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3 years ago

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How much energy is received by earth per second per unit area ?

1 answer:

Hunter-Best [27]3 years ago

8 0

Answer:

The approximate average value cited, 1.3608 ± 0.0005 kW/m², which is 81.65 kJ/m² per minute, is equivalent to approximately 1.951 calories per minute per square centimeter, or 1.951 langleys per minute.

Explanation:

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An object accelerates from rest to 85m/s over a distance of 36m. What acceleration did it experience?

Marizza181 [45]

Suvat
we have s, u, v and we want a
the suvat equation with these values in is: v^2 = u^2 - 2as
so a = (-v^2 + u^2)/-2s
plug numbers in
a = (-85^2 + 0^2)/-2*36 = 7225/72 = 100.3... ms^-2

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4 years ago

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nirvana33 [79]

Answer:

A

Explanation:

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3 years ago

The smallest separation resolvable by a microscope is of the order of magnitude of the wavelength used. What energy electrons wo

hjlf

Answer: a) for 150 Angstroms 6.63 *10^-3 eV; b) for 5 Angstroms 6.02 eV

Explanation: To solve this problem we have to use the relationship given by De Broglie as:

λ =p/h where p is the momentum and h the Planck constant

if we consider the energy given by acceleration tube for the electrons given by: E: e ΔV so is equal to kinetic energy of electrons p^2/2m

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replacing we obtained the above values.

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3 years ago

Need help 8th grade science test need help

jolli1 [7]

ANSWER: d, average speed

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3 years ago

Read 2 more answers

Consider a large tank holding 1000 L of pure water into which a brine solution of salt begins to owat a constant rate of 6L/min.

lbvjy [14]

Answer:

T = 693.147 minutes

Explanation:

The tank is being continuously stirred. So let the salt concentration of the tank at some time t be x in units of kg/L.

Therefore, the total salt in the tank at time t = 1000x kg

Brine water flows into the tank at a rate of 6 L/min which has a concentration of 0.1 kg/L

Hence, the amount of salt that is added to the tank per minute = $(6\times0.1)kg/min=0.6kg/min$

Also, there is a continuous outflow from the tank at a rate of 6 L/min.

Hence, amount of salt subtracted from the tank per minute = 6x kg/min

Now, the rate of change of salt concentration in the tank = $\frac{dx}{dt}$

So, the rate of change of salt in the tank can be given by the following equation,

$1000\frac{dx}{dt} =0.6-6x$

or, $\int\limits^{0.05}_0 {\frac{1000}{0.6-6x} } \, dx =\int\limits^T_0 {} \, dt$

or, T = 693.147 min (time taken for the tank to reach a salt concentration

of 0.05 kg/L)

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4 years ago