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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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Answer:

2/R*sqrt (g*s*sin(θ)) = w

Explanation:

Assume:

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Solution:

- Using energy principle at top and bottom of the slope. The exchange of gravitational potential energy at height h, and kinetic energy at the bottom of slope.

                                         ΔPE = ΔKE

- The change in gravitational potential energy is given as m*g*h.

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We have:

                              m*g*s*sin(θ) = 0.25*m*R^2*w^2

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3 0
4 years ago
If a basball is project upwards from the ground level with an initial velovaity of 32 feet per second, then it's height is a fun
inessss [21]

Answer:

Maximum height reached by the ball is 32 meters.

Explanation:

It is given that,

If a baseball is project upwards from the ground level with an initial velocity of 32 feet per second, then it's height is a function of time. The equation is given as :

s=-8t^2+32t...........(1)

t is the time taken

s is the height attained as a function of time.

Maximum height achieved can be calculated as :

\dfrac{ds}{dt}=0

\dfrac{d(-8t^2+32t)}{dt}=0

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Put the value of t in equation (1) as :

s=-8(2)^2+32(2)

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A neutron consists of one "up" quark of charge +2e/3 and two "down" quarks each having charge -e/3. If we assume that the down q
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Answer:

The magnitude of the electrostatic force is 120.85 N

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In scalar form, Coulomb's law states that for charges q_1 and q_2 separated by a distance d, the magnitude of the electrostatic force F between them is:

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Taking the values:

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q_1 = q_2 = - \frac{e}{3} = - \frac{1.6 \ 10^{-19} \ C}{3}

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Taking all this in consideration:

F = 8.99 \ 10 ^{9} \frac{N m^2}{C^2} \frac{ (- \frac{1.6 \ 10^{-19} \ C}{3} ) ^2}{(4.6 \ 10^{-15} m)^2}

F = 120.85  \ N

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