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Natalija [7]
3 years ago
12

It is weigh-in time for the local under 85 kg rugby team. The bathroom scale used to assess eligibility can be described by Hook

e's law, which is depressed 0.75 cm for its maximum load of 115 kg. What is the spring's effective spring constant?
Physics
1 answer:
Grace [21]3 years ago
4 0
15 m/n is the answer
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The imaginary line from the tip of the football to the football to the sideline is called
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The imaginary line from the tip of the football to the football to the sideline is called Line of Scrimmage. I believe.
8 0
4 years ago
Please help me!!!
dezoksy [38]
<h2><em>The Answer to Youre Question:</em></h2>

<em>I think it is B.</em>

<em>I hope this helps.</em>

7 0
3 years ago
A block with a mass of 0.600 kg is connected to a spring, displaced in the positive direction a distance of 50.0 cm from equilib
tankabanditka [31]

Answer:

Explanation:

The amplitude of the oscillation under SHM will be .5 m and the equation of

SHM can be written as follows

x = .5 sin(ωt + π/2) , here the initial phase is π/2 because when t = 0 , x = A ( amplitude) , ω is angular frequency.

x = .5 cosωt

given , when t = .2 s , x = .35 m

.35 = .5 cos ωt

ωt = .79

ω = .79 / .20

= 3.95 rad /s

period of oscillation

T = 2π / ω

= 2 x 3.14 / 3.95

= 1.6 s

b )

ω = \sqrt{\frac{k}{m} }

ω² = k / m

k = ω² x m

= 3.95² x .6

= 9.36 N/s

c )

v = ω\sqrt{(a^2-x^2)}

At t = .2 , x = .35

v = 3.95 \sqrt{.5^2-.35^2}

= 3.95 x .357

= 1.41 m/ s

d )

Acceleration at x

a = ω² x

= 3.95 x .35

= 1.3825 m s⁻²

7 0
3 years ago
15
PtichkaEL [24]
<h3>Option A can be correct because when the meat is kept the rump settles down and the meat is prevented from being messy everywhere.</h3>
4 0
3 years ago
On a spending spree in Malaysia, you buy an ox with a weight of 28.9 piculs in the local unit of weights: 1 picul = 100 gins, 1
Talja [164]

Answer:

1747.41 kg

Explanation:

If          1 piculs = 100 gins,

Then     28.9 piculs = (28.9 × 100) gins = 2890 gins.

Also,

If         1 gin = 16 tahils

then   2890 gins =( 16 × 2890 ) = 46240 tahils

Also,

If         1 tahil = 10 chees,

Then    46240 tahils = 46240 × 10 chees = 462400 chees.

Also,

If           1 chee = 10 hoons

Then      462400 chees =( 10 × 462400) hoons = 4624000 hoons.

Also,

If            1 hoon = 0.3779 g,

Then     4624000 hoons = 0.3779 × 4624000 = 1747409.6 g.

The mass in kg that i will declare in the shipping manifest = (1747409.6/100 )kg = 1747.41 kg

6 0
3 years ago
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