Answer:
12 N/cm²
Explanation:
From the question given above, the following data were obtained:
Weight (W) of block = 240 N
Area (A) = 20 cm²
Pressure (P) =?
Next, we shall determine the force exerted by the block. This can be obtained as follow:
Weight (W) of block = 240 N
Force (F) =.?
Weight and force has the same unit of measurement. Thus, we force applied is equivalent to the weight of the block. Thus,
Force (F) = Weight (W) of block = 240 N
Force (F) = 240 N
Finally, we shall determine the pressure on the floor as follow:
Force (F) = 240 N
Area (A) = 20 cm²
Pressure (P) =?
P = F/A
P = 240 / 20
P = 12 N/cm²
Therefore, the pressure on the floor is 12 N/cm².
Answer:
33.33 %
Explanation:
given,
error last year = 63
error this year = 42
percent decrease in the error = ?
to find the percentage difference in the error formula used is
= 
= 
= 
= 33.33 %
Percentage decrease in the number of error is equal to 33.33%.
Answer
given,
mass of the person, m = 50 Kg
length of scaffold = 6 m
mass of scaffold, M= 70 Kg
distance of person standing from one end = 1.5 m
Tension in the vertical rope = ?
now equating all the vertical forces acting in the system.
T₁ + T₂ = m g + M g
T₁ + T₂ = 50 x 9.8 + 70 x 9.8
T₁ + T₂ = 1176...........(1)
system is equilibrium so, the moment along the system will also be zero.
taking moment about rope with tension T₂.
now,
T₁ x 6 - mg x (6-1.5) - M g x 3 = 0
'3 m' is used because the weight of the scaffold pass through center of gravity.
6 T₁ = 50 x 9.8 x 4.5 + 70 x 9.8 x 3
6 T₁ = 4263
T₁ = 710.5 N
from equation (1)
T₂ = 1176 - 710.5
T₂ = 465.5 N
hence, T₁ = 710.5 N and T₂ = 465.5 N
Answer:
-10.9 rad/s²
Explanation:
ω² = ω₀² + 2α(θ - θ₀)
Given:
ω = 13.5 rad/s
ω₀ = 22.0 rad/s
θ - θ₀ = 13.8 rad
(13.5)² = (22.0)² + 2α (13.8)
α = -10.9 rad/s²
Answer:
<em>The athlete will rise 1.10 meters off the ground</em>
Explanation:
<u>Vertical Motion</u>
If an object is launched vertically upwards at an initial speed vo, then it will reach a maximum height given by

The athlete can exert a net force upwards equal to twice his weight. It makes him accelerate upwards at

The speed at the end of his push can be computed by

Replacing the value of a obtained above:

where y is the length of this crouch


This is the initial speed of this vertical launch, thus

