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navik [9.2K]
3 years ago
11

A wooden bucket filled with water has a mass of 68 kg and is attached to a rope that is wound around a cylinder with a radius of

0.078 m. A crank with a turning radius of 0.250 m is attached to the end of the cylinder. What minimum force directed perpendicularly to the crank handle is required to raise the bucket
Physics
1 answer:
Aleonysh [2.5K]3 years ago
7 0

Answer: 210.2N

Explanation:

Assume a bucket of water with a total mass of 68kg is attached to a rope, which in turn is tied around a 0.078m radius cylinder at the top of a well. A crank with a turning radius of 0.250 m is attached to the end of the cylinder.

the minimum force directed perpendicular to the crank handle required to raise the bucket is

(Assume the rope's mass is negligible, that the cylinder turns on friction-less bearings, and that g = 9.8 m/s2

The crank handle provides a torque T=0.25F where F is the force we are looking for.

A free body diagram will show that the tension in the rope times the cylinder radius R is equal to the torque on the cylinder. But the tension in the rope is just the weight of the bucket

W=mg= 68kg

W(0.078)=T=0.25F

F=0.312W=0.312(68kg)=21.216kg= 210.2N

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An automobile rounds a curve of radius 50.0 m on a flat road.
bixtya [17]

Answer:

14m/s

Explanation:

Given parameters:

Radius of the curve  = 50m

Centripetal acceleration  = 3.92m/s²

Unknown:

Speed needed to keep the car on the curve = ?

Solution:

The centripetal acceleration is the inwardly directly acceleration needed to keep a body along a curved path.

 It is given as;

      a = \frac{v^{2} }{r}  

a is the centripetal acceleration

v is the speed

r is the radius

  Now insert the parameters and find v;

         v²   = ar

        v² = 3.92 x 50  = 196

         v  = √196 = 14m/s

6 0
3 years ago
a block of wood of density 0.8 gram per centimetre cube has a volume of 60 cm3 the mass of block will be...?​
Tamiku [17]
Density =mass/Volume

0.8=mass/60
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4 0
2 years ago
An airplane starts from rest at the end of a runway and accelerates at a constant rate. In the first second, the airplane travel
Licemer1 [7]

Answer:

v=4.44\frac{m}{s}

Explanation:

Given that the airplane starts from the rest (this is initial velocity equals to zero)  and accelerates at a constant rate, position can be described like this: x=v_{0}t +\frac{1}{2} at^{2} where x is the position, t is the time a is the acceleration and v_{0} is initial velocity. In this way acceleration can be found. a=\frac{2(x-v_{0}t) }{t^{2} } =\frac{2(1.11m-0)}{1s^{2} } =2.22\frac{m}{s^{2} }.

Now we are able to found velocity at any time with the formula: v=v_{0} +at = 0\frac{m}{s} +(2.22\frac{m}{s^{2}}.2s)=4.44\frac{m}{s}

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3 years ago
Which of the following is not a result of deforestation?
klemol [59]

Answer:

B

Explanation:

5 0
2 years ago
Read 2 more answers
The gravitational attraction between two masses of 3kg that are separated by a distance of 1cm is
Luba_88 [7]

Answer:

6.003×10¯⁶ N

Explanation:

We'll begin by converting 1 cm to m. This can be obtained as follow:

100 cm = 1 m

Therefore,

1 cm = 1 cm × 1 m / 100 cm

1 cm = 0.01 m

Finally, we shall determine the gravitational attraction. This can be obtained as follow:

Mass 1 (M₁) = 3 Kg

Mass 2 (M₂) = 3 Kg

Distance apart (r) = 0.01 m

Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²

Force of attraction (F) =?

F = GM₁M₂ / r²

F = 6.67×10¯¹¹ × 3 × / 0.01²

F = 6.003×10¯¹⁰ / 1×10¯⁴

F = 6.003×10¯⁶ N

Thus the gravitational attraction is 6.003×10¯⁶ N

6 0
3 years ago
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