Answer:
The correct answer will be "400.4 N". The further explanation is given below.
Explanation:
The given values are:
Mass of truck,
m = 600 kg
g = 9.8 m/s²
On equating torques at the point O,
⇒ 
So that,
On putting the values, we get
⇒ 
⇒ 
Answer: The overhead percentage is 7.7%.
Explanation:
We call overhead, to all those bytes that are delivered to the physical layer, that don't carry real data.
We are told that we have 700 bytes of application data, so all the other bytes are simply overhead, i.e. , 58 bytes composed by the transport layer header, the network layer header, the 14 byte header at the data link layer and the 4 byte trailer at the data link layer.
So, in order to assess the overhead percentage, we divide the overhead bytes between the total quantity of bytes sent to the physical layer, as follows:
OH % = (58 / 758) * 100 = 7.7 %
Answer:
Cc= 12.7 lb.sec/ft
Explanation:
Given that
m = 22 lb
g= 32 ft/s²
x= 4.5 in
1 in = 0.083 ft
x= 0.375 ft
Spring constant ,K
K= 58.66 lb/ft
The damper coefficient for critically damped system
Cc= 12.7 lb.sec/ft
Answer: 1.137*10^7 Btu/h.
Explanation:
Given data:
Efficiency of the plant = 4.5percent
Net power output of the plant = 150kw
Solution:
The required collection rate
QH = W/n
= 150/0.045 * 0.94782/ 1 /60 */60 Btu/h.
= 3333.333 *3412.152Btu/h.
= 11373840 Btu/h
= 1.137*10^7 Btu/h.
Explanation:
750 microvolt is your answer
please mark as brilliant
