Answer:
15 m/s or 1500 cm/s
Explanation:
Given that
Speed of the shoulder, v(h) = 75 cm/s = 0.75 m/s
Distance moved during the hook, d(h) = 5 cm = 0.05 m
Distance moved by the fist, d(f) = 100 cm = 1 m
Average speed of the fist during the hook, v(f) = ? cm/s = m/s
This can be solved by a very simple relation.
d(f) / d(h) = v(f) / v(h)
v(f) = [d(f) * v(h)] / d(h)
v(f) = (1 * 0.75) / 0.05
v(f) = 0.75 / 0.05
v(f) = 15 m/s
Therefore, the average speed of the fist during the hook is 15 m/s or 1500 cm/s
Answer: Globalization is the process of interaction and integration among people, companies, and governments worldwide.
Explanation:
Answer:
a)3.5s
b)28.57m/S
c)34.33m/S
d)44.66m/S
Explanation:
Hello!
we will solve this exercise numeral by numeral
a) to find the time the ball takes in the air we must consider that vertically the ball experiences a movement with constant acceleration whose value is gravity (9.81m / S ^ 2), that the initial vertical velocity is zero, we use the following equation for a body that moves with constant acceleration

where
Vo = Initial speed
=0
T = time
g=gravity=9.81m/s^2
y = height=60m
solving for time

T=3.5s
b)The horizontal speed remains constant since there is no horizontal acceleration.
with the value of the distance traveled (100m) and the time that lasts in the air (3.5s) we estimate the horizontal speed

c)
to find the final vertical velocity we use the equations for motion with constant velocity as follows
Vf=Vo+g.t
Vf=0+(9.81 )(3.5)=34.335m/S
d)Finally, to find the resulting velocity, we add the horizontal and vertical velocities vectorially, this is achieved by finding the square root of the sum of its squares

Answer:
The main types of nucleons are protons and neutrons. A proton, as its name suggests, has a positive electric charge, and a neutron has a neutral electric charge (meaning that it has no charge). The two in the nucleus of the atom make a positive charge, since the neutron has no charge at all.
Explanation:
Answer:
4.78 x 10^-11 J
Explanation:
A = 1.5 x 10^-4 m^2
d = 2 mm = 2 x 10^-3 m
V = 12 V
Let C be the capacitance of the capacitor
C = ε0 A / d
C = (8.854 x 10^-12 x 1.5 x 10^-4) / (2 x 10^-3)
C = 6.64 x 10^-13 F
Energy stored, U = 1/2 CV^2
U = 0.5 x 6.64 x 10^-13 x 12 x 12
U = 4.78 x 10^-11 J