow Q=3008100m

The kinetic energy of the bullet will balance the heat

K=Q\\\Rightarrow \frac{1}{2}mv^2=3008100m\\\Rightarrow v=\sqrt{2\times 3008100}\\\Rightarrow v=2452.79432\ m/s

The velocity of the ice would be 2452.79432 m/s

6 0
3 years ago
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