The magnitude of the current in wire 3 is 2.4 A and in a direction pointing in the downward direction.
- The force per unit length between two parallel thin current-carrying
and 
wires at distance ' r ' is given by 
....(1) .
- If the current is flowing in both wires in the same direction, and the force between them will be the attractive force and if the current is flowing in opposite direction in wires then the force between them will be the repulsive force.
A schematic of the information provided in the question can be seen in the image attached below.
From the image, force on wire 2 due to wire 1 = force on wire 2 due to wire 3
Using equation (1) , we get
I₃ = 2.4 A and the current is pointing in the downward direction
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Answer:
Δ v = 125 m/s
Explanation:
given,
mass of space craft = 435 Kg
thrust = 0.09 N
time = 1 week
= 7 x 24 x 60 x 60 s
change in speed of craft = ?
Assuming no external force is exerted on the space craft
now,
a = 2.068 x 10⁻⁴ m/s²
using equation of motion
Δ v = a t
Δ v = 2.068 x 10⁻⁴ x 7 x 24 x 60 x 60
Δ v = 125 m/s
Answer:
t= 1.2 hours
Explanation:
Define first di distance between the points, so
The distance is
Answer:
Rhizobium.
Explanation:
The roots of most of these plants have nitrogen-fixing bacteria known as rhizobium. It helps to change nitrogen which is present in the air into the nitrates which are plants, and that is need to synthesize proteins.
Rhizobium bacteria occupy the root hairs which are present on the plants, they help the plant to grow and keep up with other plants. When the one plant dies, the fixed nitrogen which is present in it will release, and making it available for other plants, and this process helps to fertilize the soil in the environment.
Answer:
W= -2.5 (p₁*0.0012) joules
Explanation:
Given that p₀= initial pressure, p₁=final pressure, Vi= initial volume=0 and Vf=final volume= 6/5 liters where p₁=p₀ then
In adiabatic compression, work done by mixture during compression is
W= 
where f= final volume and i =initial volume, p=pressure
p can be written as p=K/V^γ where K=p₀Vi^γ =p₁Vf^γ
W= 
W= K/1-γ ( 1/Vf^γ-1 - 1/Vi^γ-1)
W=1/1-γ (p₁Vf-p₀Vi)
W= 1/1-1.40 (p₁*6/5 -p₀*0)
W= -2.5 (p₁*6/5*0.001) changing liters to m³
W= -2.5 (p₁*0.0012) joules
