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myrzilka [38]
1 year ago
10

your teacher shows you a cell that contains filaments that can cause the cell to contract,shortening the length of the cell.to w

hich system of the body might this cell belong to?
Physics
1 answer:
nika2105 [10]1 year ago
8 0

The cell belongs to the then musculoskeletal system of the body.

<h3>What type of cells contains filaments that can cause the cell to contract, shortening the length of the cell?</h3>

The myocytes or muscle cells are special cells found in smooth, cardiac, and skeletal muscles that have differentiated for the specialized function of contraction.

Myocytes are long, cylindrical, multi-nucleated, and striated. They contain filaments that can cause the cell to contract, shortening the length of the cell.

Skeletal muscle cells contain many mitochondria required to generate sufficient ATP since skeletal muscle cells have high energy requirements.

The muscle cells form a part of the musculoskeletal system of the body.

In conclusion, muscle cells are required for muscular contraction.

Learn more about muscle cells at: brainly.com/question/15441674

#SPJ1

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How can magnet be used to produce an electric current
Bess [88]

Answer:

Magnetic fields can be used to make electricity

Moving a magnet around a coil of wire, or moving a coil of wire around a magnet, pushes the electrons in the wire and creates an electrical current. Electricity generators essentially convert kinetic energy (the energy of motion) into electrical energy

Explanation:

3 0
3 years ago
A parallel-plate air capacitor is made from two plates 0.210 m square, spaced 0.815 cm apart. it is connected to a 120 v battery
GuDViN [60]

Answer:

at the beginning: 2.3\cdot 10^{-10} F

when the plates are pulled apart: 1.1\cdot 10^{-10} F

Explanation:

The capacitance of a parallel-plate capacitor is given by

C=k \epsilon_0 \frac{A}{d}

where

k is the relative permittivity of the medium (for air, k=1, so we can omit it)

\epsilon_0 = 8.85\cdot 10^{-12} F/m is the permittivity of free space

A is the area of the plates of the capacitor

d is the separation between the plates

In this problem, we have:

A=0.210 m^2 is the area of the plates

d=0.815 cm=8.15\cdot 10^{-3} m is the separation between the plates at the beginning

Substituting into the formula, we find

C=(1)(8.85\cdot 10^{-12}F/m)\frac{0.210 m^2}{8.15\cdot 10^{-3} m}=2.3\cdot 10^{-10} F

Later, the plates are pulled apart to d=1.63 cm=0.0163 m, so the capacitance becomes

C=(1)(8.85\cdot 10^{-12}F/m)\frac{0.210 m^2}{0.0163 m}=1.1\cdot 10^{-10} F

4 0
3 years ago
A wave has a wavelength of 5 meters and a frequency of 3 hertz. What is the speed of the wave?
otez555 [7]
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8 0
3 years ago
Astronauts wear liquid cooled space suits to keep their body temperature moderate. One
Anton [14]

Answer:

D. Because they are using space technology on a shirt so people can wear it on earth as well

6 0
2 years ago
A single-phase electrical load draws 600KVA at 0.6 power factor lagging. a) Find the real and reactive power absorbed by the loa
Whitepunk [10]

Answer:

(a). The reactive power is 799.99 KVAR.

(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.

Explanation:

Given that,

Power factor = 0.6

Power = 600 kVA

(a). We need to calculate the reactive power

Using formula of reactive power

Q=P\tan\phi...(I)

We need to calculate the \phi

Using formula of \phi

\phi=\cos^{-1}(Power\ factor)

Put the value into the formula

\phi=\cos^{-1}(0.6)

\phi=53.13^{\circ}

Put the value of Φ in equation (I)

Q=600\tan(53.13)

Q=799.99\ kVAR

(b). We draw the power triangle

(c). We need to calculate the reactive power of a capacitor to be connected across the load to raise the power factor to 0.95

Using formula of reactive power

Q'=600\tan(0.95)

Q'=9.94\ KVAR

We need to calculate the difference between Q and Q'

Q''=Q-Q'

Put the value into the formula

Q''=799.99-9.94

Q''=790.05\ KVAR

Hence, (a). The reactive power is 799.99 KVAR.

(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.

8 0
3 years ago
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