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tatiyna
3 years ago
15

The force generated by a single muscle fiber can be increased by increasing is called

Physics
1 answer:
mars1129 [50]3 years ago
4 0

Answer:

The force generated by a single muscle fiber can be increased by increasing the frequency of action potentials

Explanation:

The force generated by a muscle fiber is the result of the shortening of the skeletal muscle, and this force is also know as muscle tension. The larger motor units shorten along with the smaller units to produce the muscle force. The time lapsed between the beginning of the action potential in the muscle and the beginning of the contraction is the latent period. Action potential is the result of the difference electrical potential as a result of passage of an impulse along the membrane of a muscle or nerve cell.

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3 0
3 years ago
Please help i need this by today
makkiz [27]

Answer:

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Explanation:

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A 180 cm length of string has a mass of 5.0 g. it is stretched with a tension of 8.6 n between fixed supports. (a) what is the w
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5 0
4 years ago
One pot has a water height of 8cm. Calculate the water pressure at the bottom of the pot. They give g = 10N / kg p = 1000kg / m3
natima [27]

Answer:

P = 800 Pa

Explanation:

The pressure of water at the bottom of the pot can be given by the following formula:

P = \rho g h

where,

P = Pressure at the bottom of the pot = ?

ρ = density = 1000 kg/m³

h = height of water = 8 cm = 0.08 m

Therefore,

P = (1000\ kg/m^3)(10\ m/s^2)(0.08\ m)

<u>P = 800 Pa </u>

5 0
3 years ago
The masses of the earth and moon are 5.98 x 1024 and 7.35 x 1022 kg, respectively. Identical amounts of charge are placed on eac
dybincka [34]

Answer:

The magnitude of charge on each is 5.707\times 10^{13} C

Solution:

As per the question:

Mass of Earth, M_{E} = 5.98\times 10^{24} kg

Mass of Moon, M_{M} = 7.35\times 10^{22} kg

Now,

The gravitational force of attraction between the earth and the moon, if 'd' be the separation distance between them is:

F_{G} = \frac{GM_{E}M_{M}}{d^{2}}        (1)

Now,

If an identical charge 'Q' be placed on each, then the Electro static repulsive force is given by:

F_{E} = \frac{1}{4\pi\epsilon_{o}}\frac{Q^{2}}{d^{2}}           (2)

Now, when the net gravitational force is zero, the both the gravitational force and electro static force mut be equal:

Equating eqn (1) and (2):

\frac{GM_{E}M_{M}}{d^{2}} = \frac{1}{4\pi\epsilon_{o}}\frac{Q^{2}}{d^{2}}

(6.67\times 10^{- 11})\times (5.98\times 10^{24})\times (7.35\times 10^{22}) = (9\times 10^{9}){Q^{2}}

\sqrt{\farc{(6.67\times 10^{- 11})\times (5.98\times 10^{24})\times (7.35\times 10^{22})}{9\times 10^{9}}} = Q

Q = \pm 5.707\times 10^{13} C

7 0
3 years ago
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