Answer:
Particle-antiparticle annihilation is the result of a direct interaction between a particle and its corresponding anti particle, like a proton and an anti-proton or an electron and a positron. The result of such interaction is the production of two powerful gamma rays, each of whom has an energy roughly equal to E = mc^2, where m is the mass of the particle.
Charge = 0.2 Ah is the correct answer...
Answer:
A. α = 94.4 rad/s
B. a = 28.32 m/s
C. N = 34N
D. α = 94.4 rad/s
a = 28.32 m/s
N = 44.4 N
Explanation:
part A:
using:
∑T = Iα
where T is the torque, I is the moment of inertia and α is the angular momentum.
firt we will find the moment of inertia I as:
I = 
Where M is the mass and R is the radius of the wheel, then:
I = 
I = 0.36 kg*m^2
Replacing on the initial equation and solving for α, we get::
∑T = Iα
Fr = Iα
34 N = 0.36α
α = 94.4 rad/s
part B
we need to use this equation :
a = αr
where a is the aceleration of the cord that has already been pulled off and r is the radius of the wheel, so replacing values, we get:
a = (94.4)(0.3 m)
a = 28.32 m/s
part C
Using the laws of newton, we know that:
N = T
where N is the force that the axle exerts on the wheel part and T is the tension of the cord
so:
N = 34N
part D
The anly answer that change is the answer of the part D, so, aplying laws of newton, it would be:
-Mg + N +T = 0
Then, solving for N, we get:
N = -T+Mg
N = -34 + (8 kg)(9.8)
N = 44.4 N
Answer:
1.9841256 kg
Explanation:
Given;
Length of the swimming pool = 25.0 ft = 7.62 m ( 1 ft = 0.3048 m )
Width of the swimming pool = 18.5 ft = 5.64 m
Depth of the pool = 9.0 ft =
Total depth of the water in the pool when filled = 9 ft - 7 inches = 2.56 m
now,
Volume of the water in the pool = Length × Width × Depth
or
Volume of the water in the pool = 7.62 × 5.64 × 2.56 = 110.2292 m³
also,
1 m³ = 1000 L
thus,
110.2292 m³ = 110229.2 L
also it is given that 18 mg of Cl is added to 1 liter of water
therefore,
In 110229.2 L of water Cl added will be = 110229.2 × 18 = 1984125.6 mg
or
= 1.9841256 kg
The net force is 12 N to the left.