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grin007 [14]
3 years ago
12

You send a traveling wave along a particular string by oscillating one end. If you increase the frequency of oscillations, does

the speed of the wave increase, decrease, or remain the same?
Physics
1 answer:
eimsori [14]3 years ago
4 0

Answer:

The speed of the wave remains the same

Explanation:

Since the speed of the wave v = √(T/μ) where T is the tension in the string and μ is the linear density of the string.

We observed that the speed, v is independent of the frequency of the  wave in the string. So, increasing the frequency of the wave has no effect on the speed of the wave in the string, since the speed of the wave in the string is only dependent on the properties of the string.

<u>So, If you increase the frequency of oscillations, the speed of the wave remains the same.</u>

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What is the relationship between mass and weight in physics?
Alex_Xolod [135]

We are aware that weight is the product of applied gravitational force and mass. W = MG thus, where W represents the weight, M the mass, and G the gravitational force. As a result, it might also mean that "an object's weight is directly proportionate to its mass."

<h3>What is mass?</h3>
  • Mass is a physical body's total amount of matter.
  • It also serves as a gauge for the body's inertia, or resistance to acceleration (change in velocity) in the presence of a net force.
  • The strength of an object's gravitational pull to other bodies is also influenced by its mass.
  • The kilogram is the primary mass unit in the SI (kg).
  • Even though weight is frequently measured using a spring scale rather than a balancing scale and directly compared with known masses, mass is not the same as weight in physics.
<h3>What is weight?</h3>
  • The force exerted on an object by gravity is known as the weight of the object in science and engineering.
  • Weight is sometimes described as a vector quantity, or the gravitational force exerted on the object, in some common textbooks.
  • Others define weight as a scalar quantity, the gravitational force's strength.
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Learn more about mass here:

brainly.com/question/19694949

#SPJ4

7 0
2 years ago
32. (FR) A mass moving at 25 m/s slides along a rough horizontal surface. The coefficient of friction is 0.30. (A) Use force and
scoray [572]

Answer:

A)s = 104.16 m

b)s= 104.16 m

Explanation:

Given that

u = 25 m/s

μ = 0.3

The friction force will act opposite to the direction of motion.

Fr= μ m g

Fr= -  m a

a=acceleration

μ m g = - m a

a= - μ g

a= - 0.3 x 10 m/s²          ( take g= 10 m/s²)

a= - 3  m/s²

The final speed of the mass is zero ,v= 0

We know that

v² = u² +2 a s

s=distance

0² = 25² - 2 x 3 x s

625 = 6 s

s = 104.16 m

By using energy conservation

Work done by all the forces =Change in the kinetic energy

- Fr.s=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2

Negative sign because force act opposite to the displacement.

- \mu\ m\ g \ s=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2

-\mu\ g \ s=\dfrac{1}{2}v^2-\dfrac{1}{2}u^2

-0.3\times 10\times \ s=\dfrac{1}{2}\times 0^2-\dfrac{1}{2}\times 25^2

- 3 x 2 x s = - 625

s= 104.16 m

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3 years ago
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dusya [7]
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Suppose a 4.0-kg projectile is launched vertically with a speed of 8.0 m/s. What is the maximum height the projectile reaches?
eduard

Answer:

h = 3.3 m (Look at the explanation below, please)

Explanation:

This question has to do with kinetic and potential energy. At the beginning (time of launch), there is no potential energy- we assume it starts from the ground. There, is, however, kinetic energy

Kinetic energy = \frac{1}{2}mv^{2}

Plug in the numbers = \frac{1}{2}(4.0)(8^{2})

Solve = 2(64) = 128 J

Now, since we know that the mechanical energy of a system always remains constant in the absence of outside forces (there is no outside force here), we can deduce that the kinetic energy at the bottom is equal to the potential energy at the top. Look at the diagram I have attached.

Potential energy = mgh = (4.0)(9.8)(h) = 39.2(h)

Kinetic energy = Potential Energy

128 J = 39.2h

h = 3.26 m

h= 3.3 m (because of significant figures)

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Organ, integument, dermis
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