According to Newton's second law, the force applied to an object is equal to the product between the mass of the object and its acceleration:
where F is the magnitude of the force, m is the mass of the object and a its acceleration.
In this problem, the object is the insect, with mass
. The acceleration of the insect is
, therefore we can calculate the force exerted by the car on the insect:
How do we find the force exerted by the insect on the car?
According to Newton's third law (known as action-reaction law), when an object A exerts a force on an object B, object B also exerts a force equal and opposite on object A. Therefore, the force exerted by the insect on the car is equal to the force exerted by the car on the object, so it is 0.01 N.
Answer:
3000N
Explanation:
divided to get answer
the force needed to accelerate the 1000kg car by 3m/s2 is 3000N
Answer: Add an incline or grade to the road track.
Explanation:
Refer to the figure shown below.
When a vehicle travels on a level road in a circular path of radius r, a centrifugal force, F, tends to make the vehicle skid away from the center of the circular path.
The magnitude of the force is
F = mv²/r
where
m = mass of the vehicle
v = linear (tangential) velocity to the circular path.
The force that resists the skidding of the vehicle is provided by tractional frictional force at the tires, of magnitude
μN = μW = μmg
where
μ = dynamic coefficient of friction.
At high speeds, the frictional force will not overcome the centrifugal force, and the vehicle will skid.
When an incline of θ degrees is added to the road track, the frictional force is augmented by the component of the weight of the vehicle along the incline.
Therefore the force that opposes the centrifugal force becomes
μN + Wsinθ = W(sinθ + μ cosθ).
<span>Answer:
The moments of inertia are listed on p. 223, and a uniform cylinder through its center is:
I = 1/2mr2
so
I = 1/2(4.80 kg)(.0710 m)2 = 0.0120984 kgm2
Since there is a frictional torque of 1.20 Nm, we can use the angular equivalent of F = ma to find the angular deceleration:
t = Ia
-1.20 Nm = (0.0120984 kgm2)a
a = -99.19 rad/s/s
Now we have a kinematics question to solve:
wo = (10,000 Revolutions/Minute)(2p radians/revolution)(1 minute/60 sec) = 1047.2 rad/s
w = 0
a = -99.19 rad/s/s
Let's find the time first:
w = wo + at : wo = 1047.2 rad/s; w = 0 rad/s; a = -99.19 rad/s/s
t = 10.558 s = 10.6 s
And the displacement (Angular)
Now the formula I want to use is only in the formula packet in its linear form, but it works just as well in angular form
s = (u+v)t/2
Which is
q = (wo+w)t/2 : wo = 1047.2 rad/s; w = 0 rad/s; t = 10.558 s
q = (125.7 rad/s+418.9 rad/s)(3.5 s)/2 = 952.9 radians
But the problem wanted revolutions, so let's change the units:
q = (5528.075087 radians)(revolution/2p radians) = 880. revolutions</span>
