Answer:
Explanation:
For the cat to stay in place on the merry go round without sliding the magnitude of maximum static friction must be equal to magnitude of centripetal force
Where the r is the radius of merry-go-round and v is the tangential speed
but
So we have
Substitute the given values
So
Answer: 13.2 seconds.
Explanation: using equation of motion; S= ut +1/2at² where u = initial velocity=0
S= distance travelled
a = acceleration due gravity
t= time.
1 foot = 0.305m so,
S= 2860 feet =872.3m
S= ut+1/2 at²
872.3 = 0×t + 1/2×10 × t²
872.3 =0 + 5t²
T²= 872.3/5
T²= 174.46
Take the square root of T we then have;
t = 13.2 seconds to one decimal place.
Answer:
c. gravitational attraction between the sun and earth
C difference in temperature
m = mass of the person = 80 kg
M = mass of earth = 5.98 x 10²⁴ kg
R = radius of earth = 6.37 x 10⁶ m
h = height above the earth's surface = 1400 km = 1.4 x 10⁶ m
r₁ = initial distance of the person from the center of earth when on surface = R = 6.37 x 10⁶ m
r₂ = final distance of the person from the center of earth when at some height = R + h = 6.37 x 10⁶ + 1.4 x 10⁶ = 7.77 x 10⁶ m
F₁ = Gravitational force of earth on the person when at surface
Gravitational force of earth on the person when at surface is given as
F₁ = G M m/r₁² eq-1
F₂ = Gravitational force of earth on the person when at some height
Gravitational force of earth on the person when at some height is given as
F₂ = G M m/r₂² eq-2
dividing eq-1 by eq-2
F₁ /F₂ = (G M m/r₁² )/(G M m/r₂²)
F₁ /F₂ = r₂²/r₁²
inserting the values
F₁ /F₂ = (7.77 x 10⁶)²/(6.37 x 10⁶)²
F₁ /F₂ = 1.49
