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Ratling [72]
3 years ago
13

7. Mac and Tosh are arguing about the track design. Mac claims that the car is moving fastest at point F because it is furthest

along the track Tosh disagrees, claiming that the car moves fastest at point F because point F is at the lowest height Who do you agree with? Make a modification of the track design for point F and gather some evidence to support one claim or the other. Then discuss what changes you made, what observations were made, and the reasoning that supports one of the claim of either Mac or Tosh.​
Physics
1 answer:
olganol [36]3 years ago
7 0

Answer:

  • Tosh is correct.
  • Swap the locations of points B and F on the track and gather speed data. If Mac is correct, the speed at F (closer to the start) should be lower. (It will not be, confirming Tosh's claim.)

Explanation:

The total energy of the car is continuously being exchanged between potential energy and kinetic energy as the car moves along the track. Neglecting energy loss due to friction, the kinetic energy will be greatest when the potential energy is least, at the lowest point on the track. As a consequence we agree with Tosh that the speed will be greatest at F because it is the lowest point.

__

If the track were modified to move the lowest point nearer the start, say by interchanging points B and F, then data could be gathered to show whose theory is supported. The evidence needed is the speed of the car at the new location of point F. Tosh's argument is supported if the speed at the new point F is substantially the same.

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Match the following. Column A 1. Torque 2. Centre of gravity 3. Plumb line Column B A. Line of centre of gravity B. Maximum cons
gregori [183]

Answer:

1. Torque → F. Study of forces

2. C.O.G → D. Point of action of weight.

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8 0
3 years ago
I cant solve this problem, and our teacher said that this would be in the test we'll have tomorrow, can someone help me?
Ad libitum [116K]

Answer:

d = 11.1 m

Explanation:

Since the inclined plane is frictionless, this is just a simple application of the conservation law of energy:

\frac{1}{2} m {v}^{2}  = mgh

Let d be the displacement along the inclined plane. Note that the height h in terms of d and the angle is as follows:

\sin(15)  =  \frac{h}{d}  \\ or \: h = d \sin(15)

Plugging this into the energy conservation equation and cancelling m, we get

{v}^{2}  = 2gd \sin(15)

Solving for d,

d =  \frac{ {v}^{2} }{2g \sin(15) }  =  \frac{ {(7.5 \:  \frac{m}{s}) }^{2} }{2(9.8 \:  \frac{m}{ {s}^{2} })(0.259)}   \\ = 11.1 \: m

3 0
2 years ago
The engine of a jet airplane pushes exhaust gases from burning fuel backward
olga nikolaevna [1]
For more boost and to stop chases of fire
6 0
3 years ago
What velocity does a 2kg mass have when its kinetic energy is 16 J
alexandr402 [8]
We can use the equation for kinetic energy, K=1/2mv².
Your given variables are already in the correct units, so we can just plug in the variables and solve for v. 

K = 1/2mv²
16 = 1/2(2)v²
16 = (1)v²
√16 = v
v = 4 m/s

Therefore, the velocity of a 2 kg mass with 16 J of kinetic energy is 4 m/s.
Hope this is helpful!
7 0
2 years ago
PLEASE HELP!
Anna [14]

Answer:

Explanation:

At constant pressure , work done by gas = P x ΔV where P is pressure and ΔV is change in volume

ΔV = 9.2 - 5.6 = 3.6 L

3.6 L = 3.6 x 10⁻³ m³

ΔV = 3.6 x 10⁻³ m³

P = 3.7 x 10³ Pa

So work done

= 3.7 x 10³ x 3.6 x 10⁻³ J

= 13.32 J .

( c ) is the answer , because work is done by the gas so it will be positive.

5 0
3 years ago
Read 2 more answers
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