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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:

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Why is the answer B?

djyliett [7]

Answer:

Explanation:

The center of mass lies on a line that joins position 4 of one start with position 4 of the other star. The shortest distance between these two points will produce the largest velocity. You are using F = m v^2/R

Small R = large force.

Large Force = increased speed.

The masses don't have any effect on the outcome: they remain constant.

7 0

3 years ago

Using energy considerations, calculate the average force (in N) a 67.0 kg sprinter exerts backward on the track to accelerate fr

Illusion [34]

Answer:

$F_{sprinter}=110.4N$

Explanation:

Given data

Mass m=67.0 kg

Final Speed vf=8.00 m/s

Initial Speed vi=2.00 m/s

Distance d=25.0 m

Force F=30.0 N

From work-energy theorem we know that the work done equals the change in kinetic energy

W=ΔK=Kf-Ki=1/2mvf²-1/2mvi²

And

$W=F_{total}.d$

$W=1/2mv_{f}^2-1/2mv_{i}^2\\F_{total}=\frac{1/2mv_{f}^2-1/2mv_{i}^2}{d} \\F_{total}=\frac{1/2(67.0kg)(8.00m/s)^2-1/2(67.0kg)(2.00m/s)^2}{25.0m} \\F_{total}=80.4N$

and we know that the force the sprinter exerted Fsprinter the force of the headwind Fwind=30.0N

$F_{sprinter}=F_{total}+F_{wind}\\F_{sprinter}=80.4N+30N\\F_{sprinter}=110.4N$

7 0

3 years ago

Read 2 more answers

Despite a very strong wind, a tennis player

Gnoma [55]

Answer:

Option 5. 1 and 3

Solution:

The only forces acting on the tennis ball after it has left contact with the racquet and the instant before it touches the ground are the force of gravity in the downward direction and the force by the air exerted on the ball.

The ball after it left follows the path of trajectory and as it moves forward in the horizontal direction the force of the air acts on it.

In the whole projectile motion of the ball, the acceleration due to gravity acts on the ball thus the force of gravity acts on the ball in the downward direction before it hits the ground.

6 0

3 years ago

Which of the following statements must be true?

anyanavicka [17]

Answer:

your answer is B. The velocity could be in any direction, but the acceleration is in the direction of the resultant force

6 0

3 years ago

Please help i'm going to throw up from stress

Eddi Din [679]

Answer:

Explanation:

First of all, I used the specific heat of water as 4182 J/(kgC) and the specific heat of ethyl alcohol (EtOH) as 2440 J/(kgC); that means that we need the masses in kg, not g.

120.g = .1200 kg of ethyl alcohol. Now for the formula:

$t_f=\frac{(m_{H2O}*spheat_{H2O}*temp_{H2O})+(m_{EtOH}*spheat_{EtOH}*temp_{EtOH})}{(m_{H2O}*spheat_{H2O})+(m_{EtOH}*spheat_{EtOH})}$ where spheat is specific heat.

Filling that horrifying-looking formula in with some values:

$16.0=\frac{(x*4182*20.0)+(.1200*2440*10.0)}{(x*4182)+(.1200*2440)}$ and

$16.0=\frac{83640x+2928}{4182x+292.8}$ and

16(4182x + 292.8) = 83640x + 2928 and

66912x + 4684.8 = 83640x + 2928 and

1756.8 = 16728x so

x = .105 kg and the amount of water added is 105 g

4 0

3 years ago