Ask question

Ask question

All categories

aleksandrvk [35]

3 years ago

13

You are generating traveling waves on a stretched string by wiggling one end. If you suddenly begin to wiggle more rapidly witho

ut appreciably affecting the tension, you will cause the waves to move down the string a. faster than before.b. at the same speed as before.c. slower than before.

1 answer:

german3 years ago

6 0

Answer:

Option as B is correct At the same speed as before

Explanation:

As we know the relation between speed of the wave and tension in string

The speed of wave in stretched string

ν = $\sqrt{\frac{T}{\mu} }$

speed of wave is the directly proportional to the square root of tension as mentioned in question tension of string is unaffected when in linear mass density is constant, so we can say that the speed of wave will be the same

Option as B is correct At the same speed as before

You might be interested in

Which property of potential energy distinguishes it from kinetic energy

Bas_tet [7]

Answer:

Shape and position

Explanation:

Hope this helps! :)

7 0

2 years ago

Djejfje kdjejfjeofjod fjdidjskd.jdjdjdjdp

Mariana [72]

YUHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH gbghdwqygfywefyugsadfyuyuqwbhfbds hehrwuhuaijkfa

5 0

3 years ago

Explain different types of thermometer and their thermometry substance

Rudiy27

Answer:

Explained below

Explanation:

1) Liquid in glass thermometer: This type of thermometer is used primarily to measure the temperatures from inspection of changes in volume of liquid.

Thermometry substance is mercury or alcohol

2) Gas thermometer: This type is used to measure temperature as a result of changes in gas pressure or volume.

Thermometry substance is Gas.

3) Resistance thermometer: This type is used to measure temperature due to changes in electric resistance.

Thermometry substance is Resistance wire.

4) Thermocouple thermometer: This type is used to measure the temperature due to changes in electrical potential difference occurring between two metal junctions.

Thermometry substance is two wires that are dissimilar.

5) Bimetallic thermometer: This is a type of thermometer that measures temperature by converting temperature into mechanical displacement by making use of Bimetallic strip.

Thermometry substance is two metals that are dissimilar.

3 0

3 years ago

What must be the length of a simple pendulum if its oscillation frequency is to be equal to that of an air-track glider of mass

Anvisha [2.4K]

Answer:

the length of the simple pendulum is 0.25 m.

Explanation:

Given;

mass of the air-track glider, m = 0.25 kg

spring constant, k = 9.75 N/m

let the length of the simple pendulum = L

let the frequency of the air-track glider which is equal to frequency of simple pendulum = F

The oscillation frequency of air-track glider is calculated as;

$F = \frac{1}{2\pi } \sqrt{\frac{k}{m} } \\\\F = \frac{1}{2\pi } \sqrt{\frac{9.75}{0.25} } \\\\F = 0.994 \ Hz$

The frequency of the simple pendulum is given as;

$F = \frac{1}{2\pi} \sqrt{\frac{g}{l} } \\\\2\pi(F) = \sqrt{\frac{g}{l} } \\\\2\pi (0.994) = \sqrt{\frac{9.8}{l} } \\\\6.2455 = \sqrt{\frac{9.8}{l} } \\\\(6.2455)2 = \frac{9.8}{l} \\\\39.006 = \frac{9.8}{l} \\\\l = \frac{9.8}{39.006} \\\\l = 0.25 \ m$

Thus, the length of the simple pendulum is 0.25 m.

8 0

2 years ago

Block B is attached to a massless string of length L = 1 m and is free to rotate as a pendulum. The speed of block A after the c

Amanda [17]

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The minimum velocity of A is $v_A= 4m/s$

Explanation:

From the question we are told that

The length of the string is $L = 1m$

The initial speed of block A is $u_A$

The final speed of block A is $v_A = \frac{1}{2}u_A$

The initial speed of block B is $u_B = 0$

The mass of block A is $m_A = 7kg$ gh

The mass of block B is $m_B = 2 kg$

According to the principle of conservation of momentum

$m_A u_A + m_B u_B = m_Bv_B + m_A \frac{u_A}{2}$

Since block B at initial is at rest

$m_A u_A = m_Bv_B + m_A \frac{u_A}{2}$

$m_A u_A - m_A \frac{u_A}{2} = m_Bv_B$

$m_A \frac{u_A}{2} = m_Bv_B$

making $v_B$ the subject of the formula

$v_B =m_A \frac{u_A}{2 m_B}$

Substituting values

$v_B =\frac{7 u_A}{4}$

This $v__B$ is the velocity at bottom of the vertical circle just at the collision with mass A

Assuming that block B is swing through the vertical circle(shown on the second uploaded image ) with an angular velocity of $v__B'$ at the top of the vertical circle

The angular centripetal acceleration would be mathematically represented

$a= \frac{v^2_{B}'}{L}$

Note that this acceleration would be toward the center of the circle

Now the forces acting at the top of the circle can be represented mathematically as

$T + mg = m \frac{v^2_{B}'}{L}$

Where T is the tension on the string

According to the law of energy conservation

The energy at bottom of the vertical circle = The energy at the top of

the vertical circle

This can be mathematically represented as

$\frac{1}{2} m(v_B)^2 = \frac{1}{2} mv^2_B' + mg 2L$

From above

$(T + mg) L = m v^2_{B}'$

Substitute this into above equation

$\frac{1}{2} m(\frac{7 v_A}{4} )^2 = \frac{1}{2} (T + mg) L + mg 2L$

$\frac{49 mv_A^2}{16} = \frac{1}{2} (T + mg) L + mg 2L$

$\frac{49 mv_A^2}{16} = T + 5mgL$

The value of velocity of block A needed to cause B be to swing through a complete vertical circle is would be minimum when tension on the string due to the weight of B is zero

This is mathematically represented as

$\frac{49 mv_A^2}{16} = 5mgL$

making $v_A$ the subject

$v_A = \sqrt{\frac{80mgL}{49m} }$

substituting values

$v_A = \sqrt{\frac{80* 9.8 *1}{49} }$

$v_A= 4m/s$

6 0

3 years ago