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fenix001 [56]
3 years ago
9

A block of mass 0.1 kg is attached to a spring of spring constant 22 N/m on a frictionless track. The block moves in simple harm

onic motion with amplitude 0.29 m. While passing through the equilibrium point from left to right, the block is struck by a bullet, which stops inside the block. The velocity of the bullet immediately before it strikes the block is 41 m/s and the mass of the bullet is 1.23 g.
A) Find the speed of the block immediately before the collision.Answer in units of m/sB) If the simple harmonic motion after the collision is described by x = B sin(? t + f), what is the new amplitude B?Answer in units of mC) The collision occurred at the equilibrium position.How long will it take for the block to reach maximum amplitude after the collision?Answer in units of s
Physics
1 answer:
Tasya [4]3 years ago
6 0

Answer:

(A) v = 14.8m/s

Explanation:

(A) V = sqrt(k/m) × A = sqrt(22/0.1) × 0.29 =14.8m/s.

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5 0
3 years ago
Help please<br> It’s kinda urgent
user100 [1]

Answer:

a = - 50 [m/s²]

Explanation:

To solve this problem we simply have to replace the values supplied in the given equation.

Vf = final velocity = 0.5 [m/s]

Vi = initial velocity = 10 [m/s]

s = distance = 100 [m]

a = acceleration [m/s²]

Now replacing we have:

(0.5)^{2}-(10)^{2} = 2*a*(100)\\0.25-10000=200*a\\200*a=-9999.75\\a =-50 [m/s^{2} ]

The negative sign of acceleration means that the ship slows down its velocity in order to land.

4 0
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When the saw slices wood, the wood exerts a 104-N force on the blade, 0.128 m from the blade’s axis of rotation. If that force i
Soloha48 [4]

Answer:

Explanation:

When saw slices wood by exerting a force on the wood , wood also exerts a reaction force on the saw in opposite direction which is equal to the force of action that is 104 N.

So torque exerted by wood on the blade

= force x perpendicular distance from the axis of rotation

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Since this torque opposes the movement of blade , it turns the blade slower.

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