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Andre45 [30]
3 years ago
9

21. If the Sun's rays were at 45° to a vertical pillar, how would

Physics
1 answer:
diamong [38]3 years ago
4 0

Answer:

Let the height of the pole AB = x m. ∴ Length of shadow OB ol the pole AB = x m. Let the angle of elevation be ө, i.e. Hence, the angle of elevation of the Sun's altitude is 45°.

Explanation:

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Q|C Seawater contains 3.00 mg of uranium per cubic meter. (a) Given that the average ocean depth is about 4.00 km and water cove
kherson [118]

Seawater contains 3.00 mg of uranium per cubic meter.

Given:

The average ocean depth, d=4km=4000m

Water covers two-thirds of the Earth's surface

Then, the estimated amount of uranium dissolved in the ocean is 4.08 \times 10^{15}g.

We know, that the mass per unit volume of Uranium ism=3\times 10^{-3} g/m^{3}.

Also, the world's energy requirement is E_n=1.5\times10^{13} J/s.

The radius of the earth is x=6.371\times10^{6}.

Calculating the amount of Uranium under the seawater is given by the expression:

A_m= md ( \frac{2}{3} 4\pi x^{3})

Substituting all the values in the expression, we get

A_m=3\times 10^{-3} \times4000 ( \frac{2}{3} \times4\times\pi (6.371\times10^{6})^{3})=4.08\times 10^{15}g.

Thus, the amount of Uranium dissolved in the ocean is 4.08 \times 10^{15}g

To learn more about Uranium, refer to:

brainly.com/question/9099776

#SPJ4

6 0
2 years ago
Which of the following is a personal decision that involves science A)It is immoral to cheat on your homework? B)what movie will
Daniel [21]

Answer:

C because air purifier is very essential in dusty homes .

8 0
3 years ago
Read 2 more answers
A juggler throws a bowling pin into the air with an initial velocity . Another juggler drops a pin at the same instant.
PolarNik [594]
What is the question that you are trying to answer?
6 0
3 years ago
What does a Halogen atom give off when it gains an electrons?
Nostrana [21]
A halogen atom gives off lots of energy.
4 0
3 years ago
Platinum (pt) has the fcc crystal structure, an atomic radius of 0.1387 nm, and an atomic weight of 195.08 g/mol. what is its th
prohojiy [21]
The equation to be used is written as:

ρ = nA/VcNₐ
where
ρ is the density
n is the number of atoms in unit cell (for FCC, n=4)
A is the atomic weight
Vc is the volume of the cubic cell which is equal to a³, such that a is the side length (for FCC, a = 4r/√2, where r is the radis)\
Nₐ is Avogradro's constant equal to 6.022×10²³ atoms/mol

r = 0.1387 nm*(10⁻⁹ m/nm)*(100 cm/1m) = 1.387×10⁻⁸ cm
a = 4(1.387×10⁻⁸ cm)/√2 = 3.923×10⁻⁸ cm
V = a³ = (3.923×10⁻⁸ cm)³ = 6.0376×10⁻²³ cm³

ρ = [(4 atoms)(195.08 g/mol)]/[(6.0376×10⁻²³ cm³)(6.022×10²³ atoms/mol)]
ρ = 21.46 g/cm³
7 0
3 years ago
Read 2 more answers
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