The major shortcoming of Rutherford's model was that it was incomplete. It did not explain how the atom's negatively charged electrons are distrubuted in the space surronding its positively charged nucleus. A form of energy that exhibits wavelike behavior as it travels through space
Answer:
a) 39.6 m/s b) 4123 N
Explanation:
a) At the top of the loop, all of the forces point downwards (force of gravity and normal force).
Fnet=ma
ma=m(v^2/R) (centripetal acceleration)
mg=m(v^2/R)
m cancels out (this is why pilot feels weightless) so,
g=(v^2/R)
9.8 m/s^2 = v^2/160 m
v^2=1568 m^2/s^2
v=39.6 m/s
b) At the bottom of the loop, the normal force and the force of gravity point in opposite directions. The normal force is the weight felt.
Convert 300 km/hr to m/s
300 km/hr=83.3 m/s
Convert pilot's weight into mass:
760 N = 77.55 kg
Fnet=ma
n-mg=m(v^2/R)
n=(77.55 kg)(((83.3 m/s)^2)/160 m)+(77.55 kg)(9.8 m/s^2)
n=3363.2 N+760 N=4123 N
Answer: v = 4.4 m/s
Explanation:
In the absence of friction, the total mechanical energy will be constant
KE₀ + PE₀ = KE₁ + PE₁
0 + mg(6) = ½mv₁² + mg(5)
½mv₁² = mg(6 - 5)
v = √(2g(1)) = 4.4 m/s
The change in Potential energy of the cat is 176.4 J.
<h3 /><h3>Potential Energy:</h3>
This is the energy due to the position of a body. The S.I unit is Joules (J)
The formula for change in potential energy.
<h3 /><h3>Formula:</h3>
- ΔP.E = mg(H-h).............. Equation 1
<h3>Where:</h3>
- ΔP.E = Change in potential energy
- m = mass of the cat
- g = acceleration due to gravity
- H = First height
- h = second height.
From the question,
<h3>Given:</h3>
- m = 15 kg
- H = 2.5 m
- h = 1.3 m
- g = 9.8 m/s²
Substitute these values into equation 1
- ΔP.E = 15×9.8(2.5-1.3)
- ΔP.E = 15×9.8×1.2
- ΔP.E = 176.4 J.
Hence, The change in Potential energy of the cat is 176.4 J
Learn more about Potential energy here: brainly.com/question/1242059
