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2 years ago

5

A 500-kg golf cart is moving on level ground in a circular path of radius 10 m with a speed of 12 m/s. What is the centripetal f

orce acting on the cart?

1 answer:

Alborosie2 years ago

8 0

Answer:

7200N

Explanation:

Centripetal force is directly proportional to the product of the mass and the square of the velocity and inversely proportional to the radius given.

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A stationary hockey puck has a mass of 0.27 kg. A hockey player uses her stick to apply a 18 N force over a distance of 0.42 m.

jeka57 [31]

W=Fs
W=18(0.42)
W=7.56

Answer: 7.6 J (first option)

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kap26 [50]

S = 10 + (-4) = 6

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Why do you rotate when you hold a spinning wheel?

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When you hold a spinning wheel, the wheel and you, chair included, form a system that obeys the principle of "conservation of angular momentum". This means that any changes in angular momentum within the system must accompanied by an equal and opposite change, so the net force is zero.

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3 years ago

A double nozzle lying in a horizontal x-y plane discharges water into the atmosphere at a rate of 0.5 m3 /s. Assume the water sp

Kisachek [45]

Answer:

The force is $F= 46.25kN$

Explanation:

The diagram for this question is shown on the first uploaded image

At Equilibrium the summation of the of force on the vertical axis is zero

i.e $\sum F_y =0$

=> $F_y sin \ 60^o =\rho Q (v_2 -v_1 cos \ 30^o)$

$v_2$ is the is the speed of water at the nozzle which can be mathematically evaluated as

$v_2 = \frac{R}{A_n}$

substituting $0.5m^3/s$ for R and $\frac{\pi}{4}(12*\frac{1m}{100} )^2$ for $A_n$

$v_2 = \frac{0.5}{\frac{\pi}{4} * (12*\frac{1}{100} )^2 }$

$= 44.23 m/s$

$v_1$ is the is the speed of water at the pipe which can be mathematically evaluated as

$v_1 = \frac{R}{A_p}$

substituting $0.5m^3/s$ for R and $\frac{\pi}{4}(30*\frac{1m}{100} )^2$ for $A_p$

$v_1 = \frac{0.5}{\frac{\pi}{4} * (30*\frac{1}{100} )^2 }$

$= 7.07 m/s$

$\rho$ is he density of water with value $\rho =1000 kg /m^3$

Substituting values into the equation above

$F_ysin 60^o = 1000 (0.5) (44.23 -7.07 cos 30)$

$= 21.99kN$

At Equilibrium the summation of the of force on the horizontal axis is zero

i.e $\sum F_x =0$

=> $F_y sin \ 30^o =\rho Q (v_2 -v_1 sin \ 30^o)$

Since The speed at both A and B nozzle are the same then $v_2$ remains the same

Substituting values

$F_x sin30^o =1000 (0.5) (44.23 - 7.07*sin30)$

=> $F_x = 40.69kN$

Hence the force acting on the flange bolts required to hold the nozzle in place is

$F = \sqrt{F_x^2 + F_y^2}$

$= \sqrt{40.69 ^2 + 21.99^2}$

$F= 46.25kN$

6 0

3 years ago

Read 2 more answers

The sound of a tuba is very low. Why?

vaieri [72.5K]

Answer:

When you blow into a tuba the air vibrates very slowly.

Explanation:

Tuba is a buzz instrument ie sound is produced in it with the help of lip vibration . It is the lowest pitched musical instrument in the brass family .

Due to absence of resonance in it , it produces music of lowest pitch , So when one blows into it the air column of the instrument vibrates very slowly producing low pitched sound.

8 0

3 years ago