Answer:
Explanation:
Givens
vi = 10 m/s
a = 1.5 m/s^2
d = 600 m
vf = ?
Formula
vf^2 = vi^2 + 2*a*d
Solution
vf^2 = 10^2 + 2*1.5 * 600
vf^2 = 100 + 1800
vf^2 = 1900
sqrt(vf^2) = sqrt(1900)
vf = 43.59 m/s
Explanation:
A particular kind of matter with uniform properties..
Answer:
≈ 20.35 N [newton's of tension]
Explanation:
( (2.9 × 9.8) ÷ cos(35.6°) ) × sin (35.6°) =
( (28.42) ÷ (≈0.813) ) × (≈0.582) =
(≈34.96) × (≈0.582) = 20.3449446.... ≈ 20.35
Answer:
F = - k (x-xo) a graph of the weight or applied force against the elongation obtaining a line already proves Hooke's law.
Explanation:
The student wants to prove hooke's law which has the form
F = - k (x-xo)
To do this we hang the spring in a vertical position and mark the equilibrium position on a tape measure, to simplify the calculations we can make this point zero by placing our reference system in this position.
Now for a series of known masses let's get them one by one and measure the spring elongation, building a table of weight vs elongation,
we must be careful when hanging the weights so as not to create oscillations in the spring
we look for the mass of each weight
W = mg
m = W / g
and we write them in a new column, we make a graph of the weight or applied force against the elongation and it should give a straight line; the slope of this line is sought, which is the spring constant.
The fact of obtaining a line already proves Hooke's law.
1.96s and 1.86s. The time it takes to a spaceship hovering the surface of Venus to drop an object from a height of 17m is 1.96s, and the time it takes to the same spaceship hovering the surface of the Earth to drop and object from the same height is 1.86s.
In order to solve this problem, we are going to use the motion equation to calculate the time of flight of an object on Venus surface and the Earth. There is an equation of motion that relates the height as follow:
The initial velocity of the object before the dropping is 0, so we can reduce the equation to:
We know the height h of the spaceship hovering, and the gravity of Venus is 
. Substituting this values in the equation :
To calculate the time it takes to an object to reach the surface of Venus dropped by a spaceship hovering from a height of 17m, we have to clear t from the equation above, resulting:
Similarly, to calculate the time it takes to an object to reach the surface of the Earth dropped by a spaceship hovering from a height of 17m, and the gravity of the Earth 
.
