Answer:
F = - 2 A x - B
Explanation:
The force and potential energy are related by the expression
F = - dU / dx i ^ -dU / dy j ^ - dU / dz k ^
Where i ^, j ^, k ^ are the unit vectors on the x and z axis
The potential they give us is
U (x) = A x² + B x + C
Let's calculate the derivatives
dU / dx = A 2x + B + 0
The other derivatives are zero because the potential does not depend on these variables.
Let's calculate the strength
F = - 2 A x - B
"When we do experiments it's a good idea to do multiple trials, that is, do the same experiment lots of times. When we do multiple trials of the same experiment, we can make sure that our results are consistent and not altered by random events. Multiple trials can be done at one time."
Answer:
Small sports car.
Explanation:
Lets take
mass of the small car = m
mass of the truck = M
As we know that when car collide with the massive truck then due to change in the moment of the car both car as well as truck will feel force.We also know that from Third law of Newton's ,it states that every action have it reaction with same magnitude but in the opposite direction.
Therefore
F = m a
a=Acceleration of the car
F= M a'
a'=Acceleration of the massive truck
Here given that M > m that is why a > a'
Therefore car will experiences more acceleration.
The magnitude of e.m.f induced in the loop when t = 2 s is 31 Volts.
<h3>emf induced in the loop</h3>
The magnitude of e.m.f induced in the loop is calculated as follows;
emf = dФ/dt
Ф = 6t² + 7t
dФ/dt = 12t + 7
at t = 2 seconds
emf = dФ/dt = 12(2) + 7 = 31 V
Thus, the magnitude of e.m.f induced in the loop when t = 2 s is 31 Volts.
Learn more about emf here: brainly.com/question/24158806
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The helium may be treated as an ideal gas, so that
(p*V)/T =constant
where
p = pressure
V = volume
T = temperature.
Note that
7.5006 x 10⁻³ mm Hg = 1 Pa
1 L = 10⁻³ m³
Given:
At ground level,
p₁ = 752 mm Hg
= (752 mm Hg)/(7.5006 x 10⁻³ mm Hg/Pa)
= 1.0026 x 10⁵ Pa
V₁ = 9.47 x 10⁴ L = (9.47 x 10⁴ L)*(10⁻³ m³/L)
= 94.7 m³
T₁ = 27.8 °C = 27.8 + 273 K
= 300.8 K
At 36 km height,
P₂ = 73 mm Hg = 73/7.5006 x 10⁻³ Pa
= 9.7326 x 10³ Pa
T₂ = 235 K
If the volume at 36 km height is V₂, then
V₂ = (T₂/p₂)*(p₁/T₁)*V₁
= (235/9.7326 x 10³)*(1.0026 x 10⁵/300.8)*94.7
= 762.15 m³
Answer: 762.2 m³