Answer:
Magnetic field at point having a distance of 2 cm from wire is 6.99 x 10⁻⁶ T
Explanation:
Magnetic field due to finite straight wire at a point perpendicular to the wire is given by the relation :
......(1)
Here I is current in the wire, L is the length of the wire, R is the distance of the point from the wire and μ₀ is vacuum permeability constant.
In this problem,
Current, I = 0.7 A
Length of wire, L = 0.62 m
Distance of point from wire, R = 2 cm = 2 x 10⁻² m = 0.02 m
Vacuum permeability, μ₀ = 4π x 10⁻⁷ H/m
Substitute these values in equation (1).
B = 6.99 x 10⁻⁶ T
Answer
given,
force = 94 lb
weight of crate = 220 lb
Assuming the static friction be equal = 0.47
kinetic friction = 0.36
Maximum force applied to move the object is when object is just start to move.
F = μ N
F = 0.47 x 220
F = 103.4 lb
As the frictional force is more than applied then the object will not move.
so, the friction force will be equal to the force applied on the object that is equal to 94 lb.
hence, the direction of force will left.
1. The chemical reaction produced by Carlo's fire is exergonic because energy is "going out". As the reaction proceeds, entropy increases as the energy stored in the dry wood and leaves are used up as fuel to create the fire which produces low quality light and warmth.
2. This reaction is a classic example of an exothermic reaction. Exothermic reactions are characterized with the presence of heat and light in the products. Combustion reactions are always exothermic in nature.
3. Catalyst are substances that are used to speed up reactions by lowering the activation requirement. Catalysts aren't consumed in the reaction and can still be chemically retrieved afterwards. In this situation, the leaves cannot be retrieved after the reaction ends. The leaves speed up the heating of the wood but it does not behave as a catalyst.
Answer: chicken hope this helps!
Explanation:
Answer:
The center of mass of the two-ball system is 7.05 m above ground.
Explanation:
<u>Motion of 0.50 kg ball:</u>
Initial speed, u = 0 m/s
Time = 2 s
Acceleration = 9.81 m/s²
Initial height = 25 m
Substituting in equation s = ut + 0.5 at²
s = 0 x 2 + 0.5 x 9.81 x 2² = 19.62 m
Height above ground = 25 - 19.62 = 5.38 m
<u>Motion of 0.25 kg ball:</u>
Initial speed, u = 15 m/s
Time = 2 s
Acceleration = -9.81 m/s²
Substituting in equation s = ut + 0.5 at²
s = 15 x 2 - 0.5 x 9.81 x 2² = 10.38 m
Height above ground = 10.38 m
We have equation for center of gravity
m₁ = 0.50 kg
x₁ = 5.38 m
m₂ = 0.25 kg
x₂ = 10.38 m
Substituting
The center of mass of the two-ball system is 7.05 m above ground.
