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gregori [183]
3 years ago
15

An amusement ride consists of a car moving in a vertical circle on the end of a rigid boom. The radius of the circle is 10 m. Th

e combined weight of the car and riders is 5.0 kN. At the top of the circle the car has a speed of 5.0 m/s which is not changing at that instant. What is the force of the boom on the car at the top of the circle?
Physics
2 answers:
Pie3 years ago
8 0

Answer:

Force of boom= 1275.5N

Explanation:

Combined weight of car and rider=5.0KN= 5000N.

Radius of circleR=10m

EF=ma

Fc= W + FB

FB= (mv^2/r) - W

Fc is centripetal force

Mass= Weight/ g= 5000/9.8

m=510.2Kg

If car's speed is 5m/s

FB= (510.2× 5^2/10) - 5000

FB=( 12755/10) - 5000

FB= 1275.5 - 5000

FB= -3724.5N

Fc= W + FB

Fc= 5000 + (-3724.5)

Fc= 1275.5N

AveGali [126]3 years ago
7 0

Answer:

3.7 kN (Up)

Explanation:

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Which of the following statements are true? (There may be more than one correct choice.)
Luba_88 [7]

Answer:

option (A)

Explanation:

According to the Archimede's principle, when a body s immersed partly or wholly in a liquid, it experiences an upward force which is called buoyant force. The buoyant force is equal to the loss in the weight of body.

The weight of liquid displaces by the body is equal to the loss in weight of body.

Thus, option (a) is true.

5 0
3 years ago
In a vertical relationship between an employee and supervisor, who must do most of the adjusting?
maks197457 [2]
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7 0
3 years ago
The charge per unit length on a long, straight filament is -92.0 μC/m. Find the electric field 10.0 cm above the filament.
Pepsi [2]

Answer:

E = 1.655 x 10⁷ N/C towards the filament

Explanation:

Electric field due to a line charge is given by the expression

E = [tex]\frac{\lambda}{2\pi\times\epsilon_0\times r}[/tex]

where λ is linear charge density of line charge , r is distance of given point from line charge and ε₀ is a constant called permittivity and whose value is

8.85 x 10⁻¹².

Putting the given values in the equation given above

E = \frac{92\times10^{-6}}{2\times3.14\times8.85\times10^{-12}\times10^{-1}}

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4 0
3 years ago
Can someone help me
Illusion [34]
Thomas Edison is the answer im 100% sure of it.
5 0
3 years ago
A stuffed toy with a mass of 0.900 kilograms sits on the edge of a bed at a height of 0.830 if the toy falls off the bed what wi
Olenka [21]
Mechanical energy (ME) is the sum of potential energy (PE) and kinetic energy (KE). When the toy falls, energy is converted from PE to KE, but by conservation of energy, ME (and therefore PE+KE) will remain the same.

Therefore, ME at 0.500 m is the same as ME at 0.830 m (the starting point). It's easier to calculate ME at the starting point because its just PE we need to worry about (but if we wanted to we could calculate the instantaneous PE and KE at 0.500 m too and add them to get the same answer).

At the start:

ME = PE = mgh
ME = 0.900 (9.8) (0.830)
ME = 7.32 J
8 0
2 years ago
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