At sea level, water ______________ at 100°C.
2 answers:
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Answer:
<em>The sprinter traveled a distance of 7.5 m</em>
Explanation:
<u>Motion With Constant Acceleration </u>
It's a type of motion in which the rate of change of the velocity of an object is constant.
The equation that rules the change of velocities is:
Where:
a = acceleration
vo = initial speed
vf = final speed
t = time
The distance traveled by the object is given by:
Using the equation [1] we can solve for a:
The sprinter travels from rest (vo=0) to vf=7.5 m/s in t=2 s. Computing the acceleration:
Now calculate the distance:
The sprinter traveled a distance of 7.5 m
Answer:
Given that
For A weight Wt= 22.7 N
m₁ = 22.7/10 = 2.27 kg
Force alone inclined plane
Wt₁ = m₁ g sin θ
Wt₁ =22.7 sin 17.2°
Wt₁ = 6.7 N
For B weight Wt₂= 34.5-N
m₂ = 3.45 kg
coefficient of friction ,μ= 0.219
θ = 17.2 degree
The friction force on the block A
Fr= μ m₁ g cos θ
Fr= 0.216 x 22.7 x cos 17.2°
Fr= 4.68 N
Lets take acceleration of system is a m/s²
Tension = T
From Newtons law
Wt₂ - Wt₁ - Fr = (m₁ +m₂) a
34.5 - 6.7 - 4.68 = (2.27 + 3.45 ) a
a= 4.05 m/s²
Block B
Wt₂ - T = m₂ a
T = 34.5 - 4.05 x 3.45
T= 20.52 N
