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2 years ago

Which of the following accurately describes the way in which a muscle moves?

Physics

1 answer:

Vaselesa [24]2 years ago

7 0

<h3>Answer;</h3>

B. When actin filaments are pulled toward the center of the sarcomere, the fiber shortens.

<h3>Explanation;</h3>

The events of muscle fiber shortening occurs with in the sacromeres in the fibers.
Contraction of striated muscle fibers takes place as the sacromeres shorten as myosin heads pull on the actin filaments.
Filament movement starts at the region or zone where thin and thick filaments overlap.
Myofibril contains many sacromeres along its length and thuse myofibrils and muscle cells contract as the sacromeres contract.

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A bullet of mass 0.010 kg and speed of 200 m/s is brought to rest in a wooden block after penetrating a distance of 0.10 m. The

irina [24]

Answer:

W = 200 J

Explanation:

Work will be equal to the change in kinetic energy

W = ½mv² - 0

W = ½(0.010)200²

W = 200 J

7 0

3 years ago

A steel bar that is at 10 ° c is 5 meters long, a bar for heated to 120 ° c, how long is that bar? Α = 1.2.10- ° c

svet-max [94.6K]

Answer:

1) 5.0066 m

2A) β = 3×10⁻⁷ / °C

2B) 2500.045 cm²

3A) γ = 8.1×10⁻⁵ / °C

3B) 1618.144 cm³

Explanation:

1) Linear thermal expansion is:

ΔL = α L₀ ΔT

where ΔL is the change in length,

α is the linear thermal expansion coefficient,

L₀ is the original length,

and ΔT is the change in temperature.

Given L₀ = 5 m, ΔT = 110°C, and α = 1.2×10⁻⁵ / °C:

ΔL = (1.2×10⁻⁵ / °C) (5 m) (110°C)

ΔL = 0.0066 m

The length increases by , so the new length is:

L = L₀ + ΔL

L = 5 m + 0.0066 m

L = 5.0066 m

2A) The surface expansion coefficient is:

β = 2α

β = 2 (1.5×10⁻⁷ / °C)

β = 3×10⁻⁷ / °C

2B) The change in area is:

ΔA = β A₀ ΔT

ΔA = (3×10⁻⁷ / °C) (50 cm × 50 cm) (60°C)

ΔA = 0.045 cm²

So the new area is:

A = A + ΔA

A = 2500 cm² + 0.045 cm²

A = 2500.045 cm²

3A) The volumetric expansion coefficient is:

γ = 3α

γ = 3 (2.7×10⁻⁵ / °C)

γ = 8.1×10⁻⁵ / °C

3B) The change in volume is:

ΔV = γ V₀ ΔT

ΔV = (8.1×10⁻⁵ / °C) (1600 cm³) (140°C)

ΔV = 18.144 cm³

So the new area is:

V = V + ΔV

V = 1600 cm³ + 18.144 cm³

V = 1618.144 cm³

6 0

3 years ago

A 190 g rock at 20°C is completely immersed in 600 g of water at 80°C. Which one of the following statements is true right after

Umnica [9.8K]

Rock is completely immersed in hot water. By the second law of thermodynamics, thermal energy or heat is transferred from substance with higher temperature to substance with lower temperature until they come to thermal equilibrium i.e. both at same temperature.

It is given here that rock is at 20°C which is at lower temperature than water at 80°C. ∴Heat or thermal energy flows from water to rock. So, right choice is-

A. The water gives the rock thermal energy and gets no thermal energy in return.

8 0

3 years ago

What causes an impact crater to form?

Scrat [10]

Answer:

it is the impact of a meteorite, volcanic activity, or an explosion.

4 0

3 years ago

Estimate how far apart the rays of deepest red and deepest violet light are as they exit the bottom surface. assume nred = 1.57

Harlamova29_29 [7]

We begin by noting that the angle of incidence is the one that's taken with respect to the normal to the surface in question. In this case the angle of incidence is 30. The material is Flint Glass according to the original question. The refractive indez of air n1=1, the refractive index of red in flint glass is nred=1.57, finally for violet in the glass medium is nviolet=1.60. Snell's Law dictates:
$n_1sin(\theta_1)=n_2sin(\theta_2)$
Where $\theta_2$ differs for each wavelenght, that means violet and red will have different refractive indices in the glass.
In the second figure provided details are given on which are the angles in question, $\Delta x$ is the distance between both rays.
$\theta_{2red}=Asin(\frac{sin(30)}{1.57})\approx 18.5705$
$\theta_{2violet}=Asin(\frac{sin(30)}{1.60})\approx 18.21$
At what distance d from the incidence normal will the beams land at the bottom?
For violet we have:
$d_{violet}=h.tan(\theta_{2violet})\approx 0.0132m$
For red we have:
$d_{red}=h.tan(\theta_{2red})\approx 0.0134m$
We finally have:
$\Delta x=d_{red}-d_{violet}\approx2.8\times10^{-4}m$

6 0

3 years ago