When the object is at half its amplitude from equilibrium, is the magnitude of its acceleration at half its maximum value?

3. A custom car built for pretentious people has two brass horns that are supposed to produce the same frequency but actually em

Ugo [173]

When two waves with different frequencies combine, two new waves are created, with frequencies at the sum and difference of the two original ones. The new ones are called the 'beats'.

Original two frequencies: 267.8 Hz, 266.4 Hz .

Beat at the sum = (267.8 + 266.4) = 534.2 Hz .

Beat at the difference = (267.8 - 266.4) = 1.4 Hz .

In this situation, we're typically not aware of the sum-beat at 534 Hz.
Our consciousness is flooded with the two original frequencies, and
with hearing the whole combined sound going "wah wah wah" at
1.4 times per second. But the 534-Hz note is there too ... part of
what makes the whole thing sound so ragged and discordant.

7 0

4 years ago

An electron moves in a circular path perpendicular to a magnetic field of magnitude 0.220 T. If the kinetic energy of the electr

Gwar [14]

Answer:

$971605.66\ \text{m/s}$

$25.1\ \mu\text{m}$

Explanation:

m = Mass of electron = $9.11\times 10^{-31}\ \text{kg}$

B = Magnetic field = 0.22 T

K = Kinetic energy of electron = $4.3\times 10^{-19}\ \text{J}$

q = Charge = $1.6\times 10^{-19}\ \text{C}$

v = Velocity of electron

r = Radius of curved path

Kinetic energy is given by

$K=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{\dfrac{2K}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 4.3\times 10^{-19}}{9.11\times 10^{-31}}}\\\Rightarrow v=971605.66\ \text{m/s}$

The speed of the electron is $971605.66\ \text{m/s}$

The force balance of the system is given by

$qvB=\dfrac{mv^2}{r}\\\Rightarrow r=\dfrac{mv}{qB}\\\Rightarrow r=\dfrac{9.11\times 10^{-31}\times 971605.66}{1.6\times 10^{-19}\times 0.22}\\\Rightarrow r=0.0000251=25.1\ \mu\text{m}$

The radius of the curved path is $25.1\ \mu\text{m}$

6 0

3 years ago

An object 2cm high is placed 3cm in front of a concave lens of focal length 2cm, find the magnification?

Simora [160]

Answer:

0.4

Explanation:

A concave lens is a diverging lens, so it will always have a negative focal length. Image distance is always negative for a concave lens because it forms virtual images.

From the lens formula;

1/f = 1/u+ 1/v

- 1/2 = 1/3 - 1/v

1/v = 1/3 + 1/2

v= 6/5

v= 1.2 cm

Magnification = image distance/object distance

Magnification = 1.2cm/3cm

Magnification = 0.4

8 0

3 years ago

If a seagull drops a shell from rest at a height of 12 m how fast is the shell moving when it hits the rocks

Dvinal [7]

As seagull drops a shell from rest at a height of 12 m, so we use kinematic equation of motion,

$v^{2} = u^{2} +2g h$

Here, h is the height, u is initial velocity , v is final velocity and g is acceleration due to gravity.

Given, h = 12 m.

We take, $g = 9.8 \ m/s^2$ and $u = 0$ because seagull drops a shell from rest.

Therefore, the speed of shell when it hits the rocks,

$v^{2} = 0 + 2 \times 9.8 m/s^2 \times 12 \ m = 235.2 (m/s)^2 \\\\ v = \sqrt{235.2 (m/s)^2} = 15.33 \ m/s^2$

5 0

3 years ago

8) T F A car is being towed at constant velocity on a horizontal road using a horizontal chain. The tension in the chain must be

Stels [109]

Answer:

False

Explanation:

The tension in the chain must be equal to the frictional force acting on the car, not to its weight.

In fact, we have 4 forces acting on the car

- Its weight: downward

- The normal reaction of the road on the car: upward --> this force balances the weight, so the net force along the vertical direction is zero

- The tension in the chain: forward

- The frictional force between the road's surface and the tires of the car: backward

We can consider the horizontal motion only: we are said that the car is moving at constant velocity, so the horizontal acceleration is zero. According to Newton's second law:

$\sum F = ma$

zero acceleration means that the resultant of the forces on the car is zero. But there are only 2 forces acting on the car in the horizontal direction: the tension in the chain (forward) and the frictional force (backward). Since their resultant must be zero, it means that the two forces must be equal and opposite: therefore, the tension in the chain must be equal to the frictional force.

8 0

3 years ago