Answer:
(A) Original speed= 9.22 m/s
(B) Acceleration= -1.0099 m\s^2
Explanation:
A truck covers 40m in 7.10 secs
The truck slows down at a uniform velocity of 2.05 m/s
(A) The original speed can be calculated as follows
Vo= 2(40)/7.10 - 2.05
= 80/7.10 - 2.05
= 11.2676 - 2.05
= 9.22m/s
(B) The acceleration can be calculated as follows
a= Vf-Vo/t
= 2.05-9.22/7.10
= -7.17/7.10
= -1.0099m/s^2
The answer is "B" - If there are no windows then there will be no light coming in, and therefore you don't have to worry about what time of day you do the experiment at.
As far as I know acids turn litmus paper red and bases turn it blue. on the pH scale anything under 7 is acid and above is base so I think it is A. I am not totally sure, but that is the best answer I can give you! Good Luck! Hope this helps!!
Answer:
128 m
Explanation:
From the question given above, the following data were obtained:
Horizontal velocity (u) = 40 m/s
Height (h) = 50 m
Acceleration due to gravity (g) = 9.8 m/s²
Horizontal distance (s) =?
Next, we shall determine the time taken for the package to get to the ground.
This can be obtained as follow:
Height (h) = 50 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
h = ½gt²
50 = ½ × 9.8 × t²
50 = 4.9 × t²
Divide both side by 4.9
t² = 50 / 4.9
t² = 10.2
Take the square root of both side
t = √10.2
t = 3.2 s
Finally, we shall determine where the package lands by calculating the horizontal distance travelled by the package after being dropped from the plane. This can be obtained as follow:
Horizontal velocity (u) = 40 m/s
Time (t) = 3.2 s
Horizontal distance (s) =?
s = ut
s = 40 × 3.2
s = 128 m
Therefore, the package will land at 128 m relative to the plane
Answer:
<em>113.4 J</em>
Explanation:
<u>Elastic Potential Energy</u>
Is the energy stored in an elastic material like a spring of constant k, in which case the energy is proportional to the square of the change of length Δx and the constant k.
The spring has a natural length of 0.7 m and a spring constant of k=70 N/m. When the spring is stretched to a length of 2.5 m, the change of length is
Δx = 2.5 m - 0.7 m = 1.8 m
The energy stored in the spring is:
PE = 113.4 J
