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3 years ago

Answer this question thanks

Physics

1 answer:

maria [59]3 years ago

7 0

Like they said the leopard seal will overpopulate and kill out all of the other animals. Because it would be at the top of the chain then without the killer whale. Therefore, since it is at the top of the chain it will overpopulate because it has nothing to eat/kill it so then they would eat the animals lower than them<span />

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10 PTS.

gayaneshka [121]

1.2 x (2.2 x 10⁵) = 264,000 Ω

0.8 x (2.2 x 10⁵) = 176,000 Ω

With a 'nominal' value of 220,000 Ω, it could actually be anywhere <em>between 176,000Ω and 264,000Ω</em> .

8 0

3 years ago

During a game the same batter swings at a ball thrown by the pitcher and hits a line drive. Just before the ball is hit it is tr

sergeinik [125]

Answer:

The total work on the ball is 36.25 Joules

Explanation:

There is an important principle on classical mechanics that is the work-energy principle it states that the total work on an object is equal the change on its kinetic energy, mathematically expressed as:

$W_{net}=\Delta K = K_f -K_i$ (1)

With W net the total work, Kf the final kinetic energy and Ki the initial kinetic energy. We're going to use this principle to calculate the total work on the baseball by the force exerted by the bat.

Kinetic energy is the energy related with the movement of an object and every classical object with velocity has some kinetic energy, it is defined as:

$K=\frac{mv^2}{2}$

With m the mass of the object and v its velocity, knowing this we can use on:

$W_{net}= \frac{mv_f^2 -mv_i^2}{2}=\frac{m(v_f^2 -v_i^2)}{2}$

In our case vf is the velocity just after the hit and vi the velocity just before the hit. For an average baseball its mass is 145g that is 0.145 kg, then

$W_{net}=\frac{0.145*(30.0^2 -20.0^2)}{2}$

$W_{net}=36.25 J$

8 0

4 years ago

An earth scientist who studies hydrosphere

ipn [44]

Oceanographer? I think that is what it is.

5 0

3 years ago

1. Most of the stars on the HR Diagram are classified as which type of star?

xenn [34]

Answer:

1. Main Sequence - middle life 17

2. red

3. blue

4. White dwarf stars are much hotter than Red Supergiants 15. List the color of the stars from hottest to coldest: Blue, White, Yellow, Orange, Red 16.

5. red giants

Explanation:

Main sequence stars have a Morgan-Keenan luminosity class labeled V. red giant and supergiant stars (luminosity classes I through III) occupy the region above the main sequence. They have low surface temperatures and high luminosities which, according to the Stefan-Boltzmann law, means they also have large radii. White dwarf stars are much hotter than Red Supergiants 15. List the color of the stars from hottest to coldest: Blue, White, Yellow, Orange, Red 16. The hottest stars are the blue stars. A star appears blue once its surface temperature gets above 10,000 Kelvin, or so, a star will appear blue to our eyes. The lowest temperature stars are red while the hottest stars are blue. Astronomers are able to measure the temperatures of the surfaces of stars by comparing their spectra to the spectrum of a black body. Most stars, including the sun, are "main sequence stars," fueled by nuclear fusion converting hydrogen into helium. ...

As stars begin to die, they become giants and supergiants (above the main sequence).

5 0

3 years ago

Read 2 more answers

A U-tube open at both ends is partially filled withwater. Oil

ArbitrLikvidat [17]

Answer:

a) h_w = 0.02139 m , b) v₁ = 9.74 m / s

Explanation:

For this exercise we use the pascal principle that states that the pressure at one point is the same regardless of body shape.

At the initial moment (before emptying the oil), we fix the point on the surface of the liquid, in this case the left and right sides are in balance.

P₁ = P₂ = P₀

Now we add the 5 cm (h₂ = 0.05 m) of oil, in this case the weight of the oil creates an extra pressure that pushes the water, let's look for how much the water moved (h_w). The weight of oil added is equal to the weight of displaced water

W_w = W_oil

m_w g= m_oil g

Density is defined

ρ = m / V

we replace

ρ_w V_w = ρ_oil W_oil

V = A h

ρ_w A h_w = ρ_oil A h_oil

h_w = h_oil ρ_oil / ρ_w

Now let's analyze the pressure at the initial reference height for both sides two had in U

Right side

We have the atmospheric pressure, with its decrease due to the lower height, plus the oil pressure above the reference level

h’= 0.05 cm - h_w

P = (P₀ - ρ_air g (0.05-h_w)) + ρ_oil g (0.05-h_w)

Left side

We have at the same point, the atmospheric pressure with its reduction due to the height change plus the water pressure

P = (P₀ - ρ_air h h_w) + ρ_w g h_w

As we have the same point we can equalize the pressure

(P₀ -ρ_air g (0.05-h_w)) +ρ_oil g (0.05-h_w) = (Po -ρ_air h h_w) +ρ_w g h_w

ρ_air g (h_w - (0.05-h_w)) = ρ_w g h_w - ρ_oil g (0.05-h_w)

- ρ_air g 0.05 = h_w g ( ρ_w + ρ _oil) - rho_oil g 0.05

h_w g ( ρ_w + ρ_oil) = g 0.05 ( ρ_air - ρ_oil)

calculate

h_w = 0.05 (ρ_ oil- ρ_air) / ( ρ_w + ρ_oil)

h_w = 0.05 (750 - 1.29) / (1000-750)

h_w = 0.02139 m

The amount that decreases the height on one side is equal to the amount that increases the other

b) cover the right side and blow the air on the left side, let's use Bernoulli's equation, where index 1 will be for the left side and index 2 for the right side

P₁ + 1/2 ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

Since they indicate that both sides are at the same height y₁ = y₂, the right side is protected from the wind speed therefore v₂ = 0, let's write the equation

P₁ + ½ ρ_air v₁² = P₂

v₁² = (P₂-P₁) 2 / ρ_air

Let's analyze the pressure on each side of the tube, since the new equilibrium height is the height that was added of oil distributed between the two tubes, bone

h’= 2.5 cm = 0.025 m

P₁ = P₀ - ρ_w g h’

P₂ = P₀ - ρ_oil g h ’

P₂-P₁ = g h’ (rho_w - Rho_oil)

We replace

v₁² = 2g h’ (ρ_w –ρ_oil) / ρ_air

calculate

v₁² = 2 9.8 0.025 (1000 - 750) /1.29

v₁ = √ 94.96

v₁ = 9.74 m / s

3 0

4 years ago