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pickupchik [31]
2 years ago
15

4cm cubed = ?m cubed

Physics
1 answer:
Neporo4naja [7]2 years ago
5 0

Answer:

{ \rm{1 \:  {cm}^{3} =  \frac{1}{100 {}^{3} }  \:  {m}^{3}  }} \\  \\  { \rm{4 \:  {cm}^{3}  =  \frac{4}{1000000}  \:  {m}^{3} }} \\  \\ { \boxed{ \rm{ = 0.000004 \:  {m}^{3} }}}

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A line _____ on a typical Speed vs. Time graph means an object is experiencing a constant acceleration.
Savatey [412]

Answer:

The answer should be C. slanted upward to the right.

Hope this helps. :-)

7 0
3 years ago
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A generator produces 60 A of current at 120 V. The voltage is usually stepped up to 4500 V by a transformer and transmitted thro
aalyn [17]

Answer:

The percentage power lost in the transmission line if the voltage not stepped up is 50%.

Explanation:

Given that,

Current = 60 A

Voltage = 120 V

Resistance = 1.0 ohm

We need to calculate the power

Using formula of power

P=I\times V

Where,I =current

V = voltage

Put the value into the formula

P=60\times120

P=7200\ W

We need to calculate the percentage power lost in the transmission line

If the voltage is not stepped up

Then, the power loss

P'=I^2\times R

Put the value into the formula

P'=(60)^2\times1

P'=3600\ W

The percentage power loss P''

P''=\dfrac{P'}{P}\times100=\dfrac{3600}{7200}\times100

P''=50\%

Hence, The percentage power lost in the transmission line if the voltage not stepped up is 50%.

5 0
3 years ago
Why does sound travel slower at a cold temperature?
tatyana61 [14]
<h2>Answer:</h2><h2>It is due to a refractment of light.</h2>

Sound moves faster in warmer air than colder air the way bends away from the warm air and back towards of air.

7 0
3 years ago
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Two identical cows fell into a muddy hole. One fell on its side and the other fell on its feet.Which one sank furthest into the
olga_2 [115]

Answer:

The one which falls on feet.

Explanation:

Because it exerts higher pressure as

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8 0
3 years ago
A projectile is thrown with velocity v at an angle θ with horizontal. When the projectile is at a height equal to half of the ma
raketka [301]

  • Let, the maximum height covered by projectile be \sf{H_m}

\purple{ \longrightarrow  \bf{h_m =  \dfrac{ {v}^{2} \: {sin}^{2} \theta  }{2g} }}

  • Projectile is thrown with a velocity = v
  • Angle of projection = θ

  • Velocity of projectile at a height half of the maximum height covered be \sf{v_0}

\qquad______________________________

Then –

\qquad \pink{  \longrightarrow \bf{ \dfrac{h_m}{2}  = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta  }{2g} }}

\qquad \longrightarrow \sf{ \dfrac{ {v}^{2}  \: {sin}^{2} \theta  }{2g} \times  \dfrac{1}{2}  =  \dfrac{ {v_0}^{2} \: {sin}^{2} \theta  }{2g} }

\qquad\longrightarrow  \sf{ \dfrac{ {v}^{2}  \: {sin}^{2} \theta  }{4g}  =  \dfrac{ {v_0}^{2} \: {sin}^{2} \theta  }{2g} }

\qquad\longrightarrow  \sf{ \dfrac{ {v}^{2}  \: {sin}^{2} \theta  }{2}  =   {v_0}^{2} \: {sin}^{2} \theta }

\qquad\longrightarrow  \sf{ \dfrac{ {v}^{2} }{2}  =   {v_0}^{2} }

\qquad\longrightarrow \bf{v_0 =   \sqrt{ \dfrac{ {v}^{2} }{2} } =  \dfrac{v}{ \sqrt{2} }  }

  • Now, the vertical component of velocity of projectile at the height half of \sf{h_m} will be –

\qquad \longrightarrow   \bf{v_{(y)}=v_0 \: sin \theta }

\qquad \longrightarrow \bf{v_{(y)} = \dfrac{v}{ \sqrt{2} }  \: sin \theta =  \dfrac{v \: sin \: \theta}{ \sqrt{2} }  }

Therefore, the vertical component of velocity of projectile at this height will be–

☀️\qquad\pink {\bf{ \dfrac{v \: sin \:  \theta}{ \sqrt{2} }} }

6 0
2 years ago
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