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riadik2000 [5.3K]
3 years ago
15

The melting of a glacier is an example of the interactions among which of Earth’s spheres?

Physics
1 answer:
Anuta_ua [19.1K]3 years ago
4 0
Atmosphere, geosphere, cryosphere would be the correct answer I think
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The layer of leaves that blocks most of the sunlight from reaching the ground in the rain forest is called the _____.
love history [14]
The answer should be <span>canopy.</span>
4 0
3 years ago
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If the angle of refraction is 20 degrees what is the angle of incidence​
Romashka-Z-Leto [24]

Answer:

13.1

Explanation:

that's what I'm gonna go with, but u can research more

5 0
3 years ago
A satellite circles the Earth in an orbit whose radius is twice the Earth’s radius. The Earth’s mass is 5.98 x 1024 kg, and its
gavmur [86]

Hello!

Recall the period of an orbit is how long it takes the satellite to make a complete orbit around the earth. Essentially, this is the same as 'time' in the distance = speed * time equation. For an orbit, we can define these quantities:

d = 2\pi r ← The circumference of the orbit

speed = orbital speed, we will solve for this later

time = period

Therefore:

T = \frac{2\pi r}{v}

Where 'r' is the orbital radius of the satellite.

First, let's solve for 'v' assuming a uniform orbit using the equation:
v = \sqrt{\frac{Gm}{r}}

G = Gravitational Constant (6.67 × 10⁻¹¹ Nm²/kg²)

m = mass of the earth (5.98 × 10²⁴ kg)

r = radius of orbit (1.276 × 10⁷ m)

Plug in the givens:
v = \sqrt{\frac{(6.67*10^{-11})(5.98*10^{24})}{(1.276*10^7)}} = 5590.983 m/s

Now, we can solve for the period:

T = \frac{2\pi (1.276*10^7)}{5590.983} =\boxed{ 14339.776 s}

7 0
2 years ago
A diffraction grating is illuminated simultaneously with red light of wavelength 670 nm and light of an unknown wavelength. The
marusya05 [52]

Answer:

dsin∅ = m× λ

so, dsin∅red = 3(670nm)

also, dsin∅? =5λ?

however ,if they overlap then dsin∅red = dsin∅?

3(670nm) /5 =402nm

∴λ = 402nm

Explanation:

4 0
3 years ago
honor physics The velocity of a 200 kg object is changed from 5 m/s to 25 m/s in 50 seconds by an applied constant force. a. Wha
Ksivusya [100]

Answer:

(a) 4000 kgm/s.

(b) 80 N

Explanation:

(a) Change in momentum: This can be defined as the product of the mass of a body and its change in velocity. The S.I unit of impulse is Ns or kgm/s

Mathematically, Change in momentum is expressed as

ΔM = mΔv ..................................... Equation 1

Where ΔM = change in momentum, m = mass of the object, Δv = change in velocity = v₂ - v₁

Given: m = 200 kg, Δv = v₂ - v₁ = 25-5 = 20 m/s.

Substituting into equation 1

ΔM = 200(20)

ΔM = 4000 kgm/s.

Hence the change in momentum = 4000 kgm/s

(b)

Force: This can be defined as the ratio of the change in momentum of a body to the time required for the change.

F = ΔM/t.............................. Equation 2

Where F = force, ΔM = change in momentum, t = time.

Given: ΔM  = 4000 kgm/s, t = 50 second.

Substituting into equation 2

F = 4000/50

F = 80 N.

Hence the force  = 80 N

7 0
3 years ago
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