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2 years ago

Which statement about motor development is true?

Physics

1 answer:

BigorU [14]2 years ago

4 0

Answer:

(a) Motor skills are interrelated.

Explanation:

As per the question,

<u>Motor development:</u>

It is defined as the development of movement abilities such as the development of a child's bones, muscles and ability to move around and manipulate his or her environment.

Importance of a Motor Development Model are:

(1) Helps us make sense of developmental changes

(2) Provides a framework for observing the change

(3) Helps us include important factors of motor behavior in our observations.

Hence, Only option (a) is correct.

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A circular parallel plate capacitor is constructed with a radius of 0.52 mm, a plate separation of 0.013 mm, and filled with an

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Answer:

289282

Explanation:

r = Radius of plate = 0.52 mm

d = Plate separation = 0.013 mm

A = Area = $\pi r^2$

V = Potential applied = 2 mV

k = Dielectric constant = 40

$\epsilon_0$ = Electric constant = $8.854\times 10^{-12}\ \text{F/m}$

Capacitance is given by

$C=\dfrac{k\epsilon_0A}{d}$

Charge is given by

$Q=CV\\\Rightarrow Q=\dfrac{k\epsilon_0AV}{d}\\\Rightarrow Q=\dfrac{40\times 8.854\times 10^{-12}\times\pi \times (0.52\times 10^{-3})^2\times 2\times 10^{-3}}{0.013\times 10^{-3}}\\\Rightarrow Q=4.6285\times 10^{-14}\ \text{C}$

Number of electron is given by

$n=\dfrac{Q}{e}\\\Rightarrow n=\dfrac{4.6285\times 10^{-14}}{1.6\times10^{-19}}\\\Rightarrow n=289281.25\ \text{electrons}$

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2 years ago

Three resistors are wired in parallel with a battery. Two of the resistors have resistances of 38.7 Q/ and 89.5 Q. The current i

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Answer:

$214.9 \Omega$

Explanation:

The three resistors are connected in parallel: this means that the potential difference across each resistor is the same as the voltage of the battery. This can be calculated using the information about the $38.7 \Omega$ resistor: in fact, since we know its resistance and the current flowing through it (0.155 A), we can find the potential difference across this resistor, which is equal to the voltage of the battery:

$V=IR=(0.155 A)(38.7 \Omega)=6.0 V$

We also know the total current in the circuit, 0.250 A. This means that we can find the total resistance of the circuit, using Ohm's law:

$R_{eq}=\frac{V}{I}=\frac{6.0 V}{0.250 A}=24 \Omega$

So now we now the total resistance and the resistance of two of the 3 resistors; therefore, we can find the resistance of the 3rd resistor:

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\\\frac{1}{R_3}=\frac{1}{R_{eq}}-\frac{1}{R_1}-\frac{1}{R_2}=\frac{1}{24 \Omega}-\frac{1}{38.7\Omega}-\frac{1}{89.5\Omega}=0.00465 \Omega^{-1}\\R_3=\frac{1}{0.00465 \Omega^{-1}}=214.9 \Omega$

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2 years ago