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nikklg [1K]
3 years ago
7

If you wanted to increase your muscular endurance, it would be important to participate in activities that __________.

Physics
2 answers:
anastassius [24]3 years ago
8 0

Answer:

The answer is A

Explanation:

Just took the test

Maurinko [17]3 years ago
3 0
In order to increase muscular endurance, it is important to take part in activities that increase the size of muscle cells and the amount of blood that is delivered to the cells<span>. The increased size of the muscle cells reuires them to contract less to accomplish the same task; moreover, the increased blood flow helps deliver oxygen and nutrients to the blood.</span>
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There is 4000 balls in the earth of the world why is that so hard for you
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The air pressure at the ocean surface a few miles from the shoreis most likely
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The air pressure is most likely lower.
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What is the mechanical advantage of the wheel and axle shown below?
goldenfox [79]

Answer:

B. 120

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3 years ago
A potentialiterence of 12 volts is induced across a straight wire 0.20 meters long as it is moved at a constant speed of 3.0 met
shtirl [24]

Strength of the magnetic field: 20 T

Explanation:

For a conductive wire moving perpendicular to a magnetic field, the electromotive force (voltage) induced in the wire due to electromagnetic induction is given by

\epsilon=BvL

where

B is the strength of the magnetic field

v is the speed of the wire

L is the length of the wire

For the wire in this problem, we have:

\epsilon=12 V (induced emf)

L = 0.20 m (length of the wire)

v = 3.0 m/s (speed)

Solving  for B, we find the strength of the magnetic field:

B=\frac{\epsilon}{vL}=\frac{12}{(0.20)(3.0)}=20 T

Learn more about magnetic fields:

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6 0
3 years ago
The electron gun in a television tube is used to accelerate electrons with mass 9.109 × 10−31 kg from rest to 3 × 107 m/s within
zaharov [31]

Answer:

Electric field, E = 40608.75 N/C

Explanation:

It is given that,

Mass of electrons, m=9.1\times 10^{-31}\ kg

Initial speed of electron, u = 0

Final speed of electrons, v=3\times 10^7\ m/s

Distance traveled, s = 6.3 cm = 0.063 m

Firstly, we will find the acceleration of the electron using third equation of motion as :

a=\dfrac{v^2-u^2}{2s}

a=\dfrac{(3\times 10^7)^2}{2\times 0.063}

a=7.14\times 10^{15}\ m/s^2

Now we will find the electric field required in the tube as :

ma=qE

E=\dfrac{ma}{q}

E=\dfrac{9.1\times 10^{-31}\times 7.14\times 10^{15}}{1.6\times 10^{-19}}

E = 40608.75 N/C

So, the electric field required in the tube is 40608.75 N/C. Hence, this is the required solution.

3 0
3 years ago
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