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A wheel is rotating at 30.0 rpm. The wheel then accelerates uniformly to 50.0 rpm in 10.0 seconds. Determine the – a) rate of an

Mademuasel [1]

Answer:

The angular acceleration is $0.209\ rad/s^2$

Explanation:

Given that,

Angular velocity, $\omega_{i} = 30.0\ rpm$

Angular velocity, $\omega_{f} = 50.0\ rpm$

Time t = 10.0 sec

We need to calculate the angular acceleration

Using formula of angular acceleration

$\alpha=\dfrac{\Delta \omega}{\Delta t}$

$\alpha=\dfrac{\omega_{f}-\omega_{i}}{\Delta t}$

$\alpha=\dfrac{50.0-30.0}{10.0}$

Now, we change the angular velocity in rad/s.

$\omega=20\times\dfrac{2\pi}{60}$

$\omega=2.09\ rad/s$

$\alpha=\dfrac{2.09}{10.0}$

$\alpha=0.209\ rad/s^2$

Hence, The angular acceleration is $0.209\ rad/s^2$

5 0

3 years ago

Read 2 more answers

Which of the following is a resistor? fan switch battery Thank you!

Rama09 [41]

The answer is battery:)

6 0

2 years ago

Read 2 more answers

Two long, straight parallel wires are placed 38 cm apart, one above the other. The top and bottom wires are carrying currents 4.

ElenaW [278]

Answer:

The force per unit length (N/m) on the top wire is 16.842 N/m

Explanation:

Given;

distance between the two parallel wire, d = 38 cm = 0.38 m

current in the first wire, I₁ = 4.0 kA

current in the second wire, I₂ = 8.0 kA

Force per unit length, between two parallel wires is given as;

$\frac{F}{L} = \frac{\mu_oI_1I_2 }{2\pi d }$

where;

μ₀ is constant = 4π x 10⁻⁷ T.m/A

Substitute the given values in the above equation and calculate the force per unit length

$\frac{F}{L} = \frac{\mu_oI_1I_2 }{2\pi d } = \frac{4\pi *10^{-7}*4000*8000 }{2\pi *0.38} = 16.842 \ N/m$

Therefore, the force per unit length (N/m) on the top wire is 16.842 N/m

4 0

3 years ago

Name the substance in which starch becomes dark blue in colour.

matrenka [14]

Any substance that contains starch turns blue-black in presence of iodine solution.

6 0

2 years ago

A sound is recorded at 19 decibels. What is the intensity of the sound?

sp2606 [1]

$1 \times 10^{-10.1} \mathrm{Wm}^{-2}$ is the intensity of the sound.

Answer: Option B

Explanation:

The range of sound intensity that people can recognize is so large (including 13 magnitude levels). The intensity of the weakest audible noise is called the hearing threshold. (intensity about $1 \times 10^{-12} \mathrm{Wm}^{-2}$ ). Because it is difficult to imagine numbers in such a large range, it is advisable to use a scale from 0 to 100.

This is the goal of the decibel scale (dB). Because logarithm has the property of recording a large number and returning a small number, the dB scale is based on a logarithmic scale. The scale is defined so that the hearing threshold has intensity level of sound as 0.

$\text { Intensity }(d B)=(10 d B) \times \log _{10}\left(\frac{I}{I_{0}}\right)$

Where,

I = Intensity of the sound produced

$I_{0}$ = Standard Intensity of sound of 60 decibels = $1 \times 10^{-12} \mathrm{Wm}^{-2}$

So for 19 decibels, determine I as follows,

$19 d B=(10 d B) \times \log _{10}\left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)$

$\log _{10}\left(\frac{1}{1 \times 10^{-12} \mathrm{Wm}^{-2}}\right)=\frac{19}{10}$

$\log _{10}\left(\frac{1}{1 \times 10^{-12} \mathrm{Wm}^{-2}}\right)=1.9$

When log goes to other side, express in 10 to the power of that side value,

$\left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)=10^{1.9}$

$I=1 \times 10^{-12} \mathrm{Wm}^{-2} \times 10^{1.9}=1 \times 10^{-12-1.9}=1 \times 10^{-10.1} \mathrm{Wm}^{-2}$

5 0

3 years ago