Answer:
U = 1 / r²
Explanation:
In this exercise they do not ask for potential energy giving the expression of force, since these two quantities are related
F = - dU / dr
this derivative is a gradient, that is, a directional derivative, so we must have
dU = - F. dr
the esxresion for strength is
F = B / r³
let's replace
∫ dU = - ∫ B / r³ dr
in this case the force and the displacement are parallel, therefore the scalar product is reduced to the algebraic product
let's evaluate the integrals
U - Uo = -B (- / 2r² + 1 / 2r₀²)
To complete the calculation we must fix the energy at a point, in general the most common choice is to make the potential energy zero (Uo = 0) for when the distance is infinite (r = ∞)
U = B / 2r²
we substitute the value of B = 2
U = 1 / r²
Answer:
50000 N
Explanation:
From the question given above, the following data were obtained:
Mass (m) of bullet = 0.050 kg
velocity (v) = 400 m/s
Distance (s) = 0.080 m
Force (F) =?
Next, we shall determine the acceleration of the bullet. This can be obtained as follow:
Initial velocity (u) = 0 m/s
Final velocity (v) = 400 m/s
Distance (s) = 0.080 m
Acceleration (a) =?
v² = u² + 2as
400² = 0 + (2 × a × 0.08)
160000 = 0 + 0.16a
160000 = 0.16a
Divide both side by 0.16
a = 160000 / 0.16
a = 1×10⁶ m/s²
Finally, we shall determine the force exerted by the bullet on the target. This can be obtained as follow:
Mass (m) of bullet = 0.050 kg
Acceleration (a) of bullet = 1×10⁶ m/s²
Force (F) =?
F = ma
F = 0.050 × 1×10⁶
F = 50000 N
Thus, the bullet exerted a force of 50000 N on the target.
The two flaws in
her experiment’s design are
<span>- She introduced at least one confounding variable.</span>
<span>- She tried to test multiple hypotheses at a time</span>
In the above mentioned experiment she had to have four samples to prove
four hypotheses, each one separately and not to mix two hypotheses in an alone
sample, that what it brings as consequence is the confusion.
Is halved. A 6Ω resistor connected to a voltage source which voltage is decreased from 12V to 6V the current passing through the resistor is halved.
The key to solve this problem is applying Ohm's Law V = R I, clearing I from the equation, we obtain I = V/R. Then, the current is directly proportional to the voltage and inversely proportional to the resistance.
V = 12V and R = 6Ω
I = 12V/6Ω = 2A
V = 6V and R = 6Ω
V = 6V/6Ω = 1A
As we can see the current is halved if the voltage descreased from 12V to 6V
Answer:
A. Stratosphere is said to be stable layer of the atmosphere when cool air sinks and warm air rises.Due to the fact that cool air has tendency to sink ,the air is not going fluctuating up and down in the stratosphere. This means that the air remains stationary or particles remains there for a very long duration.
B. If the lifted index is negative then the parcel temperature is warmer than the actual temperature. In addition, the parcel that is less warm than the surrounding will be less dense and will rise.
C. The water vapor come from different kinds of fronts; gust fronts from existing storms as their downdraft hits the surface, spreads and lifts air in front, upper air disturbances and surface heating by solar radiation making an unbalanced vertical profile .
D. the threshold used by storm chasers to assess if the dew point temperature is high enough to produce large thunderstorms is moisture ,the surface dew point needs to be 55 degrees fahrenheit or greater for a surface based thunderstorm to occur.
E. Wind shear is the change in wind direction or speed with height in an atmosphere.
Explanation:
