Answer:
20 cm
Explanation:
Te electric potential enery U = kq₁q₂/r were q₁ = 5 nC = 5 × 10⁻⁹ C and q₂ = -2 nC = -2 × 10⁻⁹ C and r = √(x - 2)² + (0 - 0)² +(0 - 0)² = x - 2. U = -0.5 µJ = -0.5 × 10⁻⁶ J, k = 9 × 10⁹ Nm²/C².
So r = kq₁q₂/U
x - 2 = kq₁q₂/U
x = 0.02 + kq₁q₂/U m
x = 0.02 + 9 × 10⁹ Nm²/C² × 5 × 10⁻⁹ C × -2 × 10⁻⁹ C/-0.5 × 10⁻⁶ J
x = 0.02 - 90 × 10⁻⁹ Nm²/-0.5 × 10⁻⁶ J
x = 0.02 + 0.18 = 0.2 m = 20 cm
<span>E = h x f </span>
<span>. . . then : </span>
<span>f = E / h </span>
<span>f = 4,41•10^-19 / 6,62•10^-34 </span>
<span>f = 6,66•10^14 Hz (s^-1) </span>
<span>b/ What is the wavelength of this light ? </span>
<span>- - - - - - - - - - - - - - - - - - - - - - - - - - - - </span>
<span>λ = c / f </span>
<span>λ = 3•10^8 / 6,66•10^14 </span>
<span>λ = 4,50•10^-7 m </span>
Answer:
, 
Explanation:
The jet is flying at constant velocity: this means that its acceleration is zero, so the net force acting on the jet is also zero.
Therefore, we can write:

where
is the thrust force generated by each engine of the jet
is the drag force
Solving for Fd,

The velocity of the jet is

So, the rate at which the drag force does work (which is the power) is

and substituting

we find

Converting into horsepower,

Incomplete question.The complete question is here
Determine the torque applied to the shaft of a car that transmits 225 hp and rotates at a rate of 3000 rpm.
Answer:
Torque=0.51 Btu
Explanation:
Given Data
Power=225 hp
Revolutions =3000 rpm
To find
T( torque )=?
Solution
As

As force moves an object through a distance, work is done on the object. Likewise, when a torque rotates an object through an angle, work is done.
So

Very roughly 7,700 feet ... about 1.5 miles.