Answer:
a) 5.61 rad/s
b) 16.83 rad/s
Explanation:
Distance to the floor = 79cm = 0.79m
Rotation is <1 revolution
The toast will rotate at an angular speed that is constant while it falls. The toast will also be falling with constant acceleration due to gravity. Using equation of motion, which is
S = ut + 1/2gt²
S = 0 + 1/2gt²
S = gt², where, S = d
0.79 = 9.81t²
t² = 0.79/9.81
t² = 0.081
t = 0.28s
As the toast is accidentally pushed, it rotates as it falls. It will be landing on its edges when and if it hits the ground. The smallest angle here would then be 1/4 of the revolution. This is also the smallest angular speed.
ω(min) = ΔΦ revolution /Δt
ω(min) = 1/4 * 2π / 0.28
ω(min) = 0.5π/0.28
ω(min) = 5.61 rad/s
Since 1/4 of revolution is the minimum angle, the remaining(3/4), is the maximum angle. Thus
ω(max) = 3/4 * 2π / 0.28
ω(max) = 0.75 * 2π / 0.28
ω(max) = 16.83 rad/s
The answer on Edge would be (A.)= Larger and Cooler ! I'm doing the same thing as y'all. Good luck everyone.
I assuming they are not going to be very big just small enough for the car to get over the top and not go backwards. The coasting is all the momentum thy have to get over the hill and they don't have a lot of momentum. hope this helps ☺
Answer:
The new force on the test charge is 145.02N
Explanation:
Force on a unit positive charge can be calculated using coulomb's law.
F =Kq²/r²
Where
F is the force on the charge = 321 N
K is a constant = 8.99 X10⁹ Nm²/C²
q is a test charge = 4.89 μC = 4.89 X10⁻⁶ C
r is the distance between the charge and the surface = 4.1cm
F ∝1/r²
Force on the charge is inversely proportional the square of distance between the charge and the surface.
Fr² = constant
F₁r₁² = F₂r₂²
F₂ = F₁r₁²/r₂²
If the charge is then moved 2.00 cm farther away from the surface;
The new distance becomes 4.10 cm + 2.00 cm = 6.1 cm
Therefore, r₂ = 6.1 cm, r₁ = 4.1 cm, F₁ =321 N
F₂ = (321 X4.1²)/(6.1²)
F₂ = 145.02N
Therefore, The new force on the test charge is 145.02N
