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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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How many electrons are in the outer energy level of an atom of carbon
xz_007 [3.2K]

Answer:

4 electrons

Explanation:

Carbon, is the group 14 element, with four electrons in its outer shell. Carbon always shares electrons to reach a complete valence shell, making bonds with other atoms.

8 0
3 years ago
A distance of 0.002 m separates two objects of equal mass. If the gravitational force between them is
Alika [10]

Answer: 24.97 kg

Explanation:

The gravitational force between two objects of masses M1, and M2 respectively, and separated by a distance R, is:

F = G*(M1*M2)/R^2

Where G is the gravitational constant:

G = 6.67*10^-11 m^3/(kg*s^2)

In this case, we know that

R = 0.002m

F = 0.0104 N

and that M1 = M2 = M

And we want to find the value of M, then we can replace those values in the equation to get

0.0104 N = (6.67*10^-11 m^3/(kg*s^2))*(M*M)/(0.002m)^2

(0.0104 N)*(0.002m)^2/(6.67*10^-11 m^3/(kg*s^2)) = M^2

623.69 kg^2 = M^2

√(623.69 kg^2) = M = 24.97 kg

This means that the mass of each object is 24.97 kg

6 0
3 years ago
Which of the following are Electromagnetic Waves?
earnstyle [38]

Answer:

b

Explanation:

8 0
3 years ago
A cyclist rides 6.2 km east, then 9.28 km in a direction 27.27 degrees west of north, then 7.99 km west. A. What is the magnitud
Pani-rosa [81]

Answer:

Explanation:

given,

cyclist ride  6.2 km east and then 9.28 km in the direction of 27.27° west of north and then 7.99 km west.

vertical component = 9.28 cos∅

                                = 9.28 cos 27.27°

                                = 8.24 km

horizontal axis component = 9.28 sin ∅

                                             = 9.28 sin 27.27°

                                             = 4.5 km

distance of the final point from the origin

                            = 7.99 -(6.2-4.5)

                            = 6.29 km

displacement

d = \sqrt{6.29^2+8.24^2}

d = 10.37 km

b) tan \theta = \dfrac{6.29}{8.24}

θ = 37.36°

4 0
3 years ago
A screen is placed 1.20m behind a single slit. The central maximum in the resulting diffraction pattern on the screen is 1.40cm
andrew11 [14]

Answer:

2.8 cm

Explanation:

y_1 = Separation between two first order diffraction minima = 1.4 cm

D = Distance of screen = 1.2 m

m = Order

Fringe width is given by

\beta_1=\dfrac{y_1}{2}\\\Rightarrow \beta_1=\dfrac{1.4}{2}\\\Rightarrow \beta_1=0.7\ cm

Fringe width is also given by

\beta_1=\dfrac{m_1\lambda D}{d}\\\Rightarrow d=\dfrac{m_1\lambda D}{\beta_1}

For second order

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Distance between two second order minima is given by

y_2=2\beta_2

\\\Rightarrow y_2=2\dfrac{m_2}{m_1}\beta_1\\\Rightarrow y_2=2\dfrac{2}{1}\times 0.7\\\Rightarrow y_2=2.8\ cm

The distance between the two second order minima is 2.8 cm

8 0
3 years ago
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