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Free_Kalibri [48]
3 years ago
7

A ball with a mass of 275 g is dropped from rest, hits the floor and rebounds upward. If the ball hits the floor with a speed of

3.30 m/s and rebounds with a speed of 1.60 m/s, determine the following. (a) magnitude of the change in the ball's momentum in kg · m/s (Let up be in the positive direction.)
Physics
1 answer:
pochemuha3 years ago
8 0

Answer:

\Delta p=1.3475\ kg-m/s

Explanation:

The computation of magnitude of the change in the ball's momentum in kg · m/s is shown below:-

We represent

The ball mass =  m = 275 g = 0.275 kg

Thus it goes to the floor and resurfaces upward.

The ball hits the ground at 3.30 m/s speed that is

u = -3.30 m/s which represents the Negative since the ball hits the ground)

It rebounds at a speed of 1.60 m / s i.e. v = 1.60 m/s (positive as the ball rebounds upstream)

\Delta p=p_f-p_i

\Delta p=m(v-u)

\Delta p=0.275\ kg(1.60\ m/s-(-3.30\ m/s))

\Delta p=1.3475\ kg-m/s

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A 5 cm spring is suspended with a mass of 3.8589 g attached to it which extends the spring by 1.5747 cm. The same spring is plac
lana66690 [7]

Answer:

charges of the beads is 1.173 ×10^{-15} C

Explanation:

given data

mass = 3.8589 g = 0.003859 kg

spring length = 5 cm = 0.05 m

extend spring x = 1.5747 cm = 0.15747 m

spring's extension = 0.0116 m

to find out

charges of the beads

solution

we know that force is

force = mass × g

force = 0.003859 × 9.8

force = 0.03782 N

so we know  force for mass

force  = -kx

so k = force / x

put here force and x value

k = -0.03782 / 0.1575

k = -0.24 N/m

and

force for spring's extension

force = -kx

force = -0.24 ( 0.0116) = 0.002784 N

so here

total length L = 0.05 + 0.0116 = 0.0616

so charges of the beads = force × L² / ke

charges of the beads = 0.002784 × (0.0616)² / (9 ×10^{9} )

so charges of the beads = 1.173 ×10^{-15} C

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Triss [41]

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