A 24-V battery is powering a light bulb with a resistance of 3.0 ohms. What is the current flowing through the bulb? A) 7.20 A B
1 answer:
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Answer: East of North
Explanation:
We have the following data:
Speed of the wind from East to West:
Speed of the bee relative to the air:
If we graph these speeds (which in fact are velocities because are vectors) in a vector diagram, we will have a right triangle in which the airspeed of the bee (its speed relative to te air) is the hypotense and the two sides of the triangle will be the <u>Speed of the wind from East to West</u> (in the horintal part) and the <u>speed due North relative to the ground</u> (in the vertical part).
Now, we need to find the direction the bee should fly directly to the flower (due North):
Clearing :
Answer:
a) the magnitude of the force is
F= Q() and where k = 1/4πε₀
F = Qqs/4πε₀r³
b) the magnitude of the torque on the dipole
τ = Qqs/4πε₀r²
Explanation:
from coulomb's law
E =
where k = 1/4πε₀
the expression of the electric field due to dipole at a distance r is
E(r) = , where p = q × s
E(r) = where r>>s
a) find the magnitude of force due to the dipole
F=QE
F= Q()
where k = 1/4πε₀
F = Qqs/4πε₀r³
b) b) magnitude of the torque(τ) on the dipole is dependent on the perpendicular forces
τ = F sinθ × s
θ = 90°
note: sin90° = 1
τ = F × r
recall F = Qqs/4πε₀r³
∴ τ = (Qqs/4πε₀r³) × r
τ = Qqs/4πε₀r²
(a) 0.165 m/s
The total initial momentum of the astronaut+capsule system is zero (assuming they are both at rest, if we use the reference frame of the capsule):
The final total momentum is instead:
where
is the mass of the astronaut
is the velocity of the astronaut
is the mass of the capsule
is the velocity of the capsule
Since the total momentum must be conserved, we have
so
Solving the equation for , we find
(negative direction means opposite to the astronaut)
So, the change in speed of the capsule is 0.165 m/s.
(b) 520.8 N
We can calculate the average force exerted by the capsule on the man by using the impulse theorem, which states that the product between the average force and the time of the collision is equal to the change in momentum of the astronaut:
The change in momentum of the astronaut is
And the duration of the push is
So re-arranging the equation we find the average force exerted by the capsule on the astronaut:
And according to Newton's third law, the astronaut exerts an equal and opposite force on the capsule.
(c) 25.9 J, 390.6 J
The kinetic energy of an object is given by:
where
m is the mass
v is the speed
For the astronaut, m = 125 kg and v = 2.50 m/s, so its kinetic energy is
For the capsule, m = 1900 kg and v = 0.165 m/s, so its kinetic energy is