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Lesechka [4]
3 years ago
9

Three balls are thrown off the top of a building, all with the same speed but with different launch angles (position is given in

meters and time is given in seconds). The components of the initial velocities are given.
The blue ball has an initial velocity of (6 m/s, 8 m/s).
The green ball has an initial velocity of (10 m/s, 0 m/s).
The red ball has an initial velocity of (8 m/s, -6 m/s).

a. Rank the three balls according to which one hits the ground first.
b. Rank the three balls according to which one has the greatest speed the instant before impact with the ground.
c. Calculate the speed of each of the balls the instant before impact with the ground.
Physics
1 answer:
jek_recluse [69]3 years ago
4 0

Answer:

A. Red, Green, Blue

B. None. They all hit the ground with the same speed.

C. Final Kinetic Energy = Initial Gravitational Potential Energy

Explanation:

A. This one is actually just common sense. If you think about it, the object that is being thrown downward will reach the ground first. The object thrown upwards will hit the ground the last. If you don't believe me, use just the vertical component of velocity and use acceleration = 9.8 m/s^2. Find the time it takes for an arbitrary displacement each ball will travel. You will find that the one pointed downward will reach the ground in the shortest amount of time. Just do the same math over for the rest of the balls and you can find the time it takes for each ball to hit the ground.

B. This one is actually really simple if you think about it. Using the Work-Energy theorem, using the system consisting a ball and the earth there will be no external forces doing work on the system. Since each ball is at some height "h" and the same initial speed, they will all have the same amount of energy in the beginning. If we set the ground as our reference point for gravitational potential energy, then there will be no gravitational energy as the height at that point will be 0. They all must have the same kinetic energy at the ground.

C.

Mechanical Energy Final = Mechanical Energy Initial + Net Work from non-conservative forces

Kinetic Energy Final + Gravitational Potential Energy Final = Kinetic Energy Initial + Gravitational Potential Energy Initial

I am using the symbol " ' " to indicate final speeds and heights

0.5*m*v'^2 + mgh' = 0.5*m*v^2 + mgh

0.5*m*v'^2 + 0 = 0.5*m*v^2 + mgh

(cancel out the mass)

0.5*v'^2 = 0.5*v^2 +gh

Since the "v" is the same initially for all three balls and "g" and "h" are the same for all three balls. The final speed v' must be the same for all three situations regardless of mass.

If you have any questions feel free to comment again. I'll try to clarify.

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A 40 kg girl and an 8.4 kg sled are on the surface of a frozen lake, 15 m apart. By means of a rope, the girl exerts a 5.2 N for
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Answer:

(a) a_s=0.62\frac{m}{s^2}

(b) a_s=0.13\frac{m}{s^2}

(c) x_f=2.6m

Explanation:

(a) According to Newton's second law, the acceleration of a body is directly proportional to the force exerted on it and inversely proportional to it's mass.

a_s=\frac{F}{m_s}\\a_s=\frac{5.2N}{8.4kg}\\a_s=0.62\frac{m}{s^2}

(b) According to Newton's third law, the force that the sled exerts on the girl is equal in magnitude but opposite in the direction of the force that the girl exerts on the sled:

a_g=\frac{F}{m_g}\\a_g=\frac{5.2N}{40kg}\\a_g=0.13\frac{m}{s^2}

(c) Using the kinematics equation:

x_f=x_0+v_0t \pm  \frac{at^2}{2}

For the girl, we have x_0=0 and v_0=0. So:

x_f_g=\frac{a_gt^2}{2}(1)

For the sled, we have v_0=0. So:

x_f_s=x_0_s-\frac{a_st^2}{2}(2)

When they meet, the final positions are the same. So, equaling (1) and (2) and solving for t:

x_0_s-\frac{a_st^2}{2}=\frac{a_st^2}{2}\\t^2(a_g+a_s)=2x_0_s\\t=\sqrt{\frac{2x_s_0}{a_g+a_s}}\\t=\sqrt{\frac{2(15m)}{0.13\frac{m}{s^2}+0.62\frac{m}{s^2}}}\\t=6.32s

Now, we solve (1) for x_f_g

x_f_g=\frac{0.13\frac{m}{s^2}(6.32s)^2}{2}\\x_f_g=2.6m\\x_f=2.6m

5 0
3 years ago
Water flows over a section of Niagara Falls at the rate of 1.4 × 106 kg/s and falls 49.8 m. How much power is generated by the f
Ad libitum [116K]

Answer:

Power= 6.84×10⁸ W

Explanation:

Given Data

Niagara falls at rate of=1.4×10⁶ kg/s

falls=49.8 m

To find

Power Generated

Solution

Regarding this problem

GPE (gravitational potential energy) declines each second is given from that you will  find much the kinetic energy of the falling water is increasing each second.

So power can be found by follow

Power= dE/dt = d/dt (mgh)

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Power= 1.4×10⁶ kg/s × 9.81 m/s² × 49.8 m

Power= 6.84×10⁸ W  

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