All categories

3 years ago

Please Help ASAP!!!

Physics

1 answer:

const2013 [10]3 years ago

5 0

Answer:

The Forces of Flight

At any given time, there are four forces acting upon an aircraft.

These forces are lift, weight (or gravity), drag and thrust. Lift is

the key aerodynamic force that keeps objects in the air. It is the

force that opposes weight; thus, lift helps to keep an aircraft in

the air. Weight is the force that works vertically by pulling all

objects, including aircraft, toward the center of the Earth. In order

to fly an aircraft, something (lift) needs to press it in the opposite

direction of gravity. The weight of an object controls how strong

the pressure (lift) will need to be. Lift is that pressure. Drag is a

mechanical force generated by the interaction and contract of a

solid body, such as an airplane, with a fluid (liquid or gas). Finally,

the thrust is the force that is generated by the engines of an

aircraft in order for the aircraft to move forward.

Explanation:

You might be interested in

(m = 4.0 kg) the cockroach rides on the rim (at radius R) of a disk (M = 6.0 kg) that turns about its center like a merry-go-rou

kvasek [131]

Answer:

ω' = 2.5 rad/s

Explanation:

mass of cockroach, m = 4 kg

mass of disk, M = 6 kg

Radius of disc= R

initial angular velocity, ω = 2 rad/s

Let the final angular velocity is ω'

As no external torque is applied, so the angular momentum is constant.

Angular momentum = Moment of inertia x angular velocity

I ω = I' ω'

$\frac{1}{2}\left ( M+m \right )R^{2}\times {2} = \left (\frac{1}{2}MR^{2}+m(0.5R)^{2}) \right )\omega '$

$10R^{2} = 4R^{2}\omega '$

ω' = 2.5 rad/s

5 0

4 years ago

In each of the four situations below an object is experiencing (nearly) uniform circular motion. State what force is providing t

Stells [14]

Answer:

Explanation:

Given that a centripetal force is a form of force that gives rise or causes a body to move in a curved path.

Hence;

1. When a car is being driven around a track, it is the FORCE OF FRICTION that is acting upon the turned wheels of the vehicle, which transforms into the centripetal force required for circular motion.

2. When a ball being is swung on the end of a string, TENSION FORCE acts upon the ball, which transforms the centripetal force required for circular motion.

3. When the moon is orbiting the earth, it is the FORCE OF GRAVITY acting upon the moon, which transforms the centripetal force required for circular motion.

4. A rotating wheel on the other hand has NO centripetal force because centripetal force is pull towards the center of a motion. However the speed of the object is tangent to the circle, while the direction of the force is also perpendicular to the direction of the rotating wheel.

4 0

3 years ago

A simple hydraulic lift is made by fitting a piston attached to a handle into a 3.0-cm diameter cylinder. The cylinder is connec

stiv31 [10]

Answer:

Approximately $3.1 \times 10^4 \; \rm N$ (assuming that the acceleration due to gravity is $g = 9.81\; \rm kg \cdot N^{-1}$ .)

Explanation:

Let $A_1$ denote the first piston's contact area with the fluid. Let $A_2$ denote the second piston's contact area with the fluid.

Similarly, let $F_1$ and $F_2$ denote the size of the force on the two pistons. Since the person is placing all her weight on the first piston:

$F_1 = W = m \cdot g = 50\; \rm kg \times 9.81 \; \rm kg \cdot N^{-1} =495\; \rm N$ .

Since both pistons fit into cylinders, the two contact surfaces must be circles. Keep in mind that the area of a square is equal to $\pi$ times its radius, squared:

$\displaystyle A_1 = \pi \times \left(\frac{1}{2} \times 3.0\right)^2 = 2.25\, \pi\;\rm cm^{2}$ .
$\displaystyle A_2 = \pi \times \left(\frac{1}{2} \times 24\right)^2 = 144\, \pi\;\rm cm^{2}$ .

By Pascal's Law, the pressure on the two pistons should be the same. Pressure is the size of normal force per unit area:

$\displaystyle P = \frac{F}{A}$ .

For the pressures on the two pistons to match:

$\displaystyle \frac{F_1}{A_1} = \frac{F_2}{A_2}$ .

$F_1$ , $A_1$ , and $A_2$ have all been found. The question is asking for $F_2$ . Rearrange this equation to obtain:

$\displaystyle F_2 = \frac{F_1}{A_1} \cdot A_2 = F_1 \cdot \frac{A_2}{A_1}$ .

Evaluate this expression to obtain the value of $F_2$ , which represents the force on the piston with the larger diameter:

$\begin{aligned}F_2 &= F_1 \cdot \frac{A_2}{A_1} \\ &= 495\; \rm N \times \frac{2.25\, \pi\; \rm cm^2}{144\, \pi \; \rm cm^2} \approx 3.1 \times 10^4\; \rm N\end{aligned}$ .