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Artemon [7]
3 years ago
9

A roller coaster has a vertical loop with radius 25.7 m. With what minimum speed should the roller-coaster car be moving at the

top of
the loop so that the passengers do not lose contact with the seats?
m/s
Physics
1 answer:
gogolik [260]3 years ago
7 0

Answer:

15.88m/s

Explanation:

At the top of the roller coaster you will have three forces acting on the roller-coaster. See the image below. Fc is the centripetal force (for an object in circular motion), Fg is the gravitational force, and Fn is the normal force. To achieve the minimum speed we assume the roller-coaster is barely touching the vertical loop and so the normal force is zero. This leaves two acting forces.

F_g = F_c\\mg = \frac{m\times v^2}{r}\\v = \sqrt{gr} = \sqrt{9.81 \times 25.7}  = 15.88 m/s

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1 point
s2008m [1.1K]

Answer:

The person has no displacement

Explanation:

The given parameters are

The location of the person = The equator

The distance covered in one revolution = Total distance around the body

The total distance around the Earth = The circumference of the Earth = 40.075 kilometres

The total distance moved by the person standing at the equator during the Earths complete revolution = 40,075 kilometres

The initial location of the person in relation to a fixed point in space outside Earth at the start of the revolution = x km

The final location of the person in relation to the fixed point in space outside Earth at the completion of the revolution = x km

The displacement = Change in position = Final location - Initial location  

∴ The displacement = x km - x km = 0 km.

5 0
4 years ago
Starting from rest, a small block of mass m slides frictionlessly down a circular wedge of mass M and radius R which is placed o
guapka [62]

Answer:

Part a)

V = \sqrt{\frac{2\frac{m}{M}gR}{(\frac{M}{m} + 1)}}

v = \frac{M}{m}\sqrt{\frac{2\frac{m}{M}gR}{(\frac{M}{m} + 1)}}

Part b)

Since on the block wedge system there is no external force in horizontal direction so the Center of mass will not move in horizontal direction but in vertical direction it will move

so displacement in Y direction is given as

y_{cm} = \frac{mR}{m + M}

Explanation:

PART A)

As we know that there is no external force on the system of two masses in horizontal direction

So here the two masses will have its momentum conserved in horizontal direction

So we have

mv + MV = 0

Also we know that here no friction force on the system so total energy will always remains conserved

So we have

\frac{1}{2}mv^2 + \frac{1}{2}MV^2 = mgR

now we have

\frac{1}{2}m(\frac{MV}{m})^2 + \frac{1}{2}MV^2 = mgR

\frac{1}{2}MV^2(\frac{M}{m} + 1) = mgR

so we have

V = \sqrt{\frac{2\frac{m}{M}gR}{(\frac{M}{m} + 1)}}

and another block has speed

v = \frac{M}{m}\sqrt{\frac{2\frac{m}{M}gR}{(\frac{M}{m} + 1)}}

Part b)

Since on the block wedge system there is no external force in horizontal direction so the Center of mass will not move in horizontal direction but in vertical direction it will move

so displacement in Y direction is given as

y_{cm} = \frac{mR}{m + M}

7 0
3 years ago
What is the solution set of the equation 3|5 -x| + 2 = 29?
Flura [38]
D. is the answer to that
3 0
3 years ago
Read 2 more answers
The angle between incident ray and reflected ray is 130.what is the value of angle of incidence​
Artyom0805 [142]

Answer:

65

Explanation:

as i = r , so i + i = 130

so , i = 130/2 =65

7 0
3 years ago
Read 2 more answers
The greater the difference in electronegativity between two covalently bonded<br><br> atoms
katrin [286]

Answer:

The greater the difference in electronegativity between two covalently bonded atoms, the greater the bond's percentage of ionic character.

Explanation:

Bond polarity (i.e the separation of electric charge along a bond) and ionic character (amount of electron sharing) increase with an increasing difference in electronegativity.

Therefore, we can say that, the greater the difference in electronegativity between two covalently bonded atoms, the greater the bond's percentage of ionic character.

7 0
3 years ago
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