The upward force exerted on the board by the support is mathematically given as
Fu= 764.8 N
<h3>What is the upward force exerted on the board by the support?</h3>
Generally, the equation for is mathematically given as
Considering that the Net Force on the system is null
The weight of the children plus the weight of the board equals the upward force imposed on the support.
The upward force
Fu= 440 + 272 + 52.8 N
Fu= 764.8 N
In conclusion, he upward force exerted on the board by the support
Fu= 764.8 N
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Answer:
"Apparent weight during the "plan's turn" is 519.4 N
Explanation:
The "plane’s altitude" is not so important, but the fact that it is constant tells us that the plane moves in a "horizontal plane" and its "normal acceleration" is 
Given that,
v = 420 m/s
R = 11000 m
Substitute the values in the above equation,



It has a horizontal direction. Furthermore, constant speed implies zero tangential acceleration, hence vector a = vector a N. The "apparent weight" of the pilot adds his "true weight" "m" "vector" "g" and the "inertial force""-m" vector a due to plane’s acceleration, vector
In magnitude,





Because vector “a” is horizontal while vector g is vertical. Consequently, the pilot’s apparent weight is vector

Which is quite heavier than his/her true weigh of 519.4 N
The star is the main sequence
Answer:
F = 25530 N
Explanation:
F(net) = ma
F(net) = 2553 kg•(10) m/s² = 25530 N