Answer:
Explanation:
we know that
as we see that
relative error
Where X_1 IS HEIGHT OF ROCK
IS THE HEIGHT OF ROAD
= uncertainity in measuring distance
Putting all value to get uncertainity in angle
solving for we get
A person walks 2 miles every day to work, leaving her front porch at 7:00 A.M. and arriving at work at 7:30 A.M. ON the way home
1 answer:
Answer:
The displacement is zero miles
Explanation:
The displacement of an object that moves from point A to point B is defined as
Where d is the displacement of the object. The displacement does not depend on the trajectory of the object. It only depends on the linear distance between the end point and the starting point.
In this case we know that the person walks from home to work and then walks from work to home. Therefore, the total displacement is the linear distance between the point where its journey begins and the point where the route ends.
The tour begins on the front porch of your house and ends on the front porch of your house (when you return from work). If we call A to the front porch of the house then the displacement is:
The displacement is zero miles, since the person finishes the journey just where it started (front porch)
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Answer:
0.001 s
Explanation:
The force applied on an object is equal to the rate of change of momentum of the object:
where
F is the force applied
is the change in momentum
is the time interval
The change in momentum can be written as
where
m is the mass
v is the final velocity
u is the initial velocity
So the original equation can be written as
In this problem:
m = 5 kg is the mass of the fist
u = 9 m/s is the initial velocity
v = 0 is the final velocity
F = -45,000 N is the force applied (negative because its direction is opposite to the motion)
Therefore, we can re-arrange the equation to solve for the time:
Answer:
(a) 1.939 m/h
(b) 0.926 m/h
(c) -0.315 m/h
(d) -1.21 m/h
Explanation:
Here, we have the water depth given by the function of time;
D(t) = 7 + 5·cos[0.503(t-6.75)]
Therefore, to find the velocity of the depth displacement with time, we differentiate the given expression with respect to time as follows;
D'(t) =
= 5×(-sin(0.503(t-6.75))×0.503
= -2.515×(-sin(0.503(t-6.75))
= -2.515×(-sin(0.503×t-3.395))
Therefore we have;
(a) at 5:00 AM = 5 - 0:00 = 5
D'(5) = -2.515×(-sin(0.503×5-3.395)) = 1.939 m/h
(b) at 6:00 AM = 6 - 0:00 = 6
D'(5) = -2.515×(-sin(0.503×6-3.395)) = 0.926 m/h
(c) at 7:00 AM = 7 - 0:00 = 7
D'(5) = -2.515×(-sin(0.503×7-3.395)) = -0.315 m/h
(d) at Noon 12:00 PM = 12 - 0:00 = 12
D'(5) = -2.515×(-sin(0.503×12-3.395)) = -1.21 m/h.