Answer:
remain the same constant
Explanation:
if a small gear is making 5 rpm turning a big gear thats making 1 rpm the bug gear is making lots of torque but if its going the opposite way its much harder to turn the big gear 1 rpm just to make the little gear do 5 rpm and it makes way less torque if thre both the same size there ging to stay the same speed
Answer:
20, 083 L
Explanation:
The mistake was the result of not using units when converting the 7862 l to Kg. They used the density in pounds hence they multiplied by 1.77 Lb/L and obtained 13597 Lb not Kg as they assumed.
To obtain the amount needed to refuel they subtracted this quantity from the 22,300 Kg required for the trip again obtaining the wrong quantity of 8703 Kg and they converted this to liters by dividing the density to get 4916 L and then placed then 5000 L of fuel
The quantity required was
7862 L * 1.77 Lb/L = 13915.74 Lb (pounds not kilos)
then converting this pounds to Kg by multiplying by 0.454 Kg/L one gets
6173 Kg on board
Amount Required
( 22,300 -6173) : 16127 Kg
16127 Kg/ 0.803 Kg/L = 20083 L
Answer:
Rate of internal heat transfer = 23.2 Btu/Ibm
mass flow rate = 21.55 Ibm/s
Explanation:
using given data to obtain values from table F7.1
Enthalpy of water at temperature of 100 F = 68.04Btu/Ibm
Enthalpy of water at temperature of 50 F = 18.05 Btu/Ibm
from table F.3
specific constant of glycerin 
<u>The rate of internal heat transfer ( change in enthalpy ) </u>
h4 - h3 = Cp ( T4 - T3 ) --------------- ( 1 )
where ; T4 = 50 F
T3 = 10 F
Cp = 0.58 Btu/Ibm-R
substitute given values into equation 1
change in enthalpy ( h4 - h3 ) = 23.2 Btu/Ibm
<u>Determine mass flow rate of glycol</u>
attached below is the detailed solution
mass flow rate of glycol = 21.55 Ibm/s
<h3><u>CSMA/CD Protocol:
</u></h3>
Carrier sensing can transmit the data at anytime only the condition is before sending the data sense carrier if the carrier is free then send the data.
But the problem is the standing at one end of channel, we can’t send the entire carrier. Because of this 2 stations can transmit the data (use the channel) at the same time resulting in collisions.
There are no acknowledgement to detect collisions, It's stations responsibility to detect whether its data is falling into collisions or not.
<u>Example:
</u>
, at time t = 10.00 AM, A starts, 10:59:59 AM B starts at time 11:00 AM collision starts.
12:00 AM A will see collisions
Pocket Size to detect the collision.

CSMA/CD is widely used in Ethernet.
<u>Efficiency of CSMA/CD:</u>
- In the previous example we have seen that in worst case
time require to detect a collision.
- There could be many collisions may happen before a successful completion of transmission of a packet.
We are given number of collisions (contentions slots)=4.
Distance = 1km = 1000m

Y = a (b)^t/p
y is total money
a is original amount
b is growth / decay factor
t is time
p is the frequency of every growth or decay
15131.76 = 11613 x 1.08^x
15131.76 / 11613 = 1.08^x
1.303… = 1.08^x
log1.303…. = xlog1.08
x = 3.43902165741 years