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Gekata [30.6K]
3 years ago
8

P.E

Physics
1 answer:
Vladimir79 [104]3 years ago
5 0

Answer:

pagtalon?

Explanation:

because that had to be the answer

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If the rate $56 per 7 hours is reduced to a unit rate, the result is dollars per hour.
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Describe at least four of the requirements (facts) that any model of solar system formation must include (according to the nebul
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4 0
3 years ago
A piece of steel is 11.5cm long at 22C. It is heated to 1221C, close to its melting point. How long is it, in cm, at the high te
Nataly [62]

Answer:

The length at the final temperature is 11.7 cm.

Explanation:

We need to use the thermal expansion equation:

\Delta L=\alpha L_{0}\Delta T

Where:

  • L(0) is the initial length
  • ΔT is the differential temperature, final temperature minus initial temperature (T(f)-T(0))
  • ΔL is the final length minus the initial length (L(f)-L(0))
  • α is the coefficient of linear expantion of steel (12.5*10⁻⁶ 1/°C)  

So, we have:

L_{f}-L_{0}=\alpha L_{0}(T_{f}-T_{0})

L_{f}=L_{0}+\alpha L_{0}(T_{f}-T_{0})

L_{f}=0.115+(12.5*10^{-6})(0.115)(1221-22)

L_{f}=0.117\: m

Therefore, the length at the final temperature is 11.7 cm.

I hope it helps you!

7 0
3 years ago
A solid disk of mass 2 kg and radius 2 m is given a horizontal push of 20N at a point .3 m above its center. a. What is the mini
Margaret [11]

Answer:

\mu_s=1.0205

Explanation:

Given:

  • mass of solid disk, m=2\ kg
  • radius of disk, r=2\ m
  • force of push applied to disk, F=20\ N
  • distance of application of force from the center, s=0.3\ m

<em>For the condition of no slip the force of  static friction must be greater than the applied force so that there is no skidding between the contact surfaces at the contact point.</em>

\therefore F

where:

f_s = static frictional force

\Rightarrow 20

\Rightarrow 20

\Rightarrow 20

\mu_s>1.0204

7 0
3 years ago
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