Answer:
C = 17 i^ - 7 j^ + 16 k^ , | C| = 24.37
Explanation:
To work the vactor component method, we add the sum in each axis
C = A + B = (Aₓ + Bₓ) i ^ + ( +
) i ^ + (
+
) k ^
Cₓ = 12+ 5 = 17
= -37 +30 = -7
= 58 -42 = 16
Resulting vector
C = 17 i ^ - 7j ^ + 16k ^
The mangitude of the vector is
| C | = √ c²
| C | = √( 17² + 7² + 16²)
| C| = 24.37
Answer:
Explanation:
The horizontal distance covered by the ball in the falling is only determined by its horizontal motion - in fact, it is given by
where
is the horizontal velocity
t is the time of flight
The time of flight, instead, is only determined by the vertical motion of the ball: however, in this problem the vertical velocity is not changed (it is zero in both cases), so the time of flight remains the same.
In the first situation, the horizontal distance covered is
in the second case, the horizontal velocity is increased to
And so the new distance travelled will be
So, the distance increases linearly with the horizontal velocity.
Answer:
a) 0.138J
b) 3.58m/S
c) (1.52J)(I)
Explanation:
a) to find the increase in the translational kinetic energy you can use the relation
where Wp is the work done by the person and Wg is the work done by the gravitational force
By replacing Wp=Fh1 and Wg=mgh2, being h1 the distance of the motion of the hand and h2 the distance of the yo-yo, m is the mass of the yo-yo, then you obtain:
the change in the translational kinetic energy is 0.138J
b) the new speed of the yo-yo is obtained by using the previous result and the formula for the kinetic energy of an object:
where vf is the final speed, vo is the initial speed. By doing vf the subject of the formula and replacing you get:
the new speed is 3.58m/s
c) in this case what you can compute is the quotient between the initial rotational energy and the final rotational energy
hence, the change in Er is about 1.52J times the initial rotational energy