Answer:
d₂ = 1.466 m
Explanation:
In this case we must use the rotational equilibrium equations
Στ = 0
τ = F r
we must set a reference system, we use with origin at the easel B and an axis parallel to the plank
, we will use that the counterclockwise ratio is positive
+ W d₁ - w_cat d₂ = 0
d₂ = W / w d₁
d₂ = M /m d₁
d₂ = 5.00 /2.9 0.850
d₂ = 1.466 m
Answer:
The magnetic force points in the positive z-direction, which corresponds to the upward direction.
Option 2 is correct, the force points in the upwards direction.
Explanation:
The magnetic force on any charge is given as the cross product of qv and B
F = qv × B
where q = charge on the ball thrown = +q (Since it is positively charged)
v = velocity of the charged ball = (+vî) (velocity is in the eastern direction)
B = Magnetic field = (+Bj) (Magnetic field is in the northern direction; pointing forward)
F = qv × B = (+qvî) × (Bj)
F =
| î j k |
| qv 0 0|
| 0 B 0
F = i(0 - 0) - j(0 - 0) + k(qvB - 0)
F = (qvB)k N
The force is in the z-direction.
We could also use the right hand rule; if we point the index finger east (direction of the velocity), the middle finger northwards (direction of the magnetic field), the thumb points in the upward direction (direction of the magnetic force). Hence, the magnetic force is acting upwards, in the positive z-direction too.
Hope this Helps!!!
What's now called "Conventional current" is thought of as the flow of positive charge, from the battery's positive terminal to its negative one.
But it turns out that positive charges don't flow. The physical flow of charge is the flow of electrons. They come out of the battery's negative terminal, and carry negative charge around the circuit to the battery's positive one.
Time it takes the projectile to hit the ground after being thrown up:
√h/1/2a
√8/(.5)(9.81)
√8/4.905
√1.630988787
= 1.277101714
= 1. 28
hope this helps :)
Answer:
Index of expansion: 4.93
Δu = -340.8 kJ/kg
q = 232.2 kJ/kg
Explanation:
The index of expansion is the relationship of pressures:
pi/pf
The ideal gas equation:
p1*v1/T1 = p2*v2/T2
p2 = p1*v1*T2/(T2*v2)
500 C = 773 K
20 C = 293 K
p2 = 35*0.1*773/(293*1.3) = 7.1 bar
The index of expansion then is 35/7.1 = 4.93
The variation of specific internal energy is:
Δu = Cv * Δt
Δu = 0.71 * (20 - 500) = -340.8 kJ/kg
The first law of thermodynamics
q = l + Δu
The work will be the expansion work
l = p2*v2 - p1*v1
35 bar = 3500000 Pa
7.1 bar = 710000 Pa
q = p2*v2 - p1*v1 + Δu
q = 710000*1.3 - 3500000*0.1 - 340800 = 232200 J/kg = 232.2 kJ/kg