Answer:
Given:
m=1000kg
u= 16.7m/s
v=0m/s
F=8000N
Required:
s=?
Solution:
F=m × a
8000N=1000kg × a
a=8m/s^2
Since it decelerate a= -8m/s^2
v^2 = u^2 + 2as
s=v^2 - u^2 / 2a
s= 0 - (16.7m/s)^2 / 2 × -8m/s^2
s= -278.89/-16
s= 17.43m
The car travels approximately 17.43m before it stops
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Answer:
at t=46/22, x=24 699/1210 ≈ 24.56m
Explanation:
The general equation for location is:
x(t) = x₀ + v₀·t + 1/2 a·t²
Where:
x(t) is the location at time t. Let's say this is the height above the base of the cliff.
x₀ is the starting position. At the base of the cliff we'll take x₀=0 and at the top x₀=46.0
v₀ is the initial velocity. For the ball it is 0, for the stone it is 22.0.
a is the standard gravity. In this example it is pointed downwards at -9.8 m/s².
Now that we have this formula, we have to write it two times, once for the ball and once for the stone, and then figure out for which t they are equal, which is the point of collision.
Ball: x(t) = 46.0 + 0 - 1/2*9.8 t²
Stone: x(t) = 0 + 22·t - 1/2*9.8 t²
Since both objects are subject to the same gravity, the 1/2 a·t² term cancels out on both side, and what we're left with is actually quite a simple equation:
46 = 22·t
so t = 46/22 ≈ 2.09
Put this t back into either original (i.e., with the quadratic term) equation and get:
x(46/22) = 46 - 1/2 * 9.806 * (46/22)² ≈ 24.56 m
Answer:
yeah physical quantities are the quantities which can be meaured
Answer:
C.
Measure the wear on his treads before and after riding a certain number of laps.
