Answer:
a) the first vector has magnitude 58 cm and the angle is 15 measured clockwise from the positive side of the x-axis
b) the second vector, the magnitude is 55.7 cm and the angle is 35 half from the negative side of the x-axis in a clockwise direction
c) the magnitude is 54.2 cm with an angle of 18 measured counterclockwise from the x-axis
Explanation:
For this exercise we draw a Cartesian coordinate system in this system: East coincides with the positive part of the x-axis and North with the positive part of the y-axis.
a) the first vector has magnitude 58 cm and the angle is 15 measured clockwise from the positive side of the x-axis
b) the second vector, the magnitude is 55.7 cm and the angle is 35 half from the negative side of the x-axis in a clockwise direction
c) the magnitude is 54.2 cm with an angle of 18 measured counterclockwise from the x-axis
In the attachment we can see the representation of the three vectors
Answer:
2,100J
Explanation:
Workdone=Force x distance
=700 x 3
=2,100 J
Answer:
100m
Explanation:
100m
s=ut+1/2at^2
s= unknown, u=0, a=2, t=10
s=0*10+1/2(2)(10)^2
s=1/2(2)(100)
s=1(100)
displacement = 100 meters
Take the object's starting direction of motion to be the positive direction, so that a stopping force acts in the opposite direction. By Newton's second law, the object undergoes an acceleration <em>a</em> such that
-15 N = (20 kg) <em>a</em>
Solve for <em>a</em> :
<em>a</em> = - (15 N) / (20 kg) = -0.75 m/s²
The object's velocity <em>v</em> at time <em>t</em> is then given by
<em>v</em> = 3 m/s + (-0.75 m/s²) <em>t</em>
so the time it takes for the object to slow to a rest is
0 = 3 m/s + (-0.75 m/s²) <em>t</em>
<em>t</em> = (3 m/s) / (0.75 m/s²) = 4.0 s