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3 years ago

The note "Middle C" is known to have a frequency of 261.6 Hz. What would

Physics

1 answer:

Gwar [14]3 years ago

8 0

Answer: frequency = 523.2 Hz

Explanation:

One octave higher = 2f of the last octave.

For instance, if C2 = 4 Hz

A octave higher will be C3 in which its frequency will be 2 × 4Hz = 8Hz

Back to the question.

One octave higher frequency will therefore equal to:

2f = 2 × 261.6 = 523.2 Hz

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A carbon rod with a radius of 1.9 mm is used to make a resistor. What length of the carbon rod should be used to make a 3.7 Ω re

vladimir2022 [97]

Answer:

Length = 2.32 m

Explanation:

Let the length required be 'L'.

Given:

Resistance of the resistor (R) = 3.7 Ω

Radius of the rod (r) = 1.9 mm = 0.0019 m [1 mm = 0.001 m]

Resistivity of the material of rod (ρ) = $1.8\times 10^{-5}\ \Omega\cdot m$

First, let us find the area of the circular rod.

Area is given as:

$A=\pi r^2=3.14\times (0.0019)^2=1.13\times 10^{-5}\ m^2$

Now, the resistance of the material is given by the formula:

$R=\rho( \frac{L}{A})$

Express this in terms of 'L'. This gives,

$\rho\times L=R\times A\\\\L=\frac{R\times A}{\rho}$

Now, plug in the given values and solve for length 'L'. This gives,

$L=\frac{3.7\ \Omega\times 1.13\times 10^{-5}\ m^2}{1.8\times 10^{-5}\ \Omega\cdot m}\\\\L=\frac{4.181}{1.8}=2.32\ m$

Therefore, the length of the material required to make a resistor of 3.7 Ω is 2.32 m.

5 0

3 years ago

An object is hung from a spring balance attached to the ceiling of an elevator cab. The balance reads 65 N when the cab is stand

lana66690 [7]

Answer:

(A) Reading will be 65 N

(B) Net force on the elevator will be 49.076 N

Explanation:

We have given the balance force = 65 N

Acceleration due to gravity $g=9.8m/sec^2$

We know that W=mg

So $65=m\times 9.8$

m = 6.632 kg

(a) In first case as the as the speed is constant so the force on the elevator will be 65 N

(B) In second case as the elevator is decelerating at a rate of $2.4m/sec^2$

So net acceleration = 9.8-2.4= $7.4m/sec^2$

So net force on elevator will be = m× net acceleration = 6.632×7.4 = 49.076 N

7 0

3 years ago

A circular curve of radius 150 m is banked at an angle of 15 degrees. A 750-kg car negotiates the curve at 85.0 km/h without ski

Crazy boy [7]

Answer: a) 7.1 * 10^3 N; b) -880 N directed out of the curve.

Explanation: In order to solve this problem we have to use the Newton laws, then we have the following:

Pcos 15°-N=0

Psin15°-f= m*ac

from the first we obtain N, the normal force

N=750Kg*9.8* cos (15°)= 7.1 *10^3 N

Then to calculate the frictional force (f) we can use the second equation

f=P sin (15°) -m*ac where ac is the centripetal acceletarion which is equal to v^2/r

f= 750 *9.8 sin(15°)-750*(85*1000/3600)^2/150= -880 N

6 0

3 years ago

A bar of length L = 8 ft and midpoint D is falling so that, when θ = 27°, ∣∣v→D∣∣=18.5 ft/s , and the vertical acceleration of p

777dan777 [17]

Answer:

alpha=53.56rad/s

a=5784rad/s^2

Explanation:

First of all, we have to compute the time in which point D has a velocity of v=23ft/s (v0=0ft/s)

$v=v_0+at\\\\t=\frac{v}{a}=\frac{(23\frac{ft}{s})}{32.17\frac{ft}{s^2}}=0.71s$

Now, we can calculate the angular acceleration (w0=0rad/s)

$\theta=\omega_0t +\frac{1}{2}\alpha t^2\\\alpha=\frac{2\theta}{t^2}$

$\alpha=\frac{27}{(0.71s)^2}=53.56\frac{rad}{s^2}$

with this value we can compute the angular velocity

$\omega=\omega_0+\alpha t\\\omega = (53.56\frac{rad}{s^2})(0.71s)=38.02\frac{rad}{s}$

and the tangential velocity of point B, and then the acceleration of point B:

$v_t=\omega r=(38.02\frac{rad}{s})(4)=152.11\frac{ft}{s}\\a_t=\frac{v_t^2}{r}=\frac{(152.11\frac{ft}{s})^2}{4ft}=5784\frac{rad}{s^2}$

hope this helps!!

6 0

3 years ago

Read 2 more answers

A constant force of 5.00 N acts on a 2.50 kg object for 10.0 s. What are the changes in the object’s momentum and velocity?

dimulka [17.4K]

Hope this answer helps, cause Idk, I might be wrong, but I still, I used the correct formulas, so I might be correct

7 0

3 years ago