What is the answer???

Specify whether the statements are true or false.

NISA [10]

Answer:

1. False

2. True

3. True

4. True

5. False

Explanation:

Moment of a force is not a free vector. There are certain quantities along the line with which force is applied.

Force can be moved in any direction along the line of the action without changing the external reaction.

The magnitude of equivalent resultant force is distributed along the centroid point.

The resultant force of a couple force system is zero as it form opposite forces which balances off each other.

6 0

3 years ago

A vacuum pump is used to drain a basement of 20 °C water (with a density of 998 kg/m3 ). The vapor pressure of water at this tem

lord [1]

Answer:

The maximum theoretical height that the pump can be placed above liquid level is $\Delta h=9.975\,m$

Explanation:

To pump the water, we need to avoid cavitation. Cavitation is a phenomenon in which liquid experiences a phase transition into the vapour phase because pressure drops below the liquid's vapour pressure at that temperature. As a liquid is pumped upwards, it's pressure drops. to see why, let's look at Bernoulli's equation:

$\frac{\Delta P}{\rho}+g\, \Delta h +\frac{1}{2} \Delta v^2 =0$

( $\rho$ stands here for density, $h$ for height)

Now, we are assuming that there aren't friction losses here. If we assume further that the fluid is pumped out at a very small rate, the velocity term would be negligible, and we get:

$\frac{\Delta P}{\rho}+g\, \Delta h =0$

$\Delta P= -g\, \rho\, \Delta h$

This means that pressure drop is proportional to the suction lift's height.

We want the pressure drop to be small enough for the fluid's pressure to be always above vapour pressure, in the extreme the fluid's pressure will be almost equal to vapour pressure.

That means:

$\Delta P = 2.34\,kPa- 100 \,kPa = -97.66 \, kPa\\$

We insert that into our last equation and get:

$\frac{ \Delta P}{ -g\, \rho\,}= \Delta h\\\Delta h=\frac{97.66 \, kPa}{998 kg/m^3 \, \, 9.81 m/s^2} \\\Delta h=9.975\,m$

And that is the absolute highest height that the pump could bear. This, assuming that there isn't friction on the suction pipe's walls, in reality the height might be much less, depending on the system's pipes and pump.

8 0

4 years ago

A light aircraft with a wing area of 200 ft^2 and a weight of 2000 lb has a lift coefficient of 0.39 and a drag coefficient of 0

Gnoma [55]

Answer: power required to maintain level flight=82.20hp

Explanation:

Given

Area = 200 ft^2

Weight = 2000 lb

Cl( Lift coefficient)= 0.39

Cd( Drag coefficient) = 0.06

The density ρ of air at standard atmospheric pressure = 2.38 X 10^-3 slugs/ft^3

For Equilibrium to be maintained during flight conditions, the lift force must be balanced by the weight of the aircraft such that

Lift force = Weight of aircraft

(1/2)ρAU²Cl= W

1/2X 2.38 X 10^-3 X 200 X U² X 0.39 = 2000

U²= 2000 X 2 / 2.38 X 10^-3 X 200 X 0.39

U= $\sqrt{21,547.08}$

Velocity, U= 146.7892ft/s

Drag force of the velocity can be deduced from the formulae

Cd= Drag force(D) /1/2 ρU²A

Drag force=1/2 ρU²ACd

D=1/2 x (2.38 X 10^-3 slugs/ft^3) x (146.7892ft/s)² x 200 ft^2 x 0.06

D=307.69

Drag force= 308lb

power required to maintain level flight is given as

P = Drag force x Velocity = D x U

=308lb X 146.7892ft/s

=45,211.0736lb.ft/s

Changing to hp we have that

1 Horsepower, hp = 550 ft lbf/s

??=45,211.0736lb.ft/s

45,211.0736lb.ft/s/ 550 ft lbf/s= 82.20hp

6 0

3 years ago

The 30-kg gear is subjected to a force of P=(20t)N where t is in seconds. Determine the angular velocity of the gear at t=4s sta

tatyana61 [14]

Answer:

$\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}$

Explanation:

Previous concepts

Angular momentum. If we consider a particle of mass m, with velocity v, moving under the influence of a force F. The angular momentum about point O is defined as the “moment” of the particle’s linear momentum, L, about O. And the correct formula is:

$H_o =r x mv=rxL$

Applying Newton’s second law to the right hand side of the above equation, we have that r ×ma = r ×F =

MO, where MO is the moment of the force F about point O. The equation expressing the rate of change of angular momentum is this one:

MO = H˙ O

Principle of Angular Impulse and Momentum

The equation MO = H˙ O gives us the instantaneous relation between the moment and the time rate of change of angular momentum. Imagine now that the force considered acts on a particle between time t1 and time t2. The equation MO = H˙ O can then be integrated in time to obtain this:

$\int_{t_1}^{t_2}M_O dt = \int_{t_1}^{t_2}H_O dt=H_0t2 -H_0t1$

Solution to the problem

For this case we can use the principle of angular impulse and momentum that states "The mass moment of inertia of a gear about its mass center is $I_o =mK^2_o =30kg(0.125m)^2 =0.46875 kgm^2$ ".