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3 years ago

10

a fish is looking at a 1.0 m high plant at the edge of a pond. Will the plant appear to the fish shorter or taller than its actu

al height? Draw a ray diagram to explain your answer.

1 answer:

rjkz [21]3 years ago

7 0

Answer: The plant will taller than its actual height.

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Which of the following describes an element?

bixtya [17]

Answer:

A substance made of two or more types of atoms

Explanation:

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3 years ago

Meg walks with a velocity of 0.9 m/s west. She does so while riding on a train that is traveling with a velocity of 2.7 m/s east

Shkiper50 [21]

<span>Velocities are vectors so we can add them!

Let's let +x be East and -x be West.

-0.9 + 2.7 = 1.8

Since our answer is positive that means East so the answer is C.</span>

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4 years ago

Individuals with favorable traits are ________ in an environment

77julia77 [94]

Succesful is correct.

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3 years ago

Calculate the pressure the fluid exerted on your diver. The acceleration of gravity is 9.8 m/s2 and the density of the water is

erica [24]

Complete question:

A diver is 10 m below the surface of water. Calculate the pressure the fluid exerted on the diver. The acceleration of gravity is 9.8 m/s2 and the density of the water is 1000 kg/m3. Answer in units of Pa. Show your work.

Answer:

Tthe pressure the fluid exerted on the diver is 1.99 x 10⁵ Pa

Explanation:

Given;

density of water, ρ = 1000 kg/m³

diver's position below the surface of the water, h = 10 m

acceleration due to gravity, g = 9.8 m/s²

Let the atmospheric pressure, P₀ = 101325 Pa

The pressure 10 m below the surface of the water is calculated as;

P = P₀ + ρgh

P = 101325 Pa + (1000 x 9.8 x 10)Pa

P = 199325 Pa

P = 1.99 x 10⁵ Pa.

Therefore, the pressure the fluid exerted on the diver is 1.99 x 10⁵ Pa

5 0

3 years ago

The velocity of a particle moving along the x-axis varies with time according to v(t) = A + Bt−1 , where A = 2 m/s, B = 0.25 m,

kondaur [170]

Answer:

$a= -2\ m/s^2$

$a=-12.5\ m/s^2$

x=2.17 m

x=8.4 m

Explanation:

Given that

$v=A+Bt^{-1}$

$v=2+0.25t^{-1}$

To find acceleration :

we know that

$a=\dfrac{dv}{dt}$

$\dfrac{dv}{dt}=0-0.5t^{-2}$

$a=-0.5t^{-2}$

Acceleration at t= 2 s

$a=-0.5\times 2^{-2}$

$a= -2\ m/s^2$

Acceleration at t= 5 s

$a=-0.5\times 5^{-2}$

$a=-12.5\ m/s^2$

We know that

$v=\dfrac{dx}{dt}$

$dx=\left(2+\dfrac{1}{4t}\right)dt$

Position at t= 2 s:

$\int_{0}^{x}dx=\int_{1}^{2} \left(2+0.25\dfrac{1}{t}\right)dt$

$x=\left [2t+0.25\ lnt \right ]_{1}^{2}$

x=2+0.25 ln2

x=2.17 m

Position at t= 5 s:

$\int_{0}^{x}dx=\int_{1}^{5} \left(2+0.25\dfrac{1}{t}\right)dt$

$x=\left [2t+0.25\ lnt \right ]_{1}^{5}$

x=8+0.25 ln5

x=8.4 m

4 0

3 years ago