When a footballer collides with the goal post, the forces at work are the action and reaction forces. The player will exert an action force on the goal post, and then a reaction force from the goal post will stop the player. The reaction force call will cause pain and even injury to the player.
Answer:
<em>a) 0.72 V</em>
<em>b) 19.2 mA</em>
<em>c) 2.304 Watts</em>
Explanation:
A transformer is used to step-up or step-down voltage and current. It uses the principle of electromagnetic induction. When the primary coil is greater than the secondary coil, the it is a step-down transformer, and when the primary coil is less than the secondary coil, the it is a step-up transformer.
number of primary turns = 
= 500 turns
input voltage = 
= 120 V
number of secondary turns = 
= 3 turns
output voltage = 
= ?
using the equation for a transformer
substituting values, we have
= 360/500 =<em> 0.72 V</em>
<em></em>
b) by law of energy conservation,
where
= input current = ?
= output voltage = 3.2 A
= output voltage = 0.72 V
= input voltage = 120 V
substituting values, we have
120 = 3.2 x 0.72
120 = 2.304
= 2.304/120 = 0.0192 A
= <em>19.2 mA</em>
<em></em>
c) power input = 
==> 0.0192 x 120 = <em>2.304 Watts</em>
The light will bend when in
Answer:
Yes i am agree with this suggestion
Explanation:
Given that we have to assume that there is no any frictional affects.
As we know that when height increases then the discharge level will decreases when discharge level decreases then the time of filling for the bucket will increase.So the bucket will fill faster if the hose lowered until knee level.
Yes i am agree with this suggestion
Answer:
F = 800N
the magnitude of the average force exerted on the wall by the ball is 800N
Explanation:
Applying the impulse-momentum equation;
Impulse = change in momentum
Ft = m∆v
F = (m∆v)/t
Where;
F = force
t = time
m = mass
∆v = v2 - v1 = change in velocity
Given;
m = 0.80 kg
t = 0.050 s
The ball strikes the wall horizontally with a speed of 25 m/s, and it bounces back with this same speed.
v2 = 25 m/s
v1 = -25 m/s
∆v = v2 - v1 = 25 - (-25) m/s = 25 +25 = 50 m/s
Substituting the values;
F = (m∆v)/t
F = (0.80×50)/0.05
F = 800N
the magnitude of the average force exerted on the wall by the ball is 800N
