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3 years ago

All objects _ energy.

Physics

1 answer:

Alekssandra [29.7K]3 years ago

8 0

Answer:

Explanation:

All objects <u>radiate</u> energy.

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You threw a ball straight up into the air at 7.34m/s. How high does the ball go?

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Answer:

7.34x

Explanation:

Depends on how long it was in the air. 7.34x, x is how many seconds.

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4 years ago

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The importance of the Miller–Urey experiment is that (a) it proved beyond doubt that life could have arisen naturally on the you

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Explanation:

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3 years ago

A satellite is in a circular orbit around the Earth at an altitude of 2.80 3 106 m. Find (a) the period of the orbit, (b) the sp

Katen [24]

Explanation:

Given that,

Altitude $h= 2.803\times10^{6}\ m$

We need to calculate the radius

$r=R+h$

Where, R = radius of the earth

h = radius of altitude

Put the value into the formula

$r=(6.38\times10^{6}+2.803\times10^{6})$

$r=9.18\times10^{6}\ m$

(a). We need to calculate the period of the orbit,

Using formula of period

$T^2=\dfrac{4\pi^2}{GM}r^3$

$T^2=\dfrac{4\pi^2}{6.67\times10^{-11}\times5.98\times10^{24}}\times(9.18\times10^{6})^3$

$T^2=76570372.9509\ sec$

$T=8750.45\ sec$

(b). We need to calculate the speed of the satellite

Using formula of speed

$v^2=\dfrac{GM}{r}$

Put the value into the formula

$v^2=\dfrac{6.67\times10^{-11}\times5.98\times10^{24}}{9.18\times10^{6}}$

$v^2=43449455.3377\ m/s$

$v=6.59\times10^{3}\ m/s$

(c). We need to calculate the acceleration of the satellite

Using formula of acceleration

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4 years ago

How much heat is required to change 200 mL of ice at -22°C (at typical Freezer temperature) into steam

Katena32 [7]

Answer:

611.064 kJ

Explanation:

Given :

m = 200 mL = 200 g

Specific heat of ice = 2.06 J/g°C

Q = mcΔt

Δt = 0 - (-22) = 22

Q1 = 200 * 2.06 * 22 = 9064 J

Q2 = Melt 0 °C solid ice into 0 °C liquid water:

Q2 = m · ΔHf ; ΔHf = heat of fusion of water = 334j/g

Q2 = 200 * 334 = 66800 J

Q3 : Heat to convert from 0°C to 100°C

Q3 = mcΔt ; c = 4.19 J/g°C ; Δt = (100 - 0) = 100

Q3 = 200 * 4.19 * 100 = 83800 J

Q4: heat required to boil water to steam

Q = m · ΔHv

Hv = heat of vaporization of water = 2257 J/g

Q4 = 200 * 2257 = 451400 J

Total Q = Q1 + Q2 + Q3 + Q4

Q = 9064 + 66800 + 83800 + 451400

Q = 611,064 Joules

Q = 611.064 kJ

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3 years ago

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