Given : Initial velocity = -1.3 m/s
Final Velocity = -6.5 m/s.
Time = 25 minutes.
To find : Average acceleration.
Solution: We are given units in meter/second (m/s).
So, we need to convert time 25 minutes in seconds.
1 minute = 60 seconds.
25 minutes = 60*25 = 1500 seconds.
Formula for average acceleration is given by,
We are not given intial time, so we can take initial time =0.
Plugging values in the above formula.
= 
= -0.003467
or 
.
From
the problem statement, this is a conversion problem. We are asked to convert
from units of grams to units of kilograms. To do this, we need a
conversion factor which would relate the different units involved. We either
multiply or divide this certain value to the original measurement depending on
what is asked. From literature, we will find that 1000 grams is equal to 1 kilogram. We use this as follows:
<span> 1.440x10^6 g ( 1 kg / 1000 g ) = 1440 kg</span><span>
</span>
The question ask to find and calculate the induced current in the loop as a function time and the best answer would be that the induced current in the loop is 0.08 amperes. I hope you are satisfied with my answer and feel free to ask for more if you have clarifications and further questions
<span>5.98 x 10^-2 ohms.
Resistance is defined as:
R = rl/A
where
R = resistance in ohms
r = resistivity (given as 1.59x10^-8)
l = length of wire.
A = Cross sectional area of wire.
So plugging into the formula, the known values, including the area of a circle being pi*r^2, gives:
R = 1.59x10^-8 * 3.00 / (pi * (5.04 x 10^-4)^2)
R = (4.77 x 10^-8) / (pi * 2.54016 x 10 ^-7)
R = (4.77 x 10^-8) / (7.98015 x 10^-7)
R = 5.98 x 10^-2 ohms
So that wire has a resistance of 5.98 x 10^-2 ohms.</span>
Answer:
Explanation:
If
-
, 
are temperatures of gasses and liquid in Kelvins,
and 
are thicknesses of gas layer and steel slab in meters,
, 
are convection coefficients gas and liquid in 
, 
is the contact resistance in 
,
- and
are thermal conductivities of gas and steel in 
,
then: part(a):
using known values:
part(b): Using the rate equation :
the surface temperature 
and 
Similarly
The temperature distribution is shown in the attached image
