Answer:
If the Earth absorbs more insolation from the sun than it radiates back, the Earth. warms
We make use of the equation: v^2=v0^2+2a Δd. We substitute v^2 equals to zero since the final state is halting the truck. Hence we get the equation -<span>v0^2/2a = Δd. F = m a from the second law of motion. Rearranging, a = F/m
</span>F = μ Fn where the force to stop the truck is the force perpendicular or normal force multiplied by the static coefficient of friction. We substitute, -v0^2/2<span>μ Fn/m</span> = Δd. This is equal to
Answer with Explanation:
We are given that
Magnetic field,B=
Length of wire,l=15 m
Current,I=19 A
a.We have to find the magnitude of magnetic force and direction of magnetic force.
Magnetic force,F=
Using the formula
Direction=
15 degree above the horizontal in the northward direction.
The tangential velocity of the car's tire is the product of the angular velocity and radius of the car's tire which is 11(r) m/s.
<h3>
Angular velocity of the tire</h3>
The angular velocity of the tire is the rate of change of angular displacement of the tire with time.
The magnitude of the angular velocity of the tire is calculated as follows;
ω = 2πN
where;
- N is the number of revolutions per second
ω = 2π x (5.25 / 3)
ω = 11 rad/s
<h3>Tangential velocity of the tire</h3>
The tangential velocity of the car's tire is the product of the angular velocity and radius of the car's tire.
The magnitude of the tangential velocity is caculated as follows;
v = ωr
where;
- r is the radius of the car's tire
v = 11r m/s
Learn more about tangential velocity here: brainly.com/question/25780931
The electric field of a very large (essentially infinitely large) plane of charge is given by:
E = σ/(2ε₀)
E is the electric field, σ is the surface charge density, and ε₀ is the electric constant.
To determine σ:
σ = Q/A
Where Q is the total charge of the sheet and A is the sheet's area. The sheet is a square with a side length d, so A = d²:
σ = Q/d²
Make this substitution in the equation for E:
E = Q/(2ε₀d²)
We see that E is inversely proportional to the square of d:
E ∝ 1/d²
The electric field at P has some magnitude E. Now we double the side length of the sheet while keeping the same amount of charge Q distributed over the sheet. By the relationship of E with d, the electric field at P must now have a quarter of its original magnitude:
