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3 years ago

11

What's the difference between mass and inertia in a tabular form

1 answer:

Mila [183]3 years ago

6 0

Answer:

to be honest i dont know

Explanation:

^^

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If a 2,000-kg car hits a tree with 500 n of force over a time of 0.5 seconds, what is the magnitude of its impulse?

Elden [556K]

Impulse = Force * time
Impulse = 500N *0.5 s =250 N*s

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3 years ago

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3) If a ball launched at an angle of 10.0 degrees above horizontal from an initial height of 1.50 meters has a final horizontal

Irina-Kira [14]

Answer:

35.6 m

Explanation:

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3 years ago

Light waves can be easily blocked but ______ waves pass through all substances? ( fill in the blank)

IRINA_888 [86]

Compressional waves can travel through all states of matter.

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3 years ago

A mass of 100 g stretches a spring 5 cm. If the mass is set in motion from its equilibrium position with a downward velocity of

Rudiy27

Answer:

Explanation:

Given

mass of spring $m=100\ gm$

extension in spring $x=5\ cm$

downward velocity $v=70\ cm/s$

Position in undamped free vibration is given by

$u(t)=A\cos \omega _0t+B\sin \omega _0t$

where $\omega _0^2=\frac{k}{m}$

also $\frac{k}{m}=\frac{g}{L}$

$\omega _0^2=\frac{k}{m}=\frac{9.8}{0.05}$

$\omega _0=14$

$u(t)=A\cos(14t)+B\sin(14t)$

it is given

$u(0)=0$

$u'(0)=70\ cm/s$

substituting values we get

$A=0$

$u(t)=B\sin (14t)$

$u'(t)=14B\cos (14t)$

$70=14B$

$B=\frac{10}{2}$

$B=5$

$u(t)=5\sin (14t)$

3 0

3 years ago

If the distance between the Earth and Moon were half what it is now, by what factor would the force of gravity between them be c

hichkok12 [17]

Answer:

4

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

$m_1$ = Mass of Earth

$m_2$ = Mass of Moon

r = Distance between Earth and Moon

Old gravitational force

$F_o=\dfrac{Gm_1m_2}{r^2}$

New gravitational force

$F_n=\dfrac{Gm_1m_2}{(\dfrac{1}{2}r)^2}$

Dividing the equations

$\dfrac{F_n}{F_o}=\dfrac{\dfrac{Gm_1m_2}{(\dfrac{1}{2}r)^2}}{\dfrac{Gm_1m_2}{r^2}}\\\Rightarrow \dfrac{F_n}{F_o}=\dfrac{\dfrac{Gm_1m_2}{\dfrac{1}{4}r^2}}{\dfrac{Gm_1m_2}{r^2}}\\\Rightarrow \dfrac{F_n}{F_o}=4$

The ratio is $\dfrac{F_n}{F_o}=4$

The new force would be 4 times the old force

7 0

3 years ago