Answer:
The value is 
Explanation:
Generally the velocity attained by the sled after t = 3.10 s is mathematically evaluated using the kinematic equation as follows
Here u = 0 \ m/s
a = 13.5 
So
=> 
The is distance it covers at this time is
=> 
=> 
Now when sled stops its the final velocity is 
while the initial velocity will be the velocity after its acceleration i.e
So
Here 
, the negative sign shows that it is deceleration
So
=>
This problem has three questions I believe:
>
How hard does the floor push on the crate?
<span>We have to find the net
vertical (normal) Fn force which results from Fp and Fg.
We know that the normal component of Fg is just Fg, which is equal to as 1110N.
From the geometry, the normal component of Fp can be calculated:
Fpn = Fp * cos(θp)
= 1016.31 N * cos(53)
= 611.63 N
The total normal force Fn then is:
Fn = Fg + Fpn
= 1110 + 611.63
=
1721.63 N</span>
> Find the friction
force on the crate
<span>We
have to look for the net horizontal force Fh which results from Fp and Fg.
Since Fg is a normal force entirely, so we can say that the
horizontal component is zero:
Fh = Fph + Fgh
= (Fp * sin(θp)) + 0
= 1016.31 N * sin(53)
=
811.66 N</span>
> What is the minimum
coefficient of static friction needed to prevent the crate from slipping on the
floor?
We just need to compute the
ratio Fh to Fn to get the minimum μs.
μs = Fh / Fn
= 811.66 N / 1721.63 N
<span>=
0.47</span>
1) Current in the wire: 0.0875 A
The current in the wire is given by:
where
Q is the charge passing a given point in the conductor
t is the time elapsed
In this problem, we have
Q = 420 C is the total charge passing through a given point in a time of
t = 80 min = 4800 s
So, the current is
2) Drift velocity of the electrons: 
The drift velocity of the electrons in the wire is given by:
where
I = 0.0875 A is the current
is the number of free electrons per cubic meter
A is the cross-sectional area
is the charge of one electron
The radius of the wire is
So the cross-sectional area is
So, the drift velocity is
Answer:
90.77%
its capacity utilization rate for the month is 90.77%
Explanation:
The capacity utilisation rate can be expressed mathematically as;
Capacity utilisation rate = capacity used/Best operating level × 100%
Given;
Total Number of production time = 205hours
Production output/capacity used = 21400 units
Best operation rate = 115units/hour
Best operation output for the month of July( at best operation level )
=115units/hour × 205 hours = 23575 units
Capacity utilisation rate = 21400/23575 × 100%
= 90.77%
The key projectile motion is that gravity allows downward only
