Answer:
none of the above
Explanation:
The actual answer is '91 protons'. In fact, the beta decay of the thorium-234 is the following:

where inside the nucleus of Thorium (90 protons), a neutron turns into an electron (the beta particle) + a proton. Therefore, the resulting nucleus (which is Protoactinium) has a total of 90+1 = 91 protons.
So, the correct answer would be '91 protons'.
Answer:
9. 45 N North East
Unbalanced
Accelerate
10. 45 N right and 6 N down
Unbalanced
Accelerate
Explanation:
Hope this helps!
The volume of the balloon is given by:
V = 4πr³/3
V = volume, r = radius
Differentiate both sides with respect to time t:
dV/dt = 4πr²(dr/dt)
Isolate dr/dt:
dr/dt = (dV/dt)/(4πr²)
Given values:
dV/dt = 72ft³/min
r = 3ft
Plug in and solve for dr/dt:
dr/dt = 72/(4π(3)²)
dr/dt = 0.64ft/min
The radius is increasing at a rate of 0.64ft/min
The surface area of the balloon is given by:
A = 4πr²
A = surface area, r = radius
Differentiate both sides with respect to time t:
dA/dt = 8πr(dr/dt)
Given values:
r = 3ft
dr/dt = 0.64ft/min
Plug in and solve for dA/dt:
dA/dt = 8π(3)(0.64)
dA/dt = 48.25ft²/min
The surface area is changing at a rate of 48.25ft²/min
Answer:
N₁ = 393.96 N and N = 197.96 N
Explanation:
In This exercise we must use Newton's second law to find the normal force. Let's use two points the lowest and the highest of the loop
Lowest point, we write Newton's second law n for the y-axis
N -W = m a
where the acceleration is ccentripeta
a = v² / r
N = W + m v² / r
N = mg + mv² / r
we can use energy to find the speed at the bottom of the circle
starting point. Highest point where the ball is released
Em₀ = U = m g h
lowest point. Stop curl down
= K = ½ m v²
Emo = Em_{f}
m g h = ½ m v²
v² = 2 gh
we substitute
N = m (g + 2gh / r)
N = mg (1 + 2h / r)
let's calculate
N₁ = 5 9.8 (1 + 2 17.6 / 5)
N₁ = 393.96 N
headed up
we repeat the calculation in the longest part of the loop
-N -W = - m v₂² / r
N = m v₂² / r - W
N = m (v₂²/r - g)
we seek speed with the conservation of energy
Em₀ = U = m g h
final point. Top of circle with height 2r
= K + U = ½ m v₂² + mg (2r)
Em₀ = Em_{f}
mgh = ½ m v₂² + 2mgr
v₂² = 2 g (h-2r)
we substitute
N = m (2g (h-2r) / r - g)
N = mg (2 (h-r) / r 1) = mg (2h/r -2 -1)
N = mg (2h/r - 3)
N = 5 9.8 (2 17.6 / 5 -3)
N = 197.96 N
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