Is a motor a load that can be found in a circuit?
1 answer:
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Complete Question
The complete question is shown on the first uploaded image
Answer:
Force on each hand is 196.22 N
Force on each foot is 95.8 N
Explanation:
In order to get a better understanding of this question let us explain some concepts
Normal Force:
We can define normal force Fn as that type of force which makes a 90 degree angle with the surface on which it is exerted.
Torque:
We can define torque as the moment of forces that tends to produce or cause rotation
From the question we are given that
Weight of body is (W) = 584 N
The normal force on both hands (Ha) = ?
The normal force on both legs (Lg) = ?
Looking at the diagram the person is at equilibrium so
584 = Ha + Lg
an also this mean that torques acting on the body is balanced
So, 0.410 Ha = 0.840 Lg
Making Lg the subject of formula in the equation above we
Lg = 0.4881 Ha
Considering the first equation and replacing Lg with this recent equation we have
584 = Ha + 0.4881 Ha
Therefore Ha = 392.44 N
This value obtained is for both hands for each hand we divide by 2
Therefore we have for each hand =196.55 N
Since we have been able to get the force on both hands we can substitute it in to the equation where we made Lg the subject of formula and we have
Lg = 0.4881 × 392.44
= 191.22 N
The value above is the force on both legs to obtain the force on each leg we have
191.22/2 = 95.8 N.
(a) -4.6 m/s
We can solve this part by using the law of conservation of momentum: in fact, the total momentum of the bullet-rifle system before and after the shot must be equal.
Before the shot, the total momentum is zero:
p = 0 (1)
After the shot, the total momentum is:
(2)
where
m = 0.0250 kg is the mass of the bullet
v = 550 m/s is the velocity of the bullet
M = 3.00 kg is the mass of the rifle
V is the recoil velocity of the rifle
Since momentum is conserved, (1) = (2), so we can solve for V:
And the negative sign means the rifle will move backward.
(b) 31.7 J
The kinetic energy of an object is given by
where
m is the mass of the object
v is its speed
The rifle has a mass of
M = 3.00 kg
And a final speed of (speed = magnitude of velocity)
V = 4.6 m/s
Therefore, the kinetic energy it has gained is
(c) -0.5 m/s
In this case, we just need to repeat the problem as in part (a), applying the law of conservation of momentum:
where in this case, the mass of the rifle is
M = 28.0 kg
while the other data are unchanged:
m = 0.0250 kg is the mass of the bullet
v = 550 m/s is the velocity of the bullet
Solving for V, we find the new recoil velocity of the rifle:
(d) 3.5 J
As in part b), we can apply the equation of the kinetic energy:
The kinetic energy of an object is given by
where in this case, we have:
M = 28.0 kg is the mass of the rifle+shoulder
V = 0.5 m/s is the recoil speed of the rifle+shoulder
Substituting into the equation,
(e) Player's momentum is larger
The momentum of the player is
where
M = 110 kg is the mass of the player
V = 8.00 m/s is his velocity
Substituting,
The momentum of the ball is
where
m = 0.410 kg is the mass of the ball
v = 25.0 m/s is the velocity of the ball
Substituting,
The player's momentum is much larger than the ball's momentum. This problem becomes similar to the previous one in the moment when the player catches the ball: at that point, in fact, the velocity of the player-ball system will change such that their total combined momentum will be equal to the total momentum of the two individual objects before the catch.