K.E=1/2mv^2 K.E=1/2multiply1multiply8^2=32joules
Answer:
970 kN
Explanation:
The length of the block = 70 mm
The cross section of the block = 50 mm by 10 mm
The tension force applies to the 50 mm by 10 mm face, F₁ = 60 kN
The compression force applied to the 70 mm by 10 mm face, F₂ = 110 kN
By volumetric stress, we have that for there to be no change in volume, the total pressure applied by the given applied forces should be equal to the pressure removed by the added applied force
The pressure due to the force F₁ = 60 kN/(50 mm × 10 mm) = 120 MPa
The pressure due to the force F₂ = 110 kN/(70 mm × 10 mm) = 157.142857 MPa
The total pressure applied to the block, P = 120 MPa + 157.142857 MPa = 277.142857 MPa
The required force, F₃ = 277.142857 MPa × (70 mm × 50 mm) = 970 kN
Rolling friction .<span> the force that slows down the movement of a rolling object</span>
sliding friction.
Sliding friction : The opposing force that comes into play when
one body is actually sliding over the surface of the other body
is called sliding friction. e.g. A flat block is moving over a
horizontal table.
Kinetic or dynamic friction: If the applied force is increased further
and sets the body in motion, the friction opposing the motion is called
kinetic friction
Answer:
= 925.92 N
≅ 926N
Explanation:
Pressure due to car = pressure due to applied force
12000/18^2 = Force / 5^2
force = 12000 * 25/ 324
= 925.92 N
For equilibrium
Pressure1 = Pressure2
A1F1 = A2F2
12000*pi*(5^2) = F2 ( pi)*(18^2)
so, F2 = Applied force to lift car = 925.92 N
Pascal's principle
Pressure1 = Pressure2
F1/A1 = F2/A2 (F=force and A=area)
A1 =Pi*(0.05)²
A2 =Pi(0.18)²
F2=12000
F1 = 12000*(0.05)² / (0.18)² = 926N
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