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zmey [24]
2 years ago
5

Please answer of this question​

Physics
1 answer:
Bogdan [553]2 years ago
8 0

Answer:

\frac{\pi }{15}m

Explanation:

At 10am, the minute hand and hour hand are ' 2 hours apart', since the minute hand is at 12pm and hour hand is at 10am.

Angle between the two hands = 2/12 * 360

= 60°

Arc Length = 2\pi r(\frac{60}{360} )

= 2\pi (0.2)(\frac{1}{6} )\\= \frac{\pi }{15} m

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A 10 kg rock that has been dropped from a 60 meter high cliff experiences an average force of air resistance of 30 N. Calculate
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Answer:

4086 J

Explanation:

The potential energy is transformed to kinetic energy less the frictional energy. Potential energy= mgh where m represent mass, g is acceleration due to gravity and h is the height of cliff

Since we have force of air resistance, work done due to air resistance will be product of force and distance

mgh-Fh= 0.5mv^{2}= KE

Substituting 10 Kg for m, 9.81 for g and 60 m for F then the kinetic energy at the bottom will be

KE= 10*9.81*60- (30*60)=4086 J

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Could a mixture be made up of only elements and no compounds?
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Yah isnt that obvious? Gasses mix everywhere in all proportions.
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What is the difference between compound, element and mixture?
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4 0
3 years ago
A 54.0 kg ice skater is gliding along the ice, heading due north at 4.10 m/s . The ice has a small coefficient of static frictio
lesya [120]

Answer:

a. 2.668 m/s

b. 0.00494

Explanation:

The computation is shown below:

a. As we know that

W = F\times d

KE = 0.5\times m\times v^2

As the wind does not move the skater to the east little work is performed in this direction. All the work goes in the direction of the N-S. And located in that direction the component of the Force.

F = 3.70 cos 45 = 2.62 N

W = F \times d = 2.62 N \times 100 m

W = 261.6 N\times m

We know that

KE1 = Initial kinetic energy

KE2 = kinetic energy following 100 m

The energy following 100 meters equivalent to the initial kinetic energy less the energy lost to the work performed by the wind on the skater.

So, the equation is

KE2 = KE1 - W

0.5 m\times v2^2 = 0.5 m\ v1^2 - W

Now solve for v2

v2 = \sqrt{v1^2 - {\frac{2W}{M}}}

= \sqrt{4.1 m/s)^2 - \frac{2 \times 261.6 N\times m}{54.0 kg}}

= 2.668 m/s

b. Now the minimum value of Ug is

As we know that

Ff = force of friction

Us = coefficient of static friction

N = Normal force = weight of skater

So,

Ff = Us\times N

Now solve for Us

= \frac{Ff}{N}

= \frac{3.70 N \times cos 45 }{54.0 kg \times 9.81 m/s^2}

= 0.00494

4 0
3 years ago
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