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sergeinik [125]
3 years ago
6

A string of length L= 1.2 m and mass m = 20g is under 400 N of tension, its two ends are fixed. a. How many nodes will you see i

n the 5th harmonic.
i. 4
ii. 5
iii. 6
iv. 7
v. None of the above.

Physics
1 answer:
ozzi3 years ago
8 0

Answer:

(iii) 6

Explanation:

Part a)

Since it is given that both ends are fixed and it is vibrating in 5th harmonic

So here it will have 5 number of loops in it

so we can draw it in following way

each loop will have 1 antinode and two nodes which means number of nodes is one more than number of anti-nodes

So there are 5 loops which means it will have 5 antinodes

and hence there will be 6 nodes in it

so correct answer will be

(iii) 6

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When considering that in the past human societies developed in greater isolation from one another than today, each of the follow
galina1969 [7]

Answer:One can easily assume a direct cause and effect relationship between a physical environment and an aspect of culture

Explanation: During the prehistoric and the earlier centuries human societies developed in isolation, there's no interconnection or much of communication between different groups and societies.

Knowledge sharing is not prominent, but in recent times people are more connected to each other,communities and countries interaction takes place through different forums, during the earlier centuries there are no mobile communication equipments,no Television set or the level of sophistication as it is today.

3 0
3 years ago
Help please!!! Physics circular motion
svp [43]

Answer:

2. 3.1415 m/s

3. 0.63m

4. 0.006 m/s^2

Explanation:

2. v=(2*pi*r))/T, put in values and solve.

3. Circumference=2*pi*radius, radius is 0.1m, plug in and solve.

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6 0
3 years ago
A hot iron ball of mass 200 g is cooled to a temperature of 22°C. 6.9 kJ of heat is lost to the surroundings during the process.
qaws [65]
Heat lost or gained, H = mc(θ₂ - θ₁) 
Where m = mass, c = Specific heat capacity, θ₂= final temperature, θ₁ = initial temperature

m = 200g, c = 0.444 J/g°C, θ₁ = 22 °C  (Since it was cooled).

H = 6.9 kj = 6.9 *1000J = 6900 J

6900 = 200*0.444* (θ₂ - 22)

6900/(200*0.444)  =  θ₂ - 22

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So initial temperature before cooling ≈ 100°C .  Option C.


5 0
3 years ago
Read 2 more answers
A factory worker pushes a 32.0 kg crate a distance of 7.0 m along a level floor at constant velocity by pushing horizontally on
g100num [7]

Answer:

(a) 81.54 N

(b) 570.75 J

(c) - 570.75 J

(d) 0 J, 0 J

(e) 0 J  

Explanation:

mass of crate, m = 32 kg

distance, s = 7 m

coefficient of friction = 0.26

(a) As it is moving with constant velocity so the force applied is equal to the friction force.

F = 0.26 x m x g = 0.26 x 32 x 9.8 = 81.54 N

(b) The work done on the crate

W = F x s = 81.54 x 7 = 570.75 J

(c) Work done by the friction

W' = - W = - 570.75 J

(d) Work done by the normal force

W'' = m g cos 90 = 0 J

Work done by the gravity

Wg = m g cos 90 = 0 J

(e) The total work done is

Wnet = W + W' + W'' + Wg = 570.75 - 570.75 + 0 = 0 J  

6 0
3 years ago
What is the charge of an atom with 21 protons and 6 electrons
Svetach [21]

Answer:

Scandium with an ion charge of +3

Explanation:

4 0
3 years ago
Read 2 more answers
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