Answer:
0.0034 sec
Explanation:
L = initial length
T = initial time period = 2.51 s
Time period is given as
L = 1.56392 m
L' = new length
ΔT = Rise in temperature = 142 °C
α = coefficient of linear expansion = 19 x 10⁻⁶ °C
New length due to rise of temperature is given as
L' = L + LαΔT
L' = 1.56392 + (1.56392) (19 x 10⁻⁶) (142)
L' = 1.56814 m
T' = New time period
New time period is given as
T' = 2.5134 sec
Change in time period is given as
ΔT = T' - T
ΔT = 2.5134 - 2.51
ΔT = 0.0034 sec
Answer:
a. t = 1.43 s
b. d = 7.88 m
Explanation:
a. The time of flight can be found using the following equation:
Where:
: is the final height = -10 m
: is the initial height = 0
: is the initial speed in the vertical direction = 0
g: is the acceleration due to gravity = 9.81 m/s²
By solving the above equation for "t" we have:
Hence, the ball will hit the ground in 1.43 s.
b. The distance in the horizontal direction can be found as follows:
Where:
x₀: is the initial position in the horizontal direction = 0
a: is the acceleration in the horizontal direction = 0 (it is moving at constant speed)
Therefore, the ball will travel 7.88 m before it hits the ground.
I hope it helps you!
Answer: The velocity at different marked time points are given as
t1 = -
t2 = +
t3 = +
t4 = -
t5 = 0
Explanation:
The slope of the tangent of the curve indicates the instantaneous velocity. So if the slope of the tangent is positive, that Is, the tangent makes a positive angle (above the horizontal axis) with the horizontal
axis, then the velocity at this point is positive, and if the slope of the tangent is negative, that is the tangent makes a negative angle with the horizontal axis (below the horizontal axis), then the velocity at this point is negative.
When the tangent of the line is parallel to the horizontal axis, the velocity is 0.
From the position-time graph attached, the sign on the instantaneous velocity for each time marked on the graph is given below
t1 = -
t2 = +
t3 = +
t4 = -
t5 = 0
QED!
