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Alex_Xolod [135]

2 years ago

15

What can be found in every skeletal muscle?

2 answers:

anastassius [24]2 years ago

7 0

You can find muscle fibers, nerves, connective tissue, and blood vessels in every skeletal system. I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

kondor19780726 [428]2 years ago

7 0

Answer:

D. tendons, nerves, blood vessels, and bones

Explanation:

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Jane is sliding down a slide. What kind of motion is she demonstrating? A. translational motion B. rotational motion C. vibratio

Andreyy89

A. translational motion

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3 years ago

(14) The graphs show the effect of force on a metal spring.

Alexeev081 [22]

Answer:

I think is D I'm sorry if I'm wrong

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2 years ago

Which statement best describes the direction of the buoyant force on any object?

poizon [28]

The buoyant force on any object acts in the direction opposite to the force of gravity. <em>(A)</em>

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2 years ago

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25% part (c) assume that d is the distance the cheetah is away from the gazelle when it reaches full speed. Derive an expression

levacccp [35]

maximum speed of cheetah is

$v_1 = v_{max}$

speed of gazelle is given as

$v_2 = v_{g}$

Now the relative speed of Cheetah with respect to Gazelle

$v_{12} = v_1 - v_2$

$v_{12} = v_{max} - v_g$

now the relative distance between Cheetah and Gazelle is given initially as "d"

now the time taken by Cheetah to catch the Gazelle is given as

$d = v_{12}* t$

so by rearranging the terms we can say

$t = \frac{d}{v_{12}}$

$t = \frac{d}{v_{max} - v_g}$

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6 0

3 years ago

A carbon resistor is to be used as a thermometer. On a winter day when the temperature is 4.00°C, the resistance of the carbon r

dusya [7]

Answer:

28 degree C

Explanation:

We are given that

$T_1=4.00^{\circ}$

$R_1=217.7 \Ohm$

$R_2=215.1\Ohm$

$\alpha=-5.00\times 10^{-4}C^{-1}$

We have to find the temperature on a spring day when resistance is 215.1 ohm.

We know that

$\alpha(T_2-T_1)=\frac{R_2}{R_2}-1$

Using the formula

$-5.00\times 10^{-4}(T_2-4)=\frac{215.1}{217.7}-1$

$-5\times 10^{-4}(T_2-4)=0.988-1=-0.012$

$T_2-4=\frac{0.012}{5\times 10^{-4}}=24$

$T_2=24+4=28^{\circ}C$

Hence, the temperature on a spring day 28 degree C.

7 0

3 years ago