The observer can conclude that the sound is moving away from them and that its speed is increasing.
Kinetic energy is greatest at the lowest point of a roller coaster and least at the highest point
Answer:
-6327.45 Joules
650.375 Joules
378.47166 N
Explanation:
h = Height the bear slides from = 15 m
m = Mass of bear = 43 kg
g = Acceleration due to gravity = 9.81 m/s²
v = Velocity of bear = 5.5 m/s
f = Frictional force
Potential energy is given by
![P=mgh\\\Rightarrow P=43\times -9.81\times 15\\\Rightarrow P=-6327.45\ J](https://tex.z-dn.net/?f=P%3Dmgh%5C%5C%5CRightarrow%20P%3D43%5Ctimes%20-9.81%5Ctimes%2015%5C%5C%5CRightarrow%20P%3D-6327.45%5C%20J)
Change that occurs in the gravitational potential energy of the bear-Earth system during the slide is -6327.45 Joules
Kinetic energy is given by
![K=\frac{1}{2}mv^2\\\Rightarrow K=\frac{1}{2}\times 43\times 5.5^2\\\Rightarrow K=650.375\ J](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B1%7D%7B2%7Dmv%5E2%5C%5C%5CRightarrow%20K%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%2043%5Ctimes%205.5%5E2%5C%5C%5CRightarrow%20K%3D650.375%5C%20J)
Kinetic energy of the bear just before hitting the ground is 650.375 Joules
Change in total energy is given by
![\Delta E=fh=-(\Delta K+\Delta P)\\\Rightarrow fh=-(650.375-6327.45)\\\Rightarrow fh=5677.075\\\Rightarrow f=\frac{5677.075}{h}\\\Rightarrow f=\frac{5677.075}{15}\\\Rightarrow f=378.47166\ N](https://tex.z-dn.net/?f=%5CDelta%20E%3Dfh%3D-%28%5CDelta%20K%2B%5CDelta%20P%29%5C%5C%5CRightarrow%20fh%3D-%28650.375-6327.45%29%5C%5C%5CRightarrow%20fh%3D5677.075%5C%5C%5CRightarrow%20f%3D%5Cfrac%7B5677.075%7D%7Bh%7D%5C%5C%5CRightarrow%20f%3D%5Cfrac%7B5677.075%7D%7B15%7D%5C%5C%5CRightarrow%20f%3D378.47166%5C%20N)
The frictional force that acts on the sliding bear is 378.47166 N
Answer:
a) vB = 10.77 ft/s
b) W = 11.30 lb*ft
Explanation:
a) W = 8 lb ⇒ m = W/g = 8 lb/32.2 ft/s² = 0.2484 slug
vA <em>lin</em> = 5 ft/s
rA = 2 ft
v <em>rad</em> = 4 ft/s
vB = ?
rB = 1 ft
W = ?
We can apply The law of conservation of angular momentum
L<em>in</em> = L<em>fin</em>
m*vA*rA = m*vB*rB ⇒ vB = vA*rA / rB
⇒ vB = (5 ft/s)*(2 ft) / (1 ft) = 10 ft/s (tangential speed)
then we get
vB = √(vB tang² + vB rad²) ⇒ vB = √((10 ft/s)² + (4 ft/s)²)
⇒ vB = 10.77 ft/s
b) W = ΔK = K<em>B</em> - K<em>A</em> = 0.5*m*vB² - 0.5*m*vA²
⇒ W = 0.5*m*(vB² - vA²) = 0.5*0.2484 slug*((10.77 ft/s)²-(5 ft/s)²)
⇒ W = 11.30 lb*ft
Answer:
orbital speed of the electrons in their orbit will increase
Explanation:
As we know that centripetal force for electrons will be due to electrostatic attraction force of electron.
So it is given as
![F_e = F_c](https://tex.z-dn.net/?f=F_e%20%3D%20F_c)
so we have
![\frac{(Ze)(e)}{4\pi \epsilon_0 r^2} = \frac{mv^2}{r}](https://tex.z-dn.net/?f=%5Cfrac%7B%28Ze%29%28e%29%7D%7B4%5Cpi%20%5Cepsilon_0%20r%5E2%7D%20%3D%20%5Cfrac%7Bmv%5E2%7D%7Br%7D)
now on the left side if the force of attraction will increase and hence there must be the change in that part of equation
So here at the same position the speed of the electron
So we can say that correct answer will be
orbital speed of the electrons in their orbit will increase