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3 years ago

9

A student is given a 5.4 g mineral sample. She determines that the volume of her sample is 2 cm3. What is the density of her min

eral

1 answer:

Xelga [282]3 years ago

7 0

Answer:

6.7

Explanation:

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Four mass–spring systems oscillate in simple harmonic motion. Rank the periods of oscillation for the mass–spring systems from l

liubo4ka [24]

The periods of oscillation for the mass–spring systems from largest to smallest is:

m = 4 kg , k = 2 N/m (T = 8.89 s)
m = 2 kg , k = 2 N/m (T = 6.28 s)
m = 2 kg , k = 4 N/m (T = 4.44 s)
m = 1 kg , k = 4 N/m (T = 3.14 s)

<h3>Explanation:</h3>

The period of oscillation in a simple harmonic motion is defined as the following formulation:

$T=2\pi \sqrt{\frac{m}{k} }$

Where:

T = period of oscillation

m = inertia mass of the oscillating body

k = spring constant

m = 2 kg , k = 2 N/m

$T=2\pi \sqrt{\frac{2}{2} }$

$T=2\pi$

T = 6.28 s

m = 2 kg , k = 4 N/m

$T=2\pi \sqrt{\frac{2}{4} }$

$T=2\pi \sqrt{\frac{1}{2} }$

T = 4.44 s

m = 4 kg , k = 2 N/m

$T=2\pi \sqrt{\frac{4}{2} }$

$T=2\pi \sqrt{2 }$

T = 8.89 s

m = 1 kg , k = 4 N/m

$T=2\pi \sqrt{\frac{1}{4} }$

$T=\pi$

T = 3.14 s

Therefore the rank the periods of oscillation for the mass–spring systems from largest to smallest is:

m = 4 kg , k = 2 N/m (T = 8.89 s)
m = 2 kg , k = 2 N/m (T = 6.28 s)
m = 2 kg , k = 4 N/m (T = 4.44 s)
m = 1 kg , k = 4 N/m (T = 3.14 s)

Learn more about simple harmonic motion brainly.com/question/13058166

#LearnWithBrainly

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A variable is defined as anything that

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Answer : Quantity or quality that varies like manipulating a variable to see what will happen to another variable

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3 years ago

Rupert (66 kg) is now in a full-pipe of radius 25 m, calculate the minimum speed at which Rupert can skate to ensure he will not

V125BC [204]

at the top most point if Rupert will not fall then normal force at the top point is almost zero for minimum speed

so here we can say

$F_n + mg = m\frac{v^2}{R}$

now if

$F_n = 0$

$mg = \frac{mv^2}{R}$

$v = \sqrt{Rg}$

$v = \sqrt{25 \times 9.8}$

$v = 15.65 m/s$

so above will be the minimum speed

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3 years ago

The greatest height reported for a jump into an airbag is 99.4 m by stuntman Dan Koko. In 1948 he jumped from rest from the top

leva [86]

Answer:

44.1613858478 m/s

Explanation:

t = Time taken

u = Initial velocity = 0

v = Final velocity

s = Displacement = 99.4

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s² = a

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 99.4+0^2}\\\Rightarrow v=44.1613858478\ m/s$

If air resistance was absent Dan Koko would strike the airbag at 44.1613858478 m/s

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during a dodge ball game a student throw a ball at another player what forces act on the ball as it flie through the air

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Gravity acts to accelerate the ball downward, and air resistance acts in a way to slow the ball along it's instantaneous velocity (no matter which way it's moving air applies a force in the opposite direction)

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4 years ago