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1 year ago

Q|C A sinusoidal wave in a string is described by the wave function

Physics

1 answer:

creativ13 [48]1 year ago

4 0

The maximum transverse force on a 1.00cm segment of the string is 0.045N.

The maximum transverse force on the segment,

l=1.00cm=1.00cm×1m/100cm=0.01m is,

F=μla

=(12.0×10^−3kg/m)×(0.01m)×(375m/s^2)

=0.045N

Hence the maximum force is 0.045N

Learn more about transverse force here:

brainly.com/question/13618693

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With what maximum precision can its position be ascertained? [hint: ?p=m?v.]

Nana76 [90]

The formula for the momentum is
p = mv
As a consequence of the conservation of energy, there is also the law of conservation of momentum.
So,
Δp = Δmv = mΔv
The maximum precision that the position can be ascertained is less than 100% because of dissipation.

5 0

3 years ago

Which describes one event that causes an eclipse?

seraphim [82]

Answer:

earth's shadow covering the moon,thats lunar eclipse

5 0

3 years ago

7. You probably had difficulty in measuring the period, yet its calculation should prove effortless at this point. Why do you su

Art [367]

Answer:

* First fight the reaction time of the person, which is approximately 1/10 s, remember that the stopwatch does not stop alone

* Problems to synchronize when starting to measure the oscillation and end point of the oscilloscope, this error needs a good control system to be decreased.

*Effect of friction with air

Explanation:

In the measurements of the oscillatory movement in general, the most difficult magnitude of the measurement is the period, by reasoning

* First fight the reaction time of the person, which is approximately 1/10 s, remember that the stopwatch does not stop alone

* Problems to synchronize when starting to measure the oscillation and end point of the oscilloscope, this error needs a good control system to be decreased.

* Effect of friction with air

5 0

3 years ago

Two carts, one twice as heavy as the other, are at rest on a horizontal track. A person pushes each cart for 8 s. Ignoring frict

GalinKa [24]

Answer:

The correct option is: B that is 1/2 K

Explanation:

Given:

Two carts of different masses, same force were applied for same duration of time.

Mass of the lighter cart = $m$

Mass of the heavier cart = $2m$

We have to find the relationship between their kinetic energy:

Let the KE of cart having mass m be "K".

and KE of cart having mass m be "K1".

As it is given regarding Force and time so we have to bring in picture the concept of momentum Δp and find a relation with KE.

Numerical analysis.

⇒ $KE = \frac{mv^2}{2}$

⇒ $KE = \frac{mv^2}{2}\times \frac{m}{m}$

⇒ $KE = \frac{m^2v^2}{2}\times \frac{1}{m}$

⇒ $KE = \frac{(mv)^2}{2}\times \frac{1}{m}$

⇒ $KE = \frac{(\triangle p)^2}{2}\times \frac{1}{m}$

⇒ $KE = \frac{(\triangle p)^2}{2m}=\frac{(F\times t)^2}{2m}$

Now,

Kinetic energies and their ratios in terms of momentum or impulse.

KE (K) of mass m.

⇒ $K=\frac{(F\times t)^2}{2m}$ ...equation (i)

KE (K1) of mass 2m.

⇒ $K_1=\frac{(F\times t)^2}{2\times 2m}$

⇒ $K_1=\frac{(F\times t)^2}{4m}$ ...equation (ii)

Lets divide K1 and K to find the relationship between the two carts's KE.

⇒ $\frac{K_1}{K} =\frac{(F\times t)^2}{4m} \times \frac{2m}{(F\times t)^2}$

⇒ $\frac{K_1}{K} =\frac{2m}{4m}$

⇒ $\frac{K_1}{K} =\frac{2}{4}$

⇒ $\frac{K_1}{K} =\frac{1}{2}$

⇒ $K_1=\frac{K}{2}$

⇒ $K_1=\frac{1}{2}K$

The kinetic energy of the heavy cart after the push compared to the kinetic energy of the light cart is 1/2 K.

7 0

2 years ago

A pressure cooker is a pot whose lid can be tightly sealed to prevent gas from entering or escaping. Even without knowing how bi

k0ka [10]

Answer:

a) F₁₂₀ = 1.34 pa A , b) F₂₀ = 0.746 pa A

Explanation:

Part. A . The definition of pressure is

P = F / A

As the air can approach an ideal gas we can use the ideal gas equation

P V = n R T

Let's write this equation for two temperatures

P₁ V = n R T₁

P₂2 V = n R T₂

P₁ / P₂ = T₁ / T₂

point 1 has a pressure of P₁ = pa and a temperature of (20 + 273) K, point 2 is at (120 + 273) K, we calculate the pressure P₂

P₂ = P₁ T₂ / T₁

P₂ = pa 393/293

P₂ = 1.34 pa

We calculate the strength

P₂ = F₁₂₀ / A

F₁₂₀ = 1.34 pa A

Part B

In this case the data is

Point 1 has a temperature of 393K and an atmospheric pressure (P₁ = pa), point 2 has a temperature of 293K, let's calculate its pressure

P₁ / P₂ = T₁ / T₂

P₂ = P₁ T₂ / T₁

P₂ = pa 293/393

P₂ = 0.746 pa

Let's calculate the force (F20), from this point

F₂₀ / A = 0.746 pa

F₂₀ = 0.746 pa A

6 0

2 years ago