From laws of motion:
Where S is the distance/displacement (as you would call it) which is unknown
v = final velocity which is 0m/s (this is because the car stops)
u = initial velocity which is 36m/s (from the data given)
t = time taken for the distance to be covered and it is 6s
Substitute the values, hence:
But this is merely the distance he travelled in the 6 seconds he was trying to stop the car.
Therefore, the distance between the car and the cows = 160-108
Distance = 52m
Answer:
a) 
Now we can replace the velocity for t=1.75 s
For t = 3.0 s we have:
b) 
And we can find the positions for the two times required like this:
And now we can replace and we got:
Explanation:
The particle position is given by:
Part a
In order to find the velocity we need to take the first derivate for the position function like this:
Now we can replace the velocity for t=1.75 s
For t = 3.0 s we have:
Part b
For this case we can find the average velocity with the following formula:
And we can find the positions for the two times required like this:
And now we can replace and we got:
Answer: 12) 1.07 m/s (right) 13) 4.05 m/s 14) 73 m/s 15) 10.9 m/s
Explanation:
12) Conservation of momentum. Momentum is the produce of mass and velocity.
13(2) + 15(-5) = 13(-5) + 15v
v = 1.06666... ≈ 1.07 m/s (right)
13) 18(9) + 22(0) = 18v + 22v
v = 18(9)/40 = 4.05 m/s
14) 0.65(35) + 0.08(0) = 0.65(26) + 0.08v
v = 73.125
15) This is a bit trickier. Let's ASSUME you jump off at 7 m/s relative to the truck. Doing this, we can assume that the reference frame is moving along with the truck at 10 m/s
the conservation of momentum equation becomes
600(0) + 80(0) = 600v + 80(-7)
v = 0.9333333... m/s
adding back the velocity of the reference frame means the truck is now traveling.
10.9333333... ≈ 10.9 m/s
It is <span>C. Low to moderate level of exertion can be sustained over long periods of time </span>
Answer:
v= 4055.08m/s
Explanation:
This is a problem that must be addressed through the laws of classical mechanics that concern Potential Gravitational Energy.
We know for definition that,
We must find the highest point and the lowest point to identify the change in energy, so
Point a)
The problem tells us that an object is dropped at a distance of h = 1.15134R over the earth.
That is to say that the energy of that object is equal to,
Point B )
We now use the average radius distance from the earth.
Then,
By the law of conservation of energy we know that,
clearing v,
Therefore the speed of the object when it strikes the Earth’s surface is 4055.08m/s
