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3 years ago

What are the three types of muscle fibers?

Physics

2 answers:

DiKsa [7]3 years ago

6 0

Smooth,cardiac,invountary

jek_recluse [69]3 years ago

3 0

Skeletal slow-twitch, fast-twitch, involuntary

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Reading glasses with a power of 1.50 diopters make reading a book comfortable for you when you wear them 1.8 cmcm from your eye.

Vlad1618 [11]

Answer:

The near point is $n =44.8 \ cm$

Explanation:

From the question we are told that

The power is $P = 1.50$

The distance from the eye is $k = 1.8 \ cm$

The distance of the book from the eye is $z = -28 \ cm$

Generally the focal length of the glasses is

$f = \frac{1}{P}$

=> $f = \frac{1}{1.50 }$

=> $f = 0.667 \ m$

=> $f = 66.7 \ cm$

The object distance is evaluated as

$u = z + k$

=> $u = -28 + 1.8$

=> $u = -26.2 \ cm$

The image distance is evaluated from lens formula as

$\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$

=> $\frac{1}{v} = \frac{1}{66.7} + \frac{1}{-26.2}$

=> $v=- \frac{1}{0.0232}$

=> $v=- 43 \ cm$

The near point is evaluated as

$n = -v + k$

=> $n =-(-43) + 1.8$

=> $n =44.8 \ cm$

3 0

3 years ago

An Olympic runner competing in a long-distance event finishes with a time of 2 hours, 45 minutes, and 35 seconds. The event has

Readme [11.4K]

Answer:

Average speed of Olympic runner will be 23.08 km/hr

Explanation:

We have given distance d = 27.3 miles

We know that 1 mile = 1.6 km

So 27.3 mile = $=27.3\times 1.6=43.68km$

Given time = 2 hour 45 minutes and 35 sec

We know that 1 hour = 60 minutes

So 45 minutes $=\frac{45}{60}=0.75hour$

We also know that 1 hour = 3600 sec

So 1 sec $=\frac{1}{3600}=2.777\times 10^{-4}hour$

So total time t = $=2+0.75+0.000277=2.750277hour$

We know that speed $=\frac{distance}{time }=\frac{63.48}{2.750277}=23.08km/hour$

6 0

3 years ago

An automobile engine takes in 4000 j of heat and performs 1100 j of mechanical work in each cycle. (a) calculate the engine's ef

Semmy [17]

(a) The efficiency of an engine is defined as the ratio between the work done by the engine and the heat it takes in:
$\eta= \frac{W}{Q_{in}}$
The engine in this problem does a work of $W=1100 J$ and it takes in $Q_{in}=4000 J$ of heat, therefore its efficiency is
$\eta= \frac{1100 J}{4000 J}=0.275 = 27.5 \%$

(b) The heat taken by the machine is 4000 J; of this amount of heat, only 1100 J are converted into useful work. This means that the rest of the heat is wasted. Therefore, the wasted heat is the difference between the heat in input and the work done by the engine:
$Q_{wasted}=Q_{in}-W=4000 J-1100 J=2900 J$

7 0

3 years ago

A hoop, a solid disk, and a solid sphere, all with the same mass and the same radius, are set rolling without slipping up an inc

nadya68 [22]

Answer:

Solution is given in the attachments