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MA_775_DIABLO [31]

3 years ago

9

Revise this statement to make it true: Only winter is caused by the tilt of the earth's axis.

1 answer:

s2008m [1.1K]3 years ago

4 0

Answer:

Only winter is caused by the tilt of the earth's axis.<u>F</u><u>a</u><u>l</u><u>s</u><u>e</u>

<u>it</u><u> </u><u>caused</u><u> </u><u>summer</u><u> </u><u>or</u><u> </u><u>autumn</u><u> </u><u>too</u>

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Why is earth's temperature just right for life?

Rudik [331]

The answer is letter b

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4 years ago

Human blood consists of blood groups A, B, and O. However, there is one more blood group, AB, which contains alleles A and B in

Arturiano [62]

The answer is Codominance, AB blood cells inherit both A and B blood types so they are codominance

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4 years ago

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a runner increases her speed from 3.1 m/s to 3.5 m/s during the last 15 seconds of her run, what was her acceleration during her

Rama09 [41]

Acceleration = velocity/time
A= 3.5m/s/15s
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4 years ago

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The x component of vector is -27.3 m and the y component is +43.6 m. (a) What is the magnitude of ? (b) What is the angle betwee

lara [203]

Answer:51.44 units

Explanation:

Given

x component of vector is $-27.3\hat{i}$

y component of vector is $43.6\hat{j}$

so position vector is

$r=-27.3\hat{i}+43.6\hat{j}$

Magnitude of vector is

$|r|=\sqrt{27.3^2+43.6^2}$

$|r|=\sqrt{2646.25}$

|r|=51.44 units

Direction

$tan\theta =\frac{43.6}{-27.3}=-1.597$

vector is in 2nd quadrant thus

$180-\theta =57.94$

$\theta =122.06^{\circ}$

4 0

3 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

4 0

3 years ago