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1 year ago

In mechanics, massless strings are often assumed. Why is that not a good assumption when discussing waves on strings?

1 answer:

6 0

In mechanics, massless strings are often assumed. but this is not a good assumption when discussing waves on strings because the speed of a wave on a massless string would be infinite.

<h3>How to explain the information?</h3>

It should be noted that waves simply means the dynamic disturbance of a quantity.

It should be noted that in mechanics, massless strings are often assumed. but this is not a good assumption when discussing waves on strings because the speed of a wave on a massless string would be infinite.

Learn more about waves in:

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Draw a schematic diagram of a circuit consisting of a battery of five 2 V cells, a 5 ohm resistor, a 10 ohm resistor, a 15 ohm r

MakcuM [25]

Answer:

Current = 0.33 A

Explanation:

For diagram refer the attachment.

It is given that five cells of 2V are connected in series, so total voltage of the battery:

$\dashrightarrow \: \: \sf V = 2 \times 5 = 10 V$

Three resistor of 5 $\Omega$ , 10 $\Omega$ , 15 $\Omega$ are connected in Series, so the net resistance:

$\dashrightarrow \: \: \sf R_{n} = R_{1} + R_{2} + R_{3}$

$\dashrightarrow \: \: \sf R = 5 + 10 + 15$

${ \pink{\dashrightarrow \sf \: \: { \underbrace{R = 30 \: \Omega}}}}$

According to ohm's law:

$\dashrightarrow \sf\: \: V = IR$

$\dashrightarrow \sf \: \: I = \dfrac{V}{R}$

On substituting resultant voltage (V) as 10 V and resultant resistant, as 30 ${\pmb{\sf{\Omega}}}$ we get:

$\dashrightarrow \sf \: \: I = \dfrac{10V}{30\Omega}$

${ \pink{\dashrightarrow \sf \: \: { \underbrace{I = 0.33 A}}}}$

$\therefore$ The electric current passing through the above circuit when the key is closed will be <u>0.33 A</u>

4 0

2 years ago

Read 2 more answers

A 21 N block rests on a horizontal surface. The coefficients of static and kinetic friction between the surface and the block ar

Georgia [21]

Answer:

a)15 N

b)12.6 N

Explanation:

Given that

Weight of block (wt)= 21 N

μs = 0.80 and μk = 0.60

We know that

Maximum value of static friction given as

Frs = μs m g = μs .wt

by putting the values

Frs= 0.8 x 21 = 16.8 N

Value of kinetic friction

Frk= μk m g = μk .wt

By putting the values

Frk= 0.6 x 21 = 12.6 N

When T = 15 N

Static friction Frs= 16.8 N

Here the value of static friction is more than tension T .It means that block will not move and the value of friction force will be equal to the tension force.

Friction force = 15 N

When T= 35 N

Here value of tension force is more than maximum value of static friction that is why block will move .We know that when body is in motion then kinetic friction will act on the body.so the value of friction force in this case will be 12.6 N

Friction force = 12.6 N

8 0

3 years ago

A tray of electronic components contains 15 components, 4 of which are defective. If 4 components are selected, what is the poss

dlinn [17]

Answer:

a) 0.0007326

b) 0.03223

c) 0.2418

d) 0.2418

Explanation:

To find different probabilities for the selection of components among eleven good and four defective components, we will use the Combination.

a) $C(4,4) = 1; C(15,4) = 1365$

$P = \frac{C(4,4)}{C(15,4)} = \frac{1}{1365} = 0.0007326$

b) $C(4,3) = 4; C(11,1) = 11$

$P = \frac{C(4,3)*C(11,1)}{C(15,4)} = \frac{4*11}{1365} = 0.03223$

c) $C(4,2) = 6; C(11,2) = 55$

$P = \frac{C(4,2)*C(11,2)}{C(15,4)} = \frac{6*55}{1365} = 0.2418$

d) $C(11,4) = 330$

$P = \frac{C(11,4)}{C(15,4)} = \frac{330}{1365} = 0.2418$

8 0

2 years ago

Which type of energy is stored in a capacitor? electric potential energy chemical energy gravitational energy heat energy

Tom [10]

Energy stored in a capacitor is Electric Potential Energy. Capacitor is device used for storing energy. The work done to charge is a capacitor is stored in it in the form of Electrical potential energy. Electrical potential energy is defined as capacity to do work due to the position change. For example, we know fans have capacitor installed in it. When we turn off the fan, it continue moving using the electrical energy stored in the capacitor.

6 0

2 years ago

How much work is required to stop an electron (m = 9.11 \times 10^ - 31 kg) which is moving with a speed of 2.10

Karo-lina-s [1.5K]

Answer:

$-2.00876\times 10^{-18}\ J$

Explanation:

v = Speed of electron = $2.1\times 10^6\ m/s$ (generally the order of magnitude is 6)

m = Mass of electron = $9.11\times 10^{-31}\ kg$

Work done would be done by

$W=K_i-K_f\\\Rightarrow W=0-\dfrac{1}{2}mv^2\\\Rightarrow W=0-\dfrac{1}{2}\times 9.11\times 10^{-31}\times (2.1\times 10^6)^2\\\Rightarrow W=-2.00876\times 10^{-18}\ J$

The work required to stop the electron is $-2.00876\times 10^{-18}\ J$

6 0

3 years ago