Answer:
Newton
Explanation:
The SI unit of force is the newton, symbol N. The base units relevant to force are: The metre, unit of length — symbol m. The kilogram, unit of mass — symbol kg.
The spring is initially stretched, and the mass released from rest (v=0). The next time the speed becomes zero again is when the spring is fully compressed, and the mass is on the opposite side of the spring with respect to its equilibrium position, after a time t=0.100 s. This corresponds to half oscillation of the system. Therefore, the period of a full oscillation of the system is

Which means that the frequency is

and the angular frequency is

In a spring-mass system, the maximum velocity of the object is given by

where A is the amplitude of the oscillation. In our problem, the amplitude of the motion corresponds to the initial displacement of the object (A=0.500 m), therefore the maximum velocity is
Answer:
1.8 s
Explanation:
Potential energy = kinetic energy + rotational energy
mgh = ½ mv² + ½ Iω²
For a thin spherical shell, I = ⅔ mr².
mgh = ½ mv² + ½ (⅔ mr²) ω²
mgh = ½ mv² + ⅓ mr²ω²
For rolling without slipping, v = ωr.
mgh = ½ mv² + ⅓ mv²
mgh = ⅚ mv²
gh = ⅚ v²
v = √(1.2gh)
v = √(1.2 × 9.81 m/s² × 4.8 m sin 39.4°)
v = 5.47 m/s
The acceleration down the incline is constant, so given:
Δx = 4.8 m
v₀ = 0 m/s
v = 5.47 m/s
Find: t
Δx = ½ (v + v₀) t
t = 2Δx / (v + v₀)
t = 2 (4.8 m) / (5.47 m/s + 0 m/s)
t = 1.76 s
Rounding to two significant figures, it takes 1.8 seconds.
Answer:
μ=0.151
Explanation:
Given that
m= 3.5 Kg
d= 0.96 m
F= 22 N
v= 1.36 m/s
Lets take coefficient of kinetic friction = μ
Friction force Fr=μ m g
Lets take acceleration of block is a m/s²
F- Fr = m a
22 - μ x 3.5 x 10 = 3.5 a ( take g =10 m/s²)
a= 6.28 - 35μ m/s²
The final speed of the block is v
v= 1.36 m/s
We know that
v²= u²+ 2 a d
u= 0 m/s given that
1.36² = 2 x a x 0.96
a= 0.963 m/s²
a= 6.28 - 35μ m/s²
6.28 - 35μ = 0.963
μ=0.151