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Delicious77 [7]
3 years ago
14

A roller coaster car is elevated to a height of 30 m and released from rest to roll along a track. At a certain time T it is at

a height of 2 m and has lost 25.000 J of energy to friction. The car has a mass of 800 kg. Answer the following questions. (a) How fast is the car going at time T? (b) How fast would the car be going at time T if the track were frictionless?
Physics
1 answer:
Naddika [18.5K]3 years ago
4 0

Explanation:

Initial energy = final energy + work done by friction

PE = PE + KE + W

mgH = mgh + 1/2 mv² + W

(800)(9.8)(30) = (800)(9.8)(2) + 1/2 (800) v² + 25000

v = 22.1 m/s

Without friction:

PE = PE + KE

mgH = mgh + 1/2 mv²

(800)(9.8)(30) = (800)(9.8)(2) + 1/2 (800) v²

v = 23.4 m/s

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Issac newton how did he shape the new worldview?
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3 0
3 years ago
Assume that the electric field E is equal to zero at a given point. Does it mean that the electric potential V must also be equa
lyudmila [28]

Answer:

  • No, this doesn't mean the electric potential equals zero.

Explanation:

In electrostatics, the electric field \vec{E} is related to the gradient of the electric potential V with :

\vec{E} (\vec{r}) = - \vec{\nabla} V (\vec{r})

This means that for constant electric potential the electric field must be zero:

V(\vec{r}) = k

\vec{E} (\vec{r}) = - \vec{\nabla} V (\vec{r}) = - \vec{\nabla} k

\vec{E} (\vec{r}) = -  (\frac{\partial}{\partial x} , \frac{\partial}{\partial y } , \frac{\partial}{\partial z}) k

\vec{E} (\vec{r}) = -  (\frac{\partial k}{\partial x} , \frac{\partial k}{\partial y } , \frac{\partial k}{\partial z})

\vec{E} (\vec{r}) = -  (0,0,0)

This is not the only case in which we would find an zero electric field, as, any scalar field with gradient zero will give an zero electric field. For example:

V(\vec{r})= (x+2)^2 (y+4)^3 (z+5)^4

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8 0
3 years ago
I don’t have a question because I posted a photoz
Scorpion4ik [409]

Part a:

Q_{1} = 56

Q_{2} = 60

Q_{3} = 63

     The quartiles are found by finding the medium of the data, and then the mediums of the two different data sets on either side of the medium. The Q_{2} is the overall medium, Q_{1} is the medium of the first half, and Q_{3} is the medium of the second half.

-> How is the medium found? When finding the medium we put the values in order least to greatest and pick the middle value.

[] See attached

Part b:

The range is 7.

The interquartile range is the range of numbers between Q_{1} and Q_{3}. In other words, it is 50% of the data, directly in the middle.

This becomes 63 - 56 = 7

Part c:

79 is an outlier.

It is an outlier because it is 1.5 above or below (in this case, above) the interquartile range.

-> 63 + (7 + \frac{7}{2}) ≤ 79

-> 63 + 10.5 ≤ 79

-> 73.5 ≤ 79

Have a nice day!

     I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly.

- Heather

7 0
2 years ago
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