Answer:
The minimum possible coefficient of static friction between the tires and the ground is 0.64.
Explanation:
if the μ is the coefficient of static friction and R is radius of the curve and v is the speed of the car then, one thing we know is that along the curve, the frictional force, f will be equal to the centripedal force, Fc and this relation is :
Fc = f
m×(v^2)/(R) = μ×m×g
(v^2)/(R) = g×μ
μ = (v^2)/(R×g)
= ((25)^2)/((100)×(9.8))
= 0.64
Therefore, the minimum possible coefficient of static friction between the tires and the ground is 0.64.
First example: book, m= 0.75 kg, h=1.5 m, g= 9.8 m/s², it has only potential energy Ep,
Ep=m*g*h=0.75*9.8*1.5=11.025 J
Second example: brick, m=2.5 kg, v=10 m/s, h=4 m, it has potential energy Ep and kinetic energy Ek,
E=Ep+Ek=m*g*h + (1/2)*m*v²=98 J + 125 J= 223 J
Third example: ball, m=0.25 kg, v= 10 m/s, it has only kinetic energy Ek
Ek=(1/2)*m*v²=12.5 J.
Fourth example: stone, m=0.7 kg, h=7 m, it has only potential energy Ep,
Ep=m*g*h=0.7*9.8*7=48.02 J
The order of examples starting with the lowest energy:
1. book, 2. ball, 3. stone, 4. brick
Answer:
Explanation:
<u>Given Data:</u>
Mass = m = 4 kg
Acceleration due to gravity = g = 9.8 m/s²
Height = h = 1 m
<u>Required:</u>
Potential Energy = P.E. = ?
<u>Formula:</u>
P.E. = mgh
<u>Solution:</u>
P.E. = (4)(9.8)(1)
P.E. = 39.2 Joules
![\rule[225]{225}{2}](https://tex.z-dn.net/?f=%5Crule%5B225%5D%7B225%7D%7B2%7D)
Hope this helped!
<h3>~AH1807</h3>
I don’t even know I’m so dumb.
Answer:
Sound waves are pushed closer together, decreasing wavelength
and increasing frequency.
