Answer:
The cross-sectional area of the larger piston is 392cm ^{2}[/tex]
Explanation:
To solve this problem we apply the following formula:
Pascal principle: F=P*A Formula (1)
F=Force applied to the piston
P: Pressure
A= Piston area
Nomenclature:
Fp= Force on the primary piston= 500N
W= car weight =m*g=2000kg*9.8m/s2= 19600N
Fs= Force on the secondary piston= W = 19600N
As= Secondary piston area=?
The pressure acting on one side is transmitted to all the molecules of the liquid because the liquid is incompressible.
In the equation (1)
P=F/A
Pp=Ps
Answer:
35°C
Explanation:
q = mCΔT
2130 J = (0.200 kg) (710 J/kg/°C) (T − 20.0°C)
T = 35°C
Answer:
5308.34 N/C
Explanation:
Given:
Surface density of each plate (σ) = 47.0 nC/m² = 
Separation between the plates (d) = 2.20 cm
We know, from Gauss law for a thin sheet of plate that, the electric field at a point near the sheet of surface density 'σ' is given as:
Now, as the plates are oppositely charged, so the electric field in the region between the plates will be in same direction and thus their magnitudes gets added up. Therefore,
Now, plug in for 'σ' and 
for 
and solve for the electric field. This gives,
Therefore, the electric field between the plates has a magnitude of 5308.34 N/C
Answer: B. Current x delivered 6.3 C more then Y
Explanation:
Answer:
1.) Magnitude = 5596 N
2.) Direction = 60 degrees
Explanation: You are given that the breakdown vehicle A is exerting a force of 4000 N at angle 45 degree to the vertical and breakdown vehicle B is exerting a force of 2000 N
Let us resolve the two forces into X and Y component
Sum of the forces in the X - component will be 4000 × cos 45 = 2828.43 N
Sum of the forces in the Y - component will be 2000 + ( 4000 × sin 45 )
= 2000 + 2828.43
= 4828.43 N
The resultant force R will be
R = sqrt ( X^2 + Y^2 )
Substitutes the forces at X component and Y component into the formula
R = sqrt ( 2828.43^2 + 4828.43^2 )
R = sqrt ( 31313752.53 )
R = 5595.87 N
The direction will be
Tan Ø = Y/X
Substitute Y and X into the formula
Tan Ø = 4828.43 / 2828.43
Tan Ø = 1.707106
Ø = tan^-1( 1.707106 )
Ø = 59.64 degree
Therefore, approximately, the magnitude and direction of the resultant force on the truck are 5596 N and 60 degree respectively.
