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QveST [7]
2 years ago
7

A string of length 3m and total mass of 12g is under a tension of 160N. A transverse harmonic wave with wavelength 25cm and ampl

itude 2cm travels to the right along the string. It is observed that the displacement at x=0 at t=0 is 0.87cm.
a) What is the wave?
b) Wrote the wave function, y(x,t)
c) Find the particle velocity at position x=0 at time t=10s. What is the maximum particle velocity?​
Physics
1 answer:
scoundrel [369]2 years ago
3 0

Answer:

Explanation:wave=wavelength×frequency,

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Definition of graph?
xeze [42]

<u>Answer:</u>

1. A graph is defined as <em>" A Diagram represents a system of connections or interrelations among two or more things by a number of different dots, lines etc".</em>

2. In simple words <em>"Graph is a representation of any object or a physical structure by dots, lines, etc.</em>

6 0
3 years ago
What is the wavelength of a wave if the wave speed is 24 m/s and the frequency is 48 Hz?
Mice21 [21]

Answer:

Wavelength is 0.5

Explanation:

To work it out, you divide Wave speed by the Frequency (24÷48=0.5)

4 0
3 years ago
Read 2 more answers
Independent practice
coldgirl [10]

Explanation:

formula: <u>Mass</u>

Density x volume

2a) m=10kg v=0.3m³

10÷0.3=33.3 kg/m

2b) m = 160 kg V=0.1m³

160÷0.1=1600 kg/m

2c) m = 220 kg V = 0.02m³

220÷0.02=11000 kg/m

A wooden post has a volume of 0.025m³ and a mass of 20kg. Calculate its density in kg/m.

density = volume ÷ mass

20÷ 0.025=800 kg/m

Challenge: A rectangular concrete slab is 0.80m long, 0.60 m wide and 0.04m thick. Calculate its volume in m³.

Formula : Length x width x height = Volume

0.80 x 0.60 x 0.04 = 0.0192m³

B) The mass of the concrete slab is 180 kg. Calculate its density in kg/m.

density = volume ÷ mass

180 ÷ 0.0192 = 9375 kg/m

4 0
2 years ago
Luis and Aisha conducted an experiment. They exerted different forces on four objects. Their results are shown in the table.
lesya692 [45]

Answer:

Object 3 has greatest acceleration.

Explanation:

Objects               Mass                                Force

1                            10 kg                               4 N              

2                           100 grams                       20 N

3                            10 grams                         4 N

4                             1 kg                                 20 N

Acceleration of object 1,

a_1=\dfrac{F_1}{m_1}\\\\a_1=\dfrac{4}{10}\\\\a_1=0.4\ m/s^2

Acceleration of object 2,

a_2=\dfrac{F_2}{m_2}\\\\a_2=\dfrac{20}{0.1}\\\\a_2=200\ m/s^2

Acceleration of object 3,

a_3=\dfrac{F_3}{m_3}\\\\a_3=\dfrac{4}{0.01}\\\\a_3=400\ m/s^2

Acceleration of object 4,

a_4=\dfrac{F_4}{m_4}\\\\a_4=\dfrac{20}{1}\\\\a_3=20\ m/s^2

It is clear that the acceleration of object 3 is 400\ m/s^2 and it is greatest of all. So, the correct option is (3).

4 0
3 years ago
Read 2 more answers
A 25,000 kg traveling east collides with a 2,000 kg truck standing still on the tracks. After the collision the train and truck
Elis [28]

Answer:

24.084 m/s

Explanation:

From the law of conservation of linear momentum

Total momentum before collision equals to the total momentum after collision

Since momentum=mv where m is mass and v is velocity

M_{truck}V_{truck}=V_{common}*(M_{truck} +M_{standing}) where M_{truck} is the mass of the truck, V_{truck} is velocity of the truck, V_{common} is the common velocity of moving and standing truck after collision and M_{standing} is the mass of the standing truck

Making V_{truck} the subject we obtain

V_{truck}=\frac { V_{common}*(M_{truck} +M_{standing})}{M_{truck}}

Substituting M_{truck} as 25000 Kg, V_{common} as 22.3 m/s, M_{standing} as 2000 Kg we obtain

V_{truck}=\frac { 22.3 m/s *(25000 Kg +2000 Kg)}{25000}= 24.084 m/s

Therefore, assuming no friction and considering that after collision they still move eastwards hence common velocity and initial truck velocities are positive

The truck was moving at 24.084 m/s

3 0
3 years ago
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