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2 years ago

11

The aorta pumps blood out of the heart to the rest of the body. Please select the best answer from the choices provided. T or f

2 answers:

Daniel [21]2 years ago

3 0

The left ventricle relaxes and fills up with blood before squeezing and pumping the oxygen-rich blood through the aortic valve into the aorta — the main artery that carries blood to your body.<u>TRUE</u>

lisov135 [29]2 years ago

3 0

Answer:

True

Explanation:

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The propeller of a WW2 fighter plane is 2.35 m in diameter

Studentka2010 [4]

A) 1 rev = 2π rad. Using this ratio, you can find the rad/s: 1160 rev/min x 2π rad/rev x 1 min/60 s = 121.5 rad/s

b) You can find linear speed from angular speed using this equation (note the radius is half the diameter given in the question): v = ωr = 121.5 rad/s x 1.175 m = 142.8 m/s

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If at 10m above the ground an object has 50J of Kinetic Energy, with 50J of Potential Energy. How high can the object travel?

steposvetlana [31]

Hi there!

$\large\boxed{\text{B) 20 meters}}$

We know that:

$E_T = U + K$

U = Potential Energy (J)

K = Kinetic Energy (J)

E = Total Energy (J)

At 10m, the total amount of energy is equivalent to:

U + K = 50 + 50 = 100 J

To find the highest point the object can travel, K = 0 J and U is at a maximum of 100 J, so:

100J = mgh

We know at 10m U = 50J, so we can solve for mass. Let g = 10 m/s².

50J = 10(10)m

m = 1/2 kg

Now, solve for height given that E = 100 J:

100J = 1/2(10)h

100J = 5h

<u>h = 20 meters</u>

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1 year ago

When a certain amount of current flows through a resistor, it uses 3.00 W of power. If the current doubles, how much power will

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2 years ago

(a) At what height above Earth’s surface is the energy required to lift a satellite to that height equal to the kinetic energy r

Nikitich [7]

Answer:

Explanation:

Gravitational Potential Energy at earth surface $U_1=\frac{GM_em}{R_e}$

Gravitational Potential Energy at height h is $U_2=\frac{GM_em}{R_e+h}$

Energy required to lift the satellite $E_1=U_1-U_2$

$E_1=\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}$

Now Energy required to orbit around the earth

$E_2=\frac{1}{2}mv_{orbit}^2=\frac{GM_2m}{2(R_e+h)}$

$\Delta E=E_1-E_2$

$\Delta E=\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}-\frac{GM_2m}{2(R_e+h)}$

$E_1=E_2$ (given)

$\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}-\frac{GM_2m}{2(R_e+h)}=0$

$\frac{1}{R_e}-\frac{3}{2(R_e+h)}=0$

$h=\frac{R_e}{2}$

$h=3.19\times 10^6\ m$

(b)For greater height $E_1$ is greater than $E_2$

thus energy to lift the satellite is more than orbiting around earth

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2 years ago

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ololo11 [35]

Rotten fruit can be consumed by decomposers.

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