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3 years ago

During exercise, the amount of ____________ carried in the blood to the muscles increases. (3 points)

Physics

2 answers:

adoni [48]3 years ago

7 0

You would need oxygen because as your workout intensifies, your body needs more oxygen to be supplied to the muscles.

Ilya [14]3 years ago

3 0

Nutrients and oxygen

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Water drops from the nozzle of a shower onto the floor 81 inches below. The drops fall at regular intervals of time, the first d

Alexxx [7]

Answer:

0.91437 m

0.22859 m

Explanation:

g = Acceleration due to gravity = 9.81 m/s² = a

$s=81\ inches=81\times 0.0254=2.0574\ m$

$s=ut+\frac{1}{2}at^2\\\Rightarrow 2.0574=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{2.0574\times 2}{9.81}}\\\Rightarrow t=0.64764\ s$

When the time intervals are equal, if four drops are falling then we have 3 time intervals.

So, the time interval is

$t'=\dfrac{t}{3}\\\Rightarrow t'=\dfrac{0.64764}{3}\\\Rightarrow t'=0.21588\ s$

For second drop time is given by

$t''=2t'\\\Rightarrow t''=2\times 0.21588\\\Rightarrow t''=0.43176\ s$

Distance from second drop

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow y''=ut''+\dfrac{1}{2}at''^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 0.43176^2\\\Rightarrow s=0.91437\ m$

Distance from second drop is 0.91437 m

Distance from third drop

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow y''=ut'+\dfrac{1}{2}at'^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 0.21588^2\\\Rightarrow s=0.22859\ m$

Distance from third drop is 0.22859 m

6 0

3 years ago

A particle moves along a straight path through displacement while force acts on it. (Other forces also act on the particle.) Wha

allsm [11]

a) c = 1.85

b) c = 0.8

c) c = 2.33

Explanation:

The displacement of the particle is given by

$d=2.2i+cj$

While the force applied on the particle is

$F=3.2i-3.8 j$

So we have a problem in 2-dimensions.

The work done on the particle is given by the scalar product between force and displacement:

$W=F\cdot d$ (1)

Here the work done on the particle is zero, so

W = 0

Therefore from eq(1) we find:

$0=(3.2i-3.8j)\cdot (2.2i+cj)=7.04-3.8c\\3.8c=7.04\\c=\frac{7.04}{3.8}=1.85$

In this problem, the work done on the particle is

$W=4.0 J$

The force and displacement are still

$d=2.2i+cj$ (displacement)

$F=3.2i-3.8 j$ (force)

Therefore, by calculting the scalar product between force and displacement and equating it to the work done (4.0 J), we find:

$W=F\cdot d$

$4.0 =(3.2i-3.8j)\cdot (2.2i+cj)=7.04-3.8c\\3.8c=3.04\\c=\frac{3.04}{3.8}=0.8$

In this problem instead, the work done on the particle is negative:

$W=-1.8 J$

As before, the force and displacement are

$d=2.2i+cj$ (displacement)

$F=3.2i-3.8 j$ (force)

And so again, we calculate the scalar product between force and displacement and we equate it to the work done on the particle, -1.8 J.

Doing so, we find:

$W=F\cdot d$

$-1.8=(3.2i-3.8j)\cdot (2.2i+c)=7.04-3.8c\\3.8c=8.84\\c=\frac{8.84}{3.8}=2.33$

7 0

3 years ago

Two bicycle tires are set rolling with the same initial speed of 4.00 m/s along a long, straight road, and the distance each tra

77julia77 [94]

Answer:

The coefficient of rolling friction for the tire under low pressure is 0.0342.

Explanation:

Two bicycle tires are set rolling with the same initial speed of 4.00 m/s

Final speed of both the bicycle, speed is reduced by half is measured, v = 2 m/s.

Here,

$u_kmg = ma\\\\a=\mu g$

Using third equation of motion as :

$v^2-u^2=2as\\\\v^2-u^2=2\mu gs\\\\\mu =\dfrac{v^2-u^2}{2gs}\\\\\mu =\dfrac{4^2-2^2}{2\times 9.8\times 17.9}\\\\\mu=0.0342$

So, the coefficient of rolling friction for the tire under low pressure is 0.0342.

5 0

3 years ago

Your friend is flying a remote control jet in your yard, as seen in the diagram above. As it flies past you, you notice that the

rjkz [21]

Answer:c

Explanation:

7 0

4 years ago

What would the charge of an atom be if it has six protons and six electrons

disa [49]

<span>Carbon atom showing 6 protons as electro positive charge in its nuclei should have same no of -ve charged electrons , that is 6 electronns in its radii to make an atom of C as a neutral element , for that each element in its atomic state should have equal no of protons and electrons , the no of neutrons which has no electric charge and almost negligible weight does play significant role except contributes to isotopes of the given atom or an element, so C atom having 6 protons should have 6 electrons</span>

3 0

4 years ago

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