Answer:
The drag force is increase by 2.9
Explanation:
The drag force on the car is:
Where p is the density, A is the area of the car facing the air, C is the drag coefficient and v is the velocity. If k=(pCA)/2, then we have:
Fdrag=kv^2
For the speed 65 km/h:
Fdrag1=k(65^2)
For the speed 110 km/h
Fdrag2=k(110^2)
The ratio is:
Ratio=Fdrag2/Fdrag1=(k*65^2)/(k*110^2)=2.9
Answer:
The frictional force 6.446 N
The acceleration of the block a = 6.04
Explanation:
Mass of the block = 3.9 kg
°
= 0.22
(a). The frictional force is given by
3.9 × 9.81 ×
29.3 N
Therefore the frictional force
0.22 × 29.3
6.446 N
(b). Block acceleration is given by
F = 30 N
= 6.446 N
= 30 - 6.446
= 23.554 N
The net force acting on the block is given by
23.554 = 3.9 × a
a = 6.04
This is the acceleration of the block.
Answer:
Explanation:
Given:
<u>Now the value of limiting frictional force offered by the contact surface tending to have a relative motion under the effect of force:</u>
where:
normal force of reaction acting on the body= weight of the body
As we know that the frictional force acting on the body is always in the opposite direction:
So, the frictional force will not be at its maximum and will be equal in magnitude to the applied external force and hence the body will not move.
so, the frictional force will be:
Answer:
☯ Question :
☯
☥ Given :
☥ To find :
☄ We know ,
where ,
Now, substitute the values and solve for v.
➺
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✑ Additional Info :
Hope I helped!
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