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Zielflug [23.3K]
2 years ago
13

(Please help 15 points the assignment is late) (multiple questions) (everything is virtual task )

Physics
1 answer:
krok68 [10]2 years ago
5 0

Answer:

i & 2

Explanation:

U alr finished it so there's no point

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If the pendulum is brought on the moon where the gravitational acceleration is about g/6, approximately what will its period now
Andru [333]

Answer:

The new period will be √6 *T

Explanation:

period ,T=2π√(L/g)       ................equation 1

where T is the period on earth

gravitational acceleration on the moon is g/6

T1 = 2π√[L/(g/6)]

T1=2π√(6L/g)               ...............equation 2

divide equation 2 by 1

T1/T =2π√(6L/g)÷2π√(L/g)

T1/T =√(6L/L)

T1/T =√6

T1 = √6 *T

5 0
3 years ago
Write down the boiling point and freezing point of mercury an alcohol​
nekit [7.7K]
Mercury has a high boiling point of 357 degrees C.
Mercury has a freezing point of −39 degrees C.
8 0
3 years ago
What accounts for most of the mass in the universe?
aev [14]
Around 80 percent  of the mass of the universe is made up material known as "Dark matter". It does not emit light or energy but the influence of it can be detected or observed gravitationally. Motions of stars and galaxy tell us how much mater there is, but somehow the speed of rotation of galaxy does not add up to its mass alone, there is a certain amount of matter really not accounted for. Dark matter maybe made up of non-baryonic matter, or perhaps what scientist called the WIMPS or (weakly interacting massive particles.)
5 0
3 years ago
Calculate the force it would take to accelerate a 50 kg bike at a rate of 3 m/s2.
Ede4ka [16]

ummmm it might be 300... i used a calculator

sorry if it is wrong

4 0
3 years ago
A uniform solid sphere has a moment of inertia I about an axis tangent to its surface. What is the moment of inertia of this sph
arsen [322]

Answer:

option E

Explanation:

given,

I is moment of inertia about an axis tangent to its surface.

moment of inertia about the center of mass

I_{CM} = \dfrac{2}{5}mR^2.....(1)

now, moment of inertia about tangent

I= \dfrac{2}{5}mR^2 + mR^2

I= \dfrac{7}{5}mR^2...........(2)

dividing equation (1)/(2)

\dfrac{I_{CM}}{I}= \dfrac{\dfrac{2}{5}mR^2}{\dfrac{7}{5}mR^2}

\dfrac{I_{CM}}{I}=\dfrac{2}{7}

I_{CM}=\dfrac{2}{7}I

the correct answer is option E

4 0
3 years ago
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