The box as shown above is moving at a constant velocity . True or false

A uniform pole AB of weight 5w and length 8a is suspended horizontally by tho strings attached to it at C and D where AC=DB=a. A

luda_lava [24]

Moment about D=0 or,T1*6a-5w*3a-9w*2a=0 T2+T1=5w+9w

4 0

4 years ago

What is the frequency of this wave? can someone pls explain and answer

Alenkasestr [34]

Answer:

f = 1 Hz

Explanation:

From the attached figure, we find that the time period of the wave is 1 second.

It is a longitudinal wave. It travel in the form of compression and rarefaction. The relation between time period T and frequency f is given by :

$f=\dfrac{1}{T}\\\\\text{As T = 1 s}\\\\f=1\ s^{-1}\\\\=1\ Hz$

Hence, the frequency of this wave is 1 Hz.

6 0

3 years ago

A turntable that spins at a constant 80.0 rpmrpm takes 3.50 ss to reach this angular speed after it is turned on. Find its angul

Veronika [31]

Answer:

The angular acceleration is 2.39 rad/s².

The number of degrees it rotates is 841.68 degrees.

Explanation:

Given:

Initial angular speed (ω₀) = 0 rad/s

Final angular speed in rpm (N) = 80.0 rpm

Time taken (t) = 3.50 s

First, let us determine the final angular speed in radians per second.

We know that,

$\omega=\frac{2\pi N}{60}\ rad/s$

Plug in the values and find the final angular speed, 'ω'. This gives,

$\omega=\frac{2\pi\times 80.0}{60}=8.38\ rad/s$

Now, using equation of motion for rotational motion, we have:

$\omega=\omega_0+\alpha t\\\\\alpha\to angular\ acceleration$

Plug in the given values and solve for α. This gives,

$8.38=0+\alpha \times 3.50\\\\\alpha=\frac{8.38}{3.50}=2.39\ rad/s^2$

Therefore, the angular acceleration is 2.39 rad/s².

Now, again using rotational equation of motion relating angular displacement, we have:

$\omega^2=\omega_0^2+2\alpha\theta\\\\\theta=\frac{\omega^2-\omega_0^2}{2\alpha }$

Plug in the given values and solve for 'θ'. This gives,

$\theta=\frac{(8.38)^2-0}{2\times 2.39}\\\\\theta=\frac{70.2244}{4.78}=14.69\ rad$

Convert radians to degrees using the conversion factor. This gives,

π radians = 180°

So, 1 radian =( 180 ÷ π ) degrees

Therefore, $14.69\ rad=14.69\times (\frac{180}{\pi})=841.68^\circ$

So, the number of degrees it rotates is 841.68 degrees.

3 0

3 years ago

The distance between the ruled lines on a diffraction grating is 1900 nm. The grating is illuminated at normal incidence with a

SashulF [63]

Answer:

3.28 degree

Explanation:

We are given that

Distance between the ruled lines on a diffraction grating, d=1900nm= $1900\times 10^{-9}m$

Where $1nm=10^{-9} m$

$\lambda_2=400nm=400\times10^{-9}m$

$\lambda_1=700nm=700\times 10^{-9}m$

We have to find the angular width of the gap between the first order spectrum and the second order spectrum.

We know that

$\theta=sin^{-1}(\frac{m\lambda}{d})$

Using the formula

m=1

$\theta_1=sin^{-1}(\frac{1\times700\times 10^{-9}}{1900\times 10^{-9}})$

$\theta=21.62^{\circ}$

Now, m=2

$\theta_2=sin^{-1}(\frac{2\times400\times 10^{-9}}{1900\times 10^{-9}})$

$\theta_2=24.90^{\circ}$

$\Delta \theta=\theta_2-\theta_1$

$\Delta \theta=24.90-21.62$

$\Delta \theta=3.28^{\circ}$

Hence, the angular width of the gap between the first order spectrum and the second order spectrum=3.28 degree

6 0

3 years ago

(06.01 LC)

Alexandra [31]

Answer:

It remains the same.

Explanation:

During physical change, the mass, number of atoms and number of molecules will remain the same.

5 0

3 years ago

Read 2 more answers