Answer:
Explanation:
The boy throw the pencil upward at a speed of 6.33 m/s
Then,
Initial velocity of throw is 6.33 m/s
u = 6.33 m/s.
Time to reach a maximum height of 1.25m
h = 1.25m
Note: at maximum height, the final velocity is zero
v = 0m/s
Acceleration due to gravity is
g = 9.81m/s²
We want to calculate time to reach maximum height
t = ?
Then, applying equation of motion
v = u + gt
But since it is against gravity, then, g is negaive
Then,
v = u - gt
0 = 6.33 - 9.81t
-6.33 = -9.81t
Then,
t = -6.33 / -9.81
t = 0.645 seconds
Answer:
If the mass of B is m and the temperature change is the same, the mass of B will be 2m.
Explanation:
Q = mcT
T = mc/Q
M = 4Q/2cT........... (1)
T = Q/mc
Plug this in equation 1.
M = 4Q/(2c × Q/mc) = 4Q ÷ 2Q/m = 4Q × m/2Q = 2m
Which data set has the largest range? A. 55, 57, 59, 60, 61, 49, 48 B. 21, 25, 14, 16, 29, 22, 20 C. 12, 15, 16, 19, 18, 15, 27
Simora [160]
Data D has the largest range.
Data A: 61-48=13
Data B: 29-14=15
Data C:27-12=15
Data D:54-31=23
Therefore, Data D has the largest range.
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
Read more about vertical weight here:
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