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Fed [463]
3 years ago
7

answer A vertical spring stretches 3.4 cm when a 8-g object is hung from it. The object is replaced with a block of mass 26 g th

at oscillates in simple harmonic motion. Calculate the period of motion.
Physics
2 answers:
victus00 [196]3 years ago
8 0

Answer:

0.695s

Explanation:

From Hooke's law, the restoring force is given has

F = -ky .......1

Where F is the force, y is the spring displacement and k force constant of the spring.

Also recall,

F=mg ............ 2

Where m is the mass of object, g is the acceleration due to gravity.

Equating 1 and 2

Ky = mg

Given that g=9.8m/s2 , y is 3.4cm and g is 8g

K×3.4/100m =8/1000kg × 9.8m/s2

K= ( 0.008kg × 9.8m/s2 ) ÷ 0.034

K= 0.0784÷0.035

K=2.24N/m

Mass ofvthe second object is 25g =0.025kg

Period of oscillation T

T=2π√m/k

T=2×3.142√0.025/2.24

T=6.284√0.0111

T=0.659seconds

notsponge [240]3 years ago
8 0

Answer:

df,jgukilhugyftdrswjdfggfjgjfjgfthf

Explanation:

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11). Determine the acceleration due to gravity if a 6kg book was on top of
DerKrebs [107]

Answer:

g = 9.8 [m/s²]

Explanation:

To solve this problem we must remember that the potential energy is defined as the product of mass by gravitational acceleration by the height or elevation with respect to a reference level.

Ep = m*g*h

where:

Ep = potential energy = 235.2 [J]

m = mass of the book = 6 [kg]

g = gravity acceleration [m/s²]

h = elevation = 4 [m]

Now replacing these values:

235.2 = 6*g*4

g = 235.2/(24)

g = 9.8 [m/s²]

6 0
3 years ago
(10. Five pupils have a mean mass of 51 kg.
sveticcg [70]

Explanation:

If the fourth pupil has a mass of x, and the fifth pupil has a mass of x+5, then the mean mass is:

51 = (49 + 53 + 52 + x + x+5) / 5

255 = 159 + 2x

x = 48

x + 5 = 53

The fourth pupil has a mass of 48 kg.

The fifth pupil has a mass of 53 kg.

3 0
3 years ago
A mercury manometer is connected on one side to a bulb containing argon, while the other end is open to atmospheric pressure, wh
belka [17]

Answer:

740, mm of Hg

Explanation:

The pressure of argon , in mm of Hg = difference in the level of mercury on either side of the manometer.

mercury column  in the open end of the manometer is 22 mm below that in the side connected to the argon and 762 mmHg,  end is open to atmospheric pressure.

therefore, The pressure of argon , in mm of Hg  =762 -22 = 740, mm of Hg

4 0
3 years ago
Please help on this one
Murljashka [212]
Just find the area of the graft

4m/s x 5s
=20m

7 0
3 years ago
Read 2 more answers
In a drag race, a car takes off from rest and covers 375 meters in 5 seconds. What is the
salantis [7]

Explanation:

s = ut + 1/2 a t^2

375 = 0 * 5 + 1/2 * a * (5)^2

375 = 1/2 * a * 25

a = 375*2/25

a = 15* 2

a = 30m/sec^2

v = u + at

v = 0 + 30 * 5

v = 150 m/sec

hope it helps you

3 0
3 years ago
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