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Rasek [7]
2 years ago
5

A ceiling fan is turned on and reaches an angular speed of 132 rev/min in 20 s. It is then turned off and coasts to a stop in an

additional 40 s. The ratio of the average angular acceleration for the first 20 s to that for the last 40 s is which of the following?
a. 2
b. -0.5
c. 0.5
d. -2
Physics
1 answer:
Mkey [24]2 years ago
8 0

Answer:

Explanation:

α = (ωf - ωi)/t

acceleration phase

ωf = 132 rev/min (2π rad/rev / 60 s/min) = 4.4π rad/s

α₁ = (4.4π - 0)/20 = 0.22π rad/s²

α₂ = (0 - 4.4π)/40 = - 0.11π rad/s²

α₁/α₂ = 0.22π/- 0.11π = -2

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25 points and i will give brainliest +5 star rating to a correct a correct answer!!!!
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Answer:

1.97 * 10^8 m/s

Explanation:

Given that:

n = 1.52

Recall : speed of light (c) = 3 * 10^8 m/s

Speed (v) of light in glass:

v = speed of light / n

v = (3 * 10^8) / 1.52

v = 1.9736 * 10^8

Hence, speed of light in glass :

v = 1.97 * 10^8 m/s

4 0
3 years ago
Identify each picture as either an inelastic collision or elastic collision
Ivan

Answer:

<u>Inelastic collision:</u>

A collision in which there is a loss of Kinetic Energy due to internal friction of the bodies colliding.

<u>Characteristics of an inelastic collision:</u>

  • <em>the momentum of the system is conserved</em>
  • <em>the momentum of the system is conservedloss of kinetic energy</em><u> </u>

<em>I</em><em>n</em><em> </em><em>a perfectly elastic collision</em><em>, the two bodies </em><em>that</em><em> </em><em>collide with each other stick together.</em>

<u>Elastic </u><u>collision</u><u>:</u>

A collision in which the kinetic energy of the two bodies, before and after the collision, remains the same.

<u>Characteristic</u><u>s</u><u> </u><u>of</u><u> </u><u>elastic</u><u> </u><u>collision</u><u>:</u>

  • <em>the</em><em> </em><em>momentum</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>system</em><em> </em><em>is</em><em> </em><em>conserved</em>
  • <em>no</em><em> </em><em>loss</em><em> </em><em>o</em><em>f</em><em> </em><em>kinetic</em><em> </em><em>energy</em>

In everyday life, no collision is perfectly elastic.

__________________

ANSWER:

<u>Given examples:</u>

  • Two cars colliding with each other form an example of inelastic collision.

<u>Reason:</u>

<em>(</em><em>T</em><em>hey</em><em> </em><em>lose</em><em> </em><em>kinetic</em><em> </em><em>energy</em><em> </em><em>and</em><em> </em><em>come</em><em> </em><em>to</em><em> </em><em>a</em><em> </em><em>stop</em><em> </em><em>after</em><em> </em><em>the</em><em> </em><em>collision</em><em>.</em><em>)</em>

  • A ball bouncing after colliding with a surface is an example of elastic collision

<u>Reason:</u>

<em>(a very less amount of kinetic energy is lost)</em>

7 0
2 years ago
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sweet [91]
Your answer is c. homologous structures
4 0
3 years ago
A point charge q is located at the center of a spherical shell of radius a that has a charge −q uniformly distributed on its sur
muminat

Answer:

a) E = 0

b) E =  \dfrac{k_e \cdot q}{ r^2 }

Explanation:

The electric field for all points outside the spherical shell is given as follows;

a) \phi_E = \oint E \cdot  dA =  \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}

From which we have;

E \cdot  A =  \dfrac{{\Sigma Q}}{\varepsilon _{0}} = \dfrac{+q + (-q)}{\varepsilon _{0}}  = \dfrac{0}{\varepsilon _{0}} = 0

E = 0/A = 0

E = 0

b) \phi_E = \oint E \cdot  dA =  \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}

E \cdot  A  = \dfrac{+q }{\varepsilon _{0}}

E  = \dfrac{+q }{\varepsilon _{0} \cdot A} = \dfrac{+q }{\varepsilon _{0} \cdot 4 \cdot \pi \cdot r^2}

By Gauss theorem, we have;

E\oint dS =  \dfrac{q}{\varepsilon _{0}}

Therefore, we get;

E \cdot (4 \cdot \pi \cdot r^2) =  \dfrac{q}{\varepsilon _{0}}

The electrical field outside the spherical shell

E =  \dfrac{q}{\varepsilon _{0} \cdot (4 \cdot \pi \cdot r^2) }= \dfrac{q}{4 \cdot \pi \cdot \varepsilon _{0} \cdot r^2 }=  \dfrac{q}{(4 \cdot \pi \cdot \varepsilon _{0} )\cdot r^2 }

k_e=  \dfrac{1}{(4 \cdot \pi \cdot \varepsilon _{0} ) }

Therefore, we have;

E =  \dfrac{k_e \cdot q}{ r^2 }

5 0
3 years ago
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