The vertical weight carried by the builder at the rear end is F = 308.1 N
<h3>Calculations and Parameters</h3>
Given that:
The weight is carried up along the plane in rotational equilibrium condition
The torque equilibrium condition can be used to solve
We can note that the torque due to the force of the rear person about the position of the front person = Torque due to the weight of the block about the position of the front person
This would lead to:
F(W*cosθ) = mgsinθ(L/2) + mgcosθ(W/2)
F(1cos20)= 197/2(3.10sin20 + 2 cos 20)
Fcos20= 289.55
F= 308.1N
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Its resistor :}...............
Answer:
The bus
Explanation:
a = (v-u)/t
Where
a = acceleration
v = final velocity
u = initial velocity
t = time taken
For truck to get its acceleration,
a = (18-0)/5.5 = 3.27 ms⁻²
For bus to get its acceleration,
a = (24-0)/6 = 4 ms⁻²
As 4 > 3.27 bus has a greater acceleration.
You can use them by analyzing the way you can solve it.
Answer:
Explanation:
Let pressure at surface of earth be P Pa.
pressure at height of 8.1 km in air can be calculated as follows .
pressure due to column of air of 8.1 km height
= h d g , h is height , d is density of air and g is acceleration due to gravity
= 8.1 x 1000 x .87 x 9.8 = 6.9 x 10⁴ Pa .
pressure at the height of 8.1 km
= P - 6.9 x 10⁴ Pa
Pressure due to column of 16 m in the sea
= h d g
16 x 1000 x 9.8
= 15.68 x 10⁴ Pa .
Pressure at depth of 16m
= P + 15.68 x 10⁴
pressure difference between points at height of 8.1 km and pressure at point 16 m deep
= P + 15.68 x 10⁴ - P + 6.9 x 10⁴ Pa
= 22.58 x 10⁴ Pa .
