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marishachu [46]
3 years ago
12

How to find coefficient of lift from coefficient of pressure?

Physics
1 answer:
Alona [7]3 years ago
4 0
It is the number in front of the equation
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You have a horizontal cathode ray tube (CRT) for which the controls have been adjusted such that the electron beam should make a
Dvinal [7]

Answer:

Explanation:

Rotate it slowly to establish whether the spot moves. If it is broken, it will stay as it is. By rotating it the electrons' path to the centre of the screen will be tilted to another position by any external disturbing field.

8 0
3 years ago
For the sun, which is the final step in core fusion?
Marina86 [1]
The answer would be stage of<span> conversion of Hydrogen to Helium.
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3 0
3 years ago
the displacement (in centimeters) of a particle moving back and forth along a straight line is given by the equation of motion s
12345 [234]

The average velocity or displacement of a particle for the first time interval is <u>Δs / Δt = 6 cm/s.</u>

Solution:

As we know that displacement is calculated in centimeters and the unit of time is second.

The average velocity for the first interval [1,2] is given

Δs / Δt = s (t2) - s (t) / t2 - t1

Δs / Δt = 2sin2  π  + 3cos 2 π -  ( 2sin π + 3cos π ) / 2 - 1

Δs / Δt = 2(0) + 3(1) - 2(0) - 3 (-1) / 1

Δs / Δt = 6 cm/s

Thus the average velocity or displacement of a particle for the first time interval is Δs / Δt = 6 cm/s

If you need to learn more about displacement click here:

brainly.com/question/28370322

#SPJ4

The complete question is:

The displacement of a particle moving back and forth along a line is given by the following equation s(t) = 2sin π t + 3cos π t. Estimate the instantaneous velocity of the particle when t = 1

4 0
2 years ago
A stone is thrown straight up. when it reaches it’s highest point, _____
Sati [7]

both it velocity and acceleration is zero.

8 0
3 years ago
(a) Calculate the force (in N) needed to bring a 1100 kg car to rest from a speed of 85.0 km/h in a distance of 125 m (a fairly
nasty-shy [4]

(a) -2451 N

We can start by calculating the acceleration of the car. We have:

u=85.0 km/h = 23.6 m/s is the initial velocity

v = 0 is the final velocity of the car

d = 125 m is the stopping distance

So we can use the following equation

v^2 - u^2 = 2ad

To find the acceleration of the car, a:

a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(125 m)}=-2.23 m/s^2

Now we can use Newton's second Law:

F = ma

where m = 1100 kg to find the force exerted on the car in order to stop it; we find:

F=(1100 kg)(-2.23 m/s^2)=-2451 N

and the negative sign means the force is in the opposite direction to the motion of the car.

(b) -1.53\cdot 10^5 N

We can use again the equation

v^2 - u^2 = 2ad

To find the acceleration of the car. This time we have

u=85.0 km/h = 23.6 m/s is the initial velocity

v = 0 is the final velocity of the car

d = 2.0 m is the stopping distance

Substituting and solving for a,

a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(2 m)}=-139.2 m/s^2

So now we can find the force exerted on the car by using again Newton's second law:

F=ma=(1100 kg)(-139.2 m/s^2)=-1.53\cdot 10^5 N

As we can see, the force is much stronger than the force exerted in part a).

8 0
3 years ago
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