We are given: Final velocity (
)=20 m/s .
Time t= 2.51 s and
distance s = 82.9 m.
We know, equation of motion

Let us plug values of final velocity, and time in above equation.


Subtracting 2.51a from both sides, we get
-----------equation(1)
Using another equation of motion

Plugging values of vi =20-2.51a, t=2.51 and distnace s=82.9 in this equation.
We get,

Now, we need to solve it for a.
20-20+2.51a=165.8a.
-163.29a=0
a=0.
So, the acceleration would be 0 m/s^2.
Answer:
The total charge on the rod is 47.8 nC.
Explanation:
Given that,
Charge = 5.0 nC
Length of glass rod= 10 cm
Force = 840 μN
Distance = 4.0 cm
The electric field intensity due to a uniformly charged rod of length L at a distance x on its perpendicular bisector
We need to calculate the electric field
Using formula of electric field intensity

Where, Q = charge on the rod
The force is on the charged bead of charge q placed in the electric field of field strength E
Using formula of force

Put the value into the formula

We need to calculate the total charge on the rod

Put the value into the formula



Hence, The total charge on the rod is 47.8 nC.
I’ve answered this before so I know the question is missing an
important given and that given is: <span>1 has an
empty trailer and the other has a fully loaded one.
So, it would be the fully loaded trailer that would take a longer distance to
stop because a lot of weight is being pulled, and when the brakes are started,
the fully loaded trailer is more like pushing against the truck.</span>
The first problem cannot be solve because you did give the distance or length of the rope, because work = distance x force. i can only solve the the second problem. since the bucket is moving up then force due to gravity is going down, then the net force is:
Fnet = F1 - Fg
where Fg = mg
g is the accelaration due to gravity ( 9.81 m/s^2)
Fnet = 57.5 N - (3.9 kg)(9.81) N
Fnet = 19.24 N
Answer:
Part(a): The angular acceleration is
.
Part(b): The angular displacement is
.
Explanation:
Part(a):
If
be the initial angular speed, final angular speed and angular acceleration of the centrifuge respectively, then from rotational kinematic equation, we can write

where '
' is the time taken by the centrifuge to increase its angular speed.
Given,
,
and
. From equation (
), the angular acceleration is given by

Part(b):
Also the angular displacement (
) can be written as
