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____ [38]
4 years ago
5

Four basic forces act on an airplane while in flight: lift, weight, drag, and thrust. These four forces and their directions are

indicated below.
If the thrust acting on an airplane is greater than the drag, then
A.
the airplane will accelerate backward.
B.
the lift will be greater then the weight.
C.
the airplane will accelerate forward.
D.
the weight will be greater than the lift.
Physics
2 answers:
Arlecino [84]4 years ago
6 0

c- the airplane will accelerate forward

sergey [27]4 years ago
5 0
C. thrust is the force pushing something forward, while drag is the wind resistance keeping something from moving.
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saad has mass 80kg when resting on the ground at the equator what will be the centripetal acceleration on saad if the radius of
oksano4ka [1.4K]
I did try to solve. I hope it is correct, below is the solution:

<span>put everything in s.i units
then the answer what u wrote is acceleration to get is divide by mass(80) G=6.011*10^-11
M=6*10^24
R=6.4*10^6
m=80
</span>
Hope it helps. 
7 0
3 years ago
A 2. 5 kg block is launched along the ground by a spring with a spring constant of 56 N/m. The spring is initially compressed 0.
OLga [1]

The speed of the spring when it is released is 3.5 m/s.

The given parameters:

  • <em>Mass of the block, m = 2.5 kg</em>
  • <em>Spring constant, k = 56 N/m</em>
  • <em>Extension of the spring, x = 0.75 m</em>

The speed of the spring when it is released is calculated by applying the principle of conservation of energy as follows;

K.E = U_x\\\\\frac{1}{2} mv^2 = \frac{1}{2} kx^2\\\\mv^2 = kx^2\\\\v^2 = \frac{kx^2}{m} \\\\v = \sqrt{\frac{kx^2}{m} } \\\\v = \sqrt{\frac{56 \times 0.75^2}{2.5} } \\\\v = 3.5  \ m/s

Thus, the speed of the spring when it is released is 3.5 m/s.

Learn more about conservation of energy here:  brainly.com/question/166559

3 0
2 years ago
Read 2 more answers
You are investigating the report of a UFO landing in an isolated portion of New Mexico, and encounter a strange object that is r
SIZIF [17.4K]

Answer: 5 m

Explanation:

We have the following data:

I_{1}=0.11 W/m^{2} is the intensity of the sound at 7.5 m from the source

r_{1}=7.5 m is the distance at which the intensity I_{1} was measured

I_{2}=1 W/m^{2} is the intensity of the sound at r_{2} from the source

We have to find r_{2}

Since the object is radiating the signal uniformly in all directions, we can use the <u>Inverse Square Law for Intensity:</u>

\frac{I_{1}}{I_{2}}=\frac{r_{2}^{2}}{r_{1}^{2}}

Isolating r_{2}:

r_{2}=r_{1}\sqrt{\frac{I_{1}}{I_{2}}}

r_{2}=7.5 m\sqrt{\frac{0.11 W/m^{2}}{1 W/m^{2}}}

r_{2}=2.48 m This is the distance at which the intensity is the "threshold of pain"

Now, we have to substract this value to r_{1} to find how much closer to the source can we move:

r_{1}-r_{2}=7.5 m - 2.48 m=5.02 m \approx 5 m

3 0
3 years ago
Read 2 more answers
In a shot-put competition, a shot moving at 15m/s has 450J of mechanical kinetic energy. What is the mass of the shot? Please he
Llana [10]

Answer:

Mass of shot (m) = 4 kg

Explanation:

Given:

Velocity (v) = 15 m/s

Mechanical kinetic energy (K.E) = 450 J

Find:

Mass of shot (m) = ?

Computation:

Mechanical kinetic energy (K.E) = 1/2mv²

Mechanical kinetic energy (K.E) = [1/2](m)(15)²

450 = [1/2](m)(15)²

900 = 225 m

Mass of shot (m) = 4 kg

5 0
3 years ago
Physics Conversion help!!
stiv31 [10]

[Assuming that you've written 3.40 kg in 'a', and not 3.90 kg]

(a) 3,400 g x <u>0.001</u> = 3.40 kg [converting grams to kilograms]

(b) 220 cm x <u>0.01</u> = <u>2.2</u> m [converting centimeters to meters]

(c) 9.42 kg x <u>1000</u> = <u>9420</u> g [converting kilograms to grams]

(d) 6.53 m x <u>100</u> = <u>653</u> cm [converting meters to centimeters]

5 0
4 years ago
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