Explanation:
The image is not on the same side of the reflecting surface as the outgoing light.
Answer:
25.06s
Explanation:
Remaining part of the question.
(A large stone sphere has a mass of 8200 kg and a radius of 90 cm and floats with nearly zero friction on a thin layer of pressurized water.)
Solution:
F = 60N
r = 90cm = 0.9m
M = 8200kg
Moment of inertia for a sphere (I) = ⅖mr²
I = ⅖ * m * r²
I = ⅖ * 8200 * (0.9)²
I = 0.4 * 8200 * 0.81
I = 2656.8 kgm²
Torque (T) = Iα
but T = Fr
Equating both equations,
Iα = Fr
α = Fr / I
α = (60 * 0.9) / 2656.8
α = 0.020rad/s²
The time it will take her to rotate the sphere,
Θ = w₀t + ½αt²
Angular displacement for one revolution is 2Π rads..
θ = 2π rads
2π = 0 + ½ * 0.02 * t²
(w₀ is equal to zero since sphere is at rest)
2π = ½ * 0.02 * t²
6.284 = 0.01 t²
t² =6.284 / 0.01
t² = 628.4
t = √(628.4)
t = 25.06s
The formula of net Force is:F = mawhere m is the mass of the objecta is the acceleration of the object
thus, if we triple the net force applied to the object:
3F = maa = 3F / m
The acceleration is also tripled since the force is directly proportional to the acceleration.
The greatest percent of mass of the universe is dark matter
Answer:
F= 4788 N
Explanation:
Because the car moves with uniformly accelerated movement we apply the following formula:
vf²=v₀²+2*a*d Formula (1)
Where:
d:displacement in meters (m)
v₀: initial speed in m/s
vf: final speed in m/s
a: acceleration in m/s²
Data
d=36.9 m
v₀=14.0 m/s m/s
vf= 0
Calculating of the acceleration of the car
We replace dta in the formula (1)
vf²=v₀²+2*a*d
(0)²=(14)²+2*a*(36.9)
-(14)²= (73.8) *a
a= - (196) / (73.8)
a= - 2.66 m/s²
Newton's second law of the car in direction horizontal (x):
∑Fx = m*ax Formula (2)
∑F : algebraic sum of the forces in direction x-axis (N)
m : mass (kg)
a : acceleration (m/s²)
Data
m=1800 Fkg
a= - 2.66 m/s²
Magnitude of the horizontal net force (F) that is required to bring the car to a halt in a distance of 36.9 m :
We replace data in the formula (2)
-F= (1800 kg) * ( -2.66 m/s²
)
F= 4788 N
to the ecliptic.