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2 years ago

Which of the following accurately describes the way in which a muscle moves?

Physics

1 answer:

Vaselesa [24]2 years ago

7 0

<h3>Answer;</h3>

B. When actin filaments are pulled toward the center of the sarcomere, the fiber shortens.

<h3>Explanation;</h3>

The events of muscle fiber shortening occurs with in the sacromeres in the fibers.
Contraction of striated muscle fibers takes place as the sacromeres shorten as myosin heads pull on the actin filaments.
Filament movement starts at the region or zone where thin and thick filaments overlap.
Myofibril contains many sacromeres along its length and thuse myofibrils and muscle cells contract as the sacromeres contract.

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Two identical small metal spheres with q1 > 0 and |q1| > |q2| attract each other with a force of magnitude 72.1 mN when se

Brrunno [24]

1) $+2.19\mu C$

The electrostatic force between two charges is given by

$F=k\frac{q_1 q_2}{r^2}$ (1)

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

When the two spheres are brought in contact with each other, the charge equally redistribute among the two spheres, such that each sphere will have a charge of

$\frac{Q}{2}$

where Q is the total charge between the two spheres.

So we can actually rewrite the force as

$F=k\frac{(\frac{Q}{2})^2}{r^2}$

And since we know that

r = 1.41 m (distance between the spheres)

$F= 21.63 mN = 0.02163 N$

(the sign is positive since the charges repel each other)

We can solve the equation for Q:

$Q=2\sqrt{\frac{Fr^2}{k}}=2\sqrt{\frac{(0.02163)(1.41)^2}{8.98755\cdot 10^9}}}=4.37\cdot 10^{-6} C$

So, the final charge on the sphere on the right is

$\frac{Q}{2}=\frac{4.37\cdot 10^{-6} C}{2}=2.19\cdot 10^{-6}C=+2.19\mu C$

2) $q_1 = +6.70 \mu C$

Now we know the total charge initially on the two spheres. Moreover, at the beginning we know that

$F = -72.1 mN = -0.0721 N$ (we put a negative sign since the force is attractive, which means that the charges have opposite signs)

r = 1.41 m is the separation between the charges

And also,

$q_2 = Q-q_1$

So we can rewrite eq.(1) as

$F=k \frac{q_1 (Q-q_1)}{r^2}$

Solving for q1,

$Fr^2=k (q_1 Q-q_1^2})\\kq_1^2 -kQ q_1 +Fr^2 = 0$

Since $Q=4.37\cdot 10^{-6} C$ , we can substituting all numbers into the equation:

$8.98755\cdot 10^9 q_1^2 -3.93\cdot 10^4 q_1 -0.141 = 0$

which gives two solutions:

$q_1 = 6.70\cdot 10^{-6} C\\q_2 = -2.34\cdot 10^{-6} C$

Which correspond to the values of the two charges. Therefore, the initial charge q1 on the first sphere is

$q_1 = +6.70 \mu C$

8 0

3 years ago

Water from a fire hose is directed horizontally against at a rate of 50.0 kg/s and a speed of 42.0 m/s.

ss7ja [257]

Answer:

Force, |F| = 2100 N

Explanation:

It is given that,

Water from a fire hose is directed horizontally against at a rate of 50.0 kg/s, $\dfrac{m}{t}=50\ kg/s$

Initial speed, v = 42 m/s

The momentum is reduced to zero, final speed, v = 0

The relation between the force and the momentum is given by :

$F=\dfrac{p}{t}$

$F=\dfrac{mv}{t}$

$F=50\ kg/s\times 42\ m/s$

|F| = 2100 N

So, the magnitude of the force exerted on the wall is 2100 N. Hence, this is the required solution.

8 0

3 years ago

• La longitud de un alambre de un sujetador de papel (clip) extendido mide 86 mm. ¿Cuántos sujetadores se pueden obtener de un r

kari74 [83]

Answer:

34883.7

Explanation:

Primero hay que convertir las dos unidades a la misma unidad-cm-

3km- 300000 cm

86mm- 8.6mm

Después hay que dividir

300000/8.6= 34883.7209302 ~ 34883.7

7 0

3 years ago

What might common French citizens have liked and disliked about the metric system when it was adopted

Gwar [14]

They would have disliked that they had to relearn how to measure.

5 0

3 years ago

You are driving 12 meters per second. What is your speed in miles per hour? (1.6km=1 mike) EXPLAIN

Blizzard [7]

1.6km = 1 mike? Wow that guy is tall.

I think you meant 1.6 km = 1 mile? Okay then if we're going 12 meters per second how much would you travel in one hour? First we need to figure out how many seconds are in an hour. There are 60 seconds in a minute and 60 minutes in an hour so:

60×60=3600 seconds in an hour

Now we will multiply that by 12 meters per second and we get:

$12\frac{m}{s} \times 3600 s=43200m$

And 43200 meters is 43.2 km (1000 meters in 1 kilometer) meaning 43.2 kilometers an hour. Since there are 1.6 km in one mile we must divide 43.2km to 1.6.

$\frac{43.2}{1.6}=27mph$

And so your speed is 27 miles per hour.

8 0

3 years ago