Answer:
v = 1.28 m/s
Explanation:
Given that,
Maximum compression of the spring, 
Spring constant, k = 800 N/m
Mass of the block, m = 0.2 kg
To find,
The velocity of the block when it first reaches a height of 0.1 m above the ground on the ramp.
Solution,
When the block is bounced back up the ramp, the total energy of the system remains conserved. Let v is the velocity of the block such that,
Initial energy = Final energy

Substituting all the values in above equation,

v = 1.28 m/s
Therefore the velocity of block when it first reaches a height of 0.1 m above the ground on the ramp is 1.28 m/s.
Answer:
Part a)
Mass of m2 is given as

Part b)
Angular acceleration is given as

Part c)
Tension in the rope is given as

Explanation:
Part a)
When m1 and m2 both connected to the cylinder then the system is at rest
so we can use torque balance here




Part b)
When block m_2 is removed then system becomes unstable
so force equation of mass m1

also we have

now we have




so angular acceleration is given as



Part c)
Tension in the rope is given as



Answer:
True
Explanation:
An electric field is a region around a charged particle or object within which a force would be exerted on other charged particles or objects.
The answer is A. <span>The component waves have different frequencies.
The magnitudes of reinforcement usually really dependent on the number of frequencies and interference is usually caused due to the difference in frequencies. So, we can conclude that if the frequencies are different and causing interference, the reinforcement will also different
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Answer:
Explanation:
The mass of the block is 0.5kg
m = 0.5kg.
The spring constant is 50N/m
k =50N/m.
When the spring is stretch to 0.3m
e=0.3m
The spring oscillates from -0.3 to 0.3m
Therefore, amplitude is A=0.3m
Magnitude of acceleration and the direction of the force
The angular frequency (ω) is given as
ω = √(k/m)
ω = √(50/0.5)
ω = √100
ω = 10rad/s
The acceleration of a SHM is given as
a = -ω²A
a = -10²×0.3
a = -30m/s²
Since we need the magnitude of the acceleration,
Then, a = 30m/s²
To know the direction of net force let apply newtons second law
ΣFnet = ma
Fnet = 0.5 × -30
Fnet = -15N
Fnet = -15•i N
The net force is directed to the negative direction of the x -axis