Answer:
F1= 122.93 N
Explanation:
Pascal´s Principle
Pascal´s Principle can be applied in the hydraulic press:
If we apply a small force (F1) on a small area piston A1, then, a pressure (P) is generated that is transmitted equally to all the particles of the liquid until it reaches a larger area piston and therefore a force (F2) can be exerted that is proportional to the area (A2) of the piston:
P=F/A
P1=P2
Formula (1)
Data
D1 = 1.85 cm : primary cylinder piston diameter
D2 = 24.5 cm : secondary cylinder piston diameter
m = 2200-kg : car mass
Piston area calculation
A1= 2.688 cm²
A2 = 471.435 cm²
Calculating of the weight of the car (W)
W = m*g = 2200-kg * 9.8 m/s² = 21560 N
Calculation of the force in Newtons to be exerted on the primary cylinder piston
Data:
A1= 2.688cm²
A2= 471.435 cm²
F2 = W= 21560 N
We replace data in the formula (1)
F1= 122.93 N
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.
Answer:
+1.46×10¯⁶ C
Explanation:
From the question given above, the following data were obtained:
Charge 1 (q₁) = +26.3 μC = +26.3×10¯⁶ C
Force (F) = 0.615 N
Distance apart (r) = 0.750 m
Electrical constant (K) = 9×10⁹ Nm²/C²
Charge 2 (q₂) =?
The value of the second charge can be obtained as follow:
F = Kq₁q₂ / r²
0.615 = 9×10⁹ × 26.3×10¯⁶ × q₂ / 0.750²
0.615 = 236700 × q₂ / 0.5625
Cross multiply
236700 × q₂ = 0.615 × 0.5625
Divide both side by 236700
q₂ = (0.615 × 0.5625) / 236700
q₂ = +1.46×10¯⁶ C
NOTE: The force between them is repulsive as stated from the question. This means that both charge has the same sign. Since the first charge has a positive sign, the second charge also has a positive sign. Thus, the value of the second charge is +1.46×10¯⁶ C
It is because the effort distance is greater than the load distance
Explanation:
As we know, Effort×effort distance = load × load distance
So when effort distance is increases,
The effort decreases
So when the spanner’s handle is long
A tight knot can easily be opened by less effrot
I hope it helped
Answer:
Work done on an object is equal to
FDcos(angle).
So, naturally, if you lift a book from the floor on top of the table you do work on it since you are applying a force through a distance.
However, I often see the example of carrying a book through a horizontal distance is not work. The reasoning given is this: The force you apply is in the vertical distance, countering gravity and thus not in the direction of motion.
But surely you must be applying a force (and thus work) in the horizontal direction as the book would stop due to air friction if not for your fingers?
Is applying a force through a distance only work if causes an acceleration? That wouldn't make sense in my mind. If you are dragging a sled through snow, you are still doing work on it, since the force is in the direction of motion. This goes even if velocity is constant due to friction.
Explanation:
