Wires 1, 2, and 3 each have current moving through them to the right. I1 = 10 A, I2 = 5 A, and I3 = 8 A. Wire 2 is 15 cm long an
1 answer:
Answer:
F_2 = 20 x 10⁻⁶ N
Explanation:
given,
current in right direction
I₁ = 10 A
I₂ = 5 A
I₃ = 8 A
Length of Wire 2 = 15 cm
distance between wire 1 and wire 2, r₁ = 5 cm
distance between wire 2 and 3 = r₂ = 17 - 5 = 12 cm
Force on wire 2 due to current in wire 1 and wire 2
μ₀ =absolute permeability of free space= 4π x 10⁻⁷ H/m
F_2 = 20 x 10⁻⁶ N
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where
- the horizontal motion is a uniform motion, with constant speed
, where 
and 
- the vertical motion is an uniformly accelerated motion, with constant acceleration
, initial position h (the height of the building) and initial vertical velocity 
(with a negative sign, since it points downwards)
The ball strikes the ground after a time t=3.00 s, so we can find the distance covered horizontally by the ball by substituting t=3.00 s into the equation of x(t):
b) To find the height from which the ball was thrown, h, we must substitute t=3.00 s into the equation of y(t), and requiring that y(3.00 s)=0 (in fact, after 3 seconds the ball reaches the ground, so its vertical position y(t) is zero). Therefore, we have:
which becomes
c) We want the ball to reach a point 10.0 meters below the level of launching, so we want to find the time t such that
If we substitute this into the equation of y(t), we have
whose solution is
(the other solution is negative, so it has no physical meaning). Therefore, the ball reaches a point 10 meters below the level of launching after 1.18 s.
where
- the horizontal motion is a uniform motion, with constant speed
- the vertical motion is an uniformly accelerated motion, with constant acceleration
The ball strikes the ground after a time t=3.00 s, so we can find the distance covered horizontally by the ball by substituting t=3.00 s into the equation of x(t):
b) To find the height from which the ball was thrown, h, we must substitute t=3.00 s into the equation of y(t), and requiring that y(3.00 s)=0 (in fact, after 3 seconds the ball reaches the ground, so its vertical position y(t) is zero). Therefore, we have:
which becomes
c) We want the ball to reach a point 10.0 meters below the level of launching, so we want to find the time t such that
If we substitute this into the equation of y(t), we have
whose solution is
Answer:
The new volume of the balloon is 8.8 L
Explanation:
Charles's Law consists of the relationship between the volume and the temperature of a certain amount of ideal gas, which is kept at a constant pressure. This law establishes that the volume is directly proportional to the temperature of the gas, that is to say that if the temperature increases, the volume of the gas increases and that on the contrary if the temperature of the gas decreases, the volume decreases.
In other words, Charles's law states that when the amount of gas and pressure are kept constant, the ratio between the volume and the temperature will always have the same value:
Having a certain volume of gas V1 that is at a temperature T1, by varying the volume of gas to a new value V2, then the temperature will change to T2, and it will be fulfilled:
In this case:
- V1= 8.5 L
- T1= 294 K
- V2= ?
- T2= 305 K
Replacing:
Solving:
V2=8.8 L
<u><em>The new volume of the balloon is 8.8 L</em></u>