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3 years ago

What is amplitude?

1 answer:

8 0

1. Frequency is the number of complete waves that pass a point in a second. 2.Wavelength is the distance between two crests or two troughs. 3.Time period <span> is the time it takes for one complete wave to pass a given point. 4. Amplitude is the height of the wave. Hence option 4 is correct. </span>

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A car accelerates from rest at 2 m/s. what is the speed after 8 sec?

beks73 [17]

Answer:

16m/s

Explanation:

$v_{f}=v_{i}+at$

$v_{f}=0+2\cdot8$

$v_{f}=16\ \frac{m}{s}$

Therefore, the speed after 8 seconds is 16m/s

6 0

2 years ago

A long cylindrical capacitor is made of a central wire of radius a = 2.50 mm surrounded by a conducting shell of radius b = 7.50

asambeis [7]

Answer:

The capacitance per unit length is $5.06\times10^{-11}\ F/m$

(b) is correct option.

Explanation:

Given that,

Radius a= 2.50 mm

Radius b=7.50 mm

Dielectric constant = 3.68

Potential difference = 120 V

We need to calculate charge per length for the capacitance

Using formula of charge per length

$\lambda=\dfrac{4\pi\epsilon_{0}\Delta V}{2 ln(\dfrac{r_{2}}{r_{1}})}$

Put the value into the formula

$\lambda=\dfrac{120}{9\times10^{9}\times2 ln(\dfrac{7.50\times10^{-3}}{2.50\times10^{-3}})}$

$\lambda=6.068\times10^{-9}\ C/m$

We know that,

$\lambda=\dfrac{Q}{L}$

We need to calculate the capacitance per unit length

Using formula of capacitance per unit length

$C=\dfrac{\dfrac{Q}{L}}{\Delta V}$

$C=\dfrac{6.068\times10^{-9}}{120}$

$C=5.06\times10^{-11}\ F/m$

Hence, The capacitance per unit length is $5.06\times10^{-11}\ F/m$

7 0

3 years ago

A 1-kg block is lifted vertically 1 m at constant speed by a boy. The work done by the boy is about:

Igoryamba

Answer:

$9.8\; {\rm J}$ , assuming that the gravitational field strength is $g = 9.8\; {\rm N \cdot kg^{-1}}$ .

Explanation:

Notice that both the speed and the direction of motion of this block are constant. In other words, the velocity of this block is constant.

By Newton's Second Law, the net force on this block would be $0$ . External forces on this block should be balanced. Thus, the magnitude of the (downward) weight of this block should be equal to the magnitude of the (upward) force that the boy applies on this block.

Let $m$ denote the mass of this block. It is given that $m = 1\; {\rm kg}$ . The weight of this block would be:

$\begin{aligned}\text{weight} &= m\, g \\ &= 1\; {\rm kg} \times 9.8\; {\rm N \cdot kg^{-1}} \\ &= 9.8\; {\rm N}\end{aligned}$ .

Hence, the force that the boy applies on this block would be upward with a magnitude of $F = 9.8\; {\rm N}$ .

The mechanical work that a force did is equal to the product of:

the magnitude of the force, and
the displacement of the object in the direction of the force.

The displacement of this block (upward by $s = 1\; {\rm m}$ ) is in the same direction as the (upward) force that this boy had applied. Thus, the work that this boy had done would be the product of: