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diamong [38]
3 years ago
6

When bones are moved by muscles due to

Physics
2 answers:
ryzh [129]3 years ago
7 0
A is the correct answer
77julia77 [94]3 years ago
6 0
Respiratory and circulatory
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A train starts at rest in a station and accelerates at a constant 0.987 m/s2 for 182 seconds. Then the train decelerates at a co
algol [13]

Answer:

\displaystyle X_T=66.6\ km

Explanation:

<u>Accelerated Motion </u>

When a body changes its speed at a constant rate, i.e. same changes take same times, then it has a constant acceleration. The acceleration can be positive or negative. In the first case, the speed increases, and in the second time, the speed lowers until it eventually stops. The equation for the speed vf at any time t is given by

\displaystyle V_f=V_o+a\ t

where a is the acceleration, and vo is the initial speed .

The train has two different types of motion. It first starts from rest and has a constant acceleration of 0.987 m/s^2 for 182 seconds. Then it brakes with a constant acceleration of -0.321 m/s^2 until it comes to a stop. We need to find the total distance traveled.

The equation for the distance is

\displaystyle X=V_o\ t+\frac{a\ t^2}{2}

Our data is

\displaystyle V_o=0,a=0.987m/s^2,\ t=182\ sec

Let's compute the first distance X1

\displaystyle X_1=0+\frac{0.987\times 182^2}{2}

\displaystyle X_1=16,346.7\ m

Now, we find the speed at the end of the first period of time

\displaystyle V_{f1}=0+0.987\times 182

\displaystyle V_{f1}=179.6\ m/s

That is the speed the train is at the moment it starts to brake. We need to compute the time needed to stop the train, that is, to make vf=0

\displaystyle V_o=179.6,a=-0.321\ m/s^2\ ,V_f=0

\displaystyle t=\frac{v_f-v_o}{a}=\frac{0-179.6}{-0.321}

\displaystyle t=559.5\ sec

Computing the second distance

\displaystyle X_2=179.6\times559.5\ \frac{-0.321\times 559.5^2}{2}

\displaystyle X_2=50,243.2\ m

The total distance is

\displaystyle X_t=x_1+x_2=16,346.7+50,243.2

\displaystyle X_t=66,589.9\ m

\displaystyle \boxed{X_T=66.6\ km}

8 0
4 years ago
Calculate the pressure exerted by a 4000N camel on the sand. The camel’s feet have a
Harrizon [31]

Answer:

The pressure exerted by camel feet is <u>2000 N/m²</u>.

Step-by-step explanation:

<h3><u>Solution</u> :</h3>

Here, we have given that ;

  • Force applied on camel feet = 4000 N
  • Total area of camel feet = 2 m²

We need to find the pressure exerted by camel feet.

As we know that :

{\longrightarrow{\pmb{\sf{Pressure= \dfrac{Area}{Force}}}}}

Substituting all the given values in the formula to find the pressure exerted by camel feet.

\begin{gathered} \begin{array}{l} {\longrightarrow{\sf{Pressure= \dfrac{Area}{Force}}}} \\  \\ {\longrightarrow{\sf{Pressure= \dfrac{4000}{2}}}}  \\  \\ {\longrightarrow{\sf{Pressure= \cancel{\dfrac{4000}{2}}}}} \\  \\ {\longrightarrow{\sf{Pressure= 2000 \: N/{m}^{2}}}} \\  \\\star \:  \small\underline{\boxed{\sf{\purple{Pressure= 2000 \: N/{m}^{2}}}}} \end{array}\end{gathered}

Hence, the pressure exerted by camel feet is 2000 N/m².

\rule{300}{2.5}

3 0
2 years ago
What are electromagnetic waves and how do they travel?
True [87]

Answer:

Propagation of an Electromagnetic Wave. Electromagnetic waves are waves which can travel through the vacuum of outer space. Mechanical waves, unlike electromagnetic waves, require the presence of a material medium in order to transport their energy from one location to another.

5 0
3 years ago
Which of the following equations illustrates the law of conservation of<br> matter?
vagabundo [1.1K]

Answer:

I see no eaquations?

It should be matter=constant

3 0
3 years ago
A 2-kg object is initially at the bottom of a long 50° inclined plane, and is beginning to slide up this inclined plane. The ini
Over [174]

Answer:

Explanation:

Given

mass of object m=2 kg

inclination \theta =50^{\circ}

\mu _k=0.3

\mu _s=0.4

initial velocity u=3 m/s

acceleration of block during upward motion

a=g\sin \theta -\mu _kg\cos \theta

a=g(\sin 50-0.3\cos 50)

a=5.617 m/s^2

using relation

v^2-u^2=2a\cdot s

where s=distance\ moved

v=final\ velocity

v=0 because block stopped after moving distance s

0-(3)^2=2\cdot (-5.617)\cdot s

s=\frac{4.5}{5.617}

s=0.801

If block stopped after s m then force acting on block is

F=mg\sin \theta =friction force f_r=\mu mg\cos \theta

F>f_r therefore block will slide back down to the bottom            

4 0
3 years ago
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