By studying meteorites which are the most ancient material in space.
Answer:
a) p = 4.96 10⁻¹⁹ kg m / s
, b) p = 35 .18 10⁻¹⁹ kg m / s
,
c) p_correst / p_approximate = 7.09
Explanation:
a) The moment is defined in classical mechanics as
p = m v
Let's calculate its value
p = 1.67 10⁻²⁷ 0.99 3. 10⁸
p = 4.96 10⁻¹⁹ kg m / s
b) in special relativity the moment is defined as
p = m v / √(1 –v² / c²)
Let's calculate
p = 1.67 10⁻²⁷ 0.99 10⁸/ √(1- 0.99²)
p = 4.96 10⁻¹⁹ / 0.141
p = 35 .18 10⁻¹⁹ kg m / s
c) the relationship between the two values is
p_correst / p_approximate = 35.18 / 4.96
p_correst / p_approximate = 7.09
We need to see what forces act on the box:
In the x direction:
Fh-Ff-Gsinα=ma, where Fh is the horizontal force that is pulling the box up the incline, Ff is the force of friction, Gsinα is the horizontal component of the gravitational force, m is mass of the box and a is the acceleration of the box.
In the y direction:
N-Gcosα = 0, where N is the force of the ramp and Gcosα is the vertical component of the gravitational force.
From N-Gcosα=0 we get:
N=Gcosα, we will need this for the force of friction.
Now to solve for Fh:
Fh=ma + Ff + Gsinα,
Ff=μN=μGcosα, this is the friction force where μ is the coefficient of friction. We put that into the equation for Fh.
G=mg, where m is the mass of the box and g=9.81 m/s²
Fh=ma + μmgcosα+mgsinα
Now we plug in the numbers and get:
Fh=6*3.6 + 0.3*6*9.81*0.8 + 6*9.81*0.6 = 21.6 + 14.1 + 35.3 = 71 N
The horizontal force for pulling the body up the ramp needs to be Fh=71 N.
Answer:
Explanation:
Given a parallel plate capacitor of
Area=A
Distance apart =d
Potential difference, =V
If the distance is reduce to d/2
What is p.d
We know that
Q=CV
Then,
V=Q/C
Then this shows that the voltage is inversely proportional to the capacitance
Therefore,
V∝1/C
So, VC=K
Now, the capacitance of a parallel plate capacitor is given as
C= εA/d
When the distance apart is d
Then,
C1=εA/d
When the distance is half d/2
C2= εA/(d/2)
C2= 2εA/d
Then, applying
VC=K
V1 is voltage of the full capacitor V1=V
V2 is the required voltage let say V'
Then,
V1C1=V2C2
V × εA/d=V' × 2εA/d
VεA/d = 2V'εA/d
Then the εA/d cancels on both sides and remains
V=2V'
Then, V'=V/2
The potential difference is half when the distance between the parallel plate capacitor was reduce to d/2