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2 years ago

How can i find Air velocity??????????????

Physics

1 answer:

melomori [17]2 years ago

3 0

Answer: By dividing airflow by duct cross section.

Explanation:

In short, air velocity in the ducts is calculated by dividing airflow by duct cross-section. Airflow is expressed as a simple number. Example: Air conditioner has a max. airflow of 600 CFM.

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What are the forces that act on the ball?

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3 years ago

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3 years ago

Read 2 more answers

In xray machines, electrons are subjected to electric fields as great as 6.0 x 10^5 N/C. Find

katrin2010 [14]

Answer:

a = 1.055 x 10¹⁷ m/s²

Explanation:

First, we will find the force on electron:

$E = \frac{F}{q}\\\\F = Eq\\$

where,

F = Force = ?

E = Electric Field = 6 x 10⁵ N/C

q = charge on electron = 1.6 x 10⁻¹⁹ C

Therefore,

$F = (6\ x\ 10^5\ N/C)(1.6\ x\ 10^{-19}\ C)\\$

F = 9.6 x 10⁻¹⁴ N

Now, we will calculate the acceleration using Newton's Second Law:

$F = ma\\a = \frac{F}{m}\\$

where,

a = acceleration = ?

m = mass of electron = 9.1 x 10⁻³¹ kg

therefore,

$a = \frac{9.6\ x\ 10^{-14}\ N}{9.1\ x\ 10^{-31}\ kg}\\\\$

5 0

3 years ago

A spring (k=15.19kN/m)is is compresses 25cm and held in place on a 36.87° incline. A block (M=10kg) is placed on the spring. Whe

Savatey [412]

Answer:

The maximum vertical displacement is 2.07 meters.

Explanation:

We can solve this problem using energy. Since there is a frictional force acting on the block, we need to consider the work done by this force. So, the initial potential energy stored in the spring is transferred to the block and it starts to move upwards. Let's name the point at which the block leaves the ramp "1" and the highest point of its trajectory in the air "2". Then, we can say that:

$E_0=E_1\\\\U_e_0=K_1+U_g_1+W_f_1$

Where $U_e_0$ is the elastic potential energy stored in the spring, $K_1$ is the kinetic energy of the block at point 1, $U_g_1$ is the gravitational potential energy of the block at point 1, and $W_f_1$ is the work done by friction at point 1.

Now, rearranging the equation we obtain:

$\frac{1}{2}kx^{2}=\frac{1}{2}mv_1^{2}+mgh_1+\mu Ns_1$

Where $k$ is the spring constant, $x$ is the compression of the spring, $m$ is the mass of the block, $v_1$ is the speed at point 1, $g$ is the acceleration due to gravity, $h_1$ is the vertical height of the block at point 1, $\mu$ is the coefficient of kinetic friction, $N$ is the magnitude of the normal force and $s_1$ is the displacement of the block along the ramp to point 1.

Since the force is in an inclined plane, the normal force is equal to:

$N=mg\cos\theta$

Where $\theta$ is the angle of the ramp.

We can find the height $h_1$ using trigonometry:

$h_1=s_1\sin\theta$

Then, our equation becomes:

$\frac{1}{2}kx^{2}=\frac{1}{2}mv_1^{2}+mgs_1\sin\theta+\mu mgs_1\cos\theta\\\\\implies v_1=\sqrt{\frac{2(\frac{1}{2}kx^{2}-mgs_1\sin\theta-\mu mgs_1\cos\theta)}{m}}=\sqrt{\frac{kx^{2}}{m}-2gs_1(\sin\theta+\mu \cos\theta)}$

Plugging in the known values, we get:

$v_1=\sqrt{\frac{(15190N/m)(0.25m)^{2}}{10kg}-2(9.8m/s^{2})(1.12m)(\sin36.87\°+(0.300) \cos36.87\°)}\\\\v_1=8.75m/s$

Now, we can obtain the height from point 1 to point 2 using the kinematics equations. We care about the vertical axis, so first we calculate the vertical component of the velocity at point 1:

$v_1_y=v_1\sin\theta=(8.75m/s)\sin36.87\°=5.25m/s$

Now, we have:

$y=\frac{v_1_y^{2}}{2g}\\\\y=\frac{(5.25m/s)^{2}}{2(9.8m/s^{2})}\\\\y=1.40m$

Finally, the maximum vertical displacement $h_2$ is equal to the height $h_1$ plus the vertical displacement $y$ :

$h_2=h_1+y=s_1\sin\theta +y\\\\h_2=(1.12m)\sin36.87\°+1.40m\\\\h_2=2.07m$

It means that the maximum vertical displacement of the block after it becomes airborne is 2.07 meters.

7 0

3 years ago