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Roman55 [17]
2 years ago
5

Which of the following statements is correct? Which of the following statements is correct? The more a muscle shortens, the more

power it generates. The number of muscle fibers best determines how powerful a muscle will be. Muscle fibers running in parallel arrangement generate more power. Multipennate muscles do not produce much power because the fibers run in many directions.
Physics
1 answer:
NeX [460]2 years ago
4 0

The correct answer are

a)  "The number of muscle fibers best determines how powerful a muscle will be"

b) "The more a muscle shortens, the more power it generates."

Reason :

Muscle fiber in longitudinal directions generate more power

Multipennate muscles do not produce much power because the tendon branches within muscle .



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A 1500-kg car locks its brakes and skids to a stop on a slippery horizontal road, leaving skid marks that are 15 m long. How muc
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Answer:

E=88200\ J

Explanation:

Given:

  • mass of car, m=1500\ kg
  • distance of skidding after the application of brakes, d=15\ m
  • coefficient of kinetic friction, \mu_k=0.4

<u>So, the energy dissipated during the skidding of car:</u>

<em>Frictional force:</em>

f=\mu_k.N

where N = normal reaction by ground on the car

f=0.4\ties 1500\times 9.8

f=5880\ N

<em>Now from the work-energy equivalence:</em>

E=f.d

E=5880\times 15

E=88200\ J is the dissipated energy.

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2 years ago
What is the value of work done on an object when a 0.1x102-newton force moves it 30 meters and the angle between the force and t
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Answer:306joules

Explanation:

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2 years ago
A certain fuse "blows" if the current in it exceeds 1.0 A, at which instant the fuse melts with a current density of What is the
Alborosie

Answer:

<em>0.45 mm</em>

Explanation:

The complete question is

a certain fuse "blows" if the current in it exceeds 1.0 A, at which instant the fuse melts with a current density of 620 A/ cm^2. What is the diameter of the wire in the fuse?

A) 0.45 mm

B) 0.63 mm

C.) 0.68 mm

D) 0.91 mm

Current in the fuse is 1.0 A

Current density of the fuse when it melts is 620 A/cm^2

Area of the wire in the fuse = I/ρ

Where I is the current through the fuse

ρ is the current density of the fuse

Area = 1/620 = 1.613 x 10^-3 cm^2

We know that 10000 cm^2 = 1 m^2, therefore,

1.613 x 10^-3 cm^2 = 1.613 x 10^-7 m^2

Recall that this area of this wire is gotten as

A = \frac{\pi d^{2} }{4}

where d is the diameter of the wire

1.613 x 10^-7 = \frac{3.142* d^{2} }{4}

6.448 x 10^-7 = 3.142 x d^{2}

d^{2} =\sqrt{ 2.05*10^-7}

d = 4.5 x 10^-4 m = <em>0.45 mm</em>

8 0
2 years ago
Two thin lenses with a focal length of magnitude 12.0cm, the first diverging and the second converging, are located 9.00cm apart
attashe74 [19]

Answer:

Explanation:

b ) First is concave lens with focal length f₁ = - 12 cm .

object distance u = - 20 cm .

Lens formula

1 / v - 1 / u = 1 / f

1 / v + 1 / 20 = -1 / 12

1 / v =  - 1 / 20  -1 / 12

= - .05 - .08333

= - .13333

v = - 1 / .13333

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first image is formed before the first lens on the side of object.

This will become object for second lens

distance from second lens = 7.5 + 9 = 16.5 cm

c )

For second lens

object distance u = - 16.5 cm

focal length f₂ = + 12 cm ( lens is convex )

image distance = v

lens formula ,

1 / v - 1 / u = 1 / f₂

1 / v + 1 / 16.5 = 1 / 12

1 / v =   1 / 12 -  1 / 16.5

= .08333- .0606

= .02273

v = 1 /  .02273

= 44 cm ( approx )

It will be formed on the other side of convex lens

distance from first lens

= 44 + 9 = 53 cm .

magnification by first lens = v / u

= -7.5 / -20 = .375 .

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= 44 / - 16.5

= - 2.67

d )

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= .375 x - 2.67

= - 1.00125

height of final image

= 2.50 mm x 1.00125

= 2.503mm

e )

The final image will be inverted with respect to object  because total magnification is negative .

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