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3 years ago

Oak trees make many acorns that can become new oak trees what is this an example of

Physics

2 answers:

Gelneren [198K]3 years ago

6 0

This is an example of Reproduction its a constant cycle that builds nature

Kisachek [45]3 years ago

4 0

Reproduction! hope this helps u.

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The MAC is 58 inches, The CG limits are from 26% to 43% MAC. If the CG is found to be

eimsori [14]

By working with percentages, we want to see how many inches is the center of gravity out of the limits. We will find that the CG is 1.45 inches out of limits.

<h3>What are the limits?</h3>

First, we need to find the limits.

We know that the MAC is 58 inches, and the limits are from 26% to 43% MAC.

So if 58 in is the 100%, the 26% and 43% of that are:

26% → (26%/100%)*58in = 0.26*58 in = 15.08 in
43% → (43%/100%)*58in = 0.43*58 in = 24.94 in.

But we know that the CG is found to be 45.5% MAC, then it measures:

(45.5%/100%)*58in = 0.455*58in = 26.39 in

We need to compare it with the largest limit, so we get:

26.39 in - 24.94 in = 1.45 in

This means that the CG is 1.45 inches out of limits.

If you want to learn more about percentages, you can read:

brainly.com/question/14345924

6 0

2 years ago

On a distant planet, a rock falls in 48.4 s from the top of a 1.10e+02 m cliff to the planet surface below. What is the accelera

pav-90 [236]

this can be solve using the formala of free fall

t = sqrt( 2y/ g)

where t is the time of fall

y is the height

g is the acceleration due to gravity

48.4 s = sqrt (2 (1.10e+02 m)/ g)

G = 0.0930 m/s2

The velocity at impact

V = sqrt(2gy)

= sqrt( 2 ( 0.0930 m/s2)( 1.10e+02 m)

V = 4.523 m/s

8 0

4 years ago

Is the use of a toaster 100% efficient? Why or why not

erastovalidia [21]

Not always. It depends on how
Many volts there are

4 0

3 years ago

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Which company has experienced censorship in China?

True [87]

Japan most of the time

4 0

4 years ago

Read 2 more answers

An object in space has altitude of 2210 km, velocity of 7000 m/s and flight path angle of 20 degrees. Find the eccentricity, sem

Dmitry [639]

Answer:

Eccentricity = 0.0557

Semi-major axis = 9,095 km

Angular momentum/mass = 60,116 $\frac{km^{2} }{Sec}$

Kinetic energies/mass = 24,500 KJ

Potential energies/mass = 21,673 KJ

Explanation:

Eccentricity

To find the eccentricity use the following formula

Eccentricity = [ Altitude from the earth x ( $Velocity^{2}$ / Gravitational parameter for the Earth ) ] - 1

Where

Altitude from the earth radius = Radius of the earth + Altitude of the earth from radius = 6,378 km + 2,210 km = 8,588 km

Velocity = 7,000 m/s

Gravitational parameter for the Earth = 3.986004418 × $10^{14}$

Eccentricity = ?

Placing values in the formula

Eccentricity = [ ( 8,588 km x 1000 ) x ( $7000^{2}$ / 3.986004418 × $10^{14}$ ) ] - 1

Eccentricity = 0.0557

Semi-major axis

Total Distance = Semi-major axis x ( 1 - Eccentricity )

Where

Total Distance = 8,588 km

Eccentricity = 0.0557

8,588 x 1,000 m = Semi-major axis x ( 1 - 0.0557 )

8,588,000 m = Semi-major axis x 0.9443

Semi-major axis = 8,588,000 m / 0.9443

Semi-major axis = 9,094,567.40 m

Semi-major axis = 9,094.56740 km

Semi-major axis = 9,095 km

Angular momentum/mass

L = MVR

L/M = VR

Where

V = Velocity = $\frac{7,000 m/s}{1000}$ = 7 km/s

R = Total Radius = Radius of Earth + Altitude = 6,378 km + 2,210 km = 8,588 km

Placing values in the formula

L/M = 7 km/s x 8,588 km = 60,116 $\frac{km^{2} }{Sec}$

Kinetic energies/mass

Ke = I $W^{2}$

Ke = $\frac{1}{2} mr^{2}$ $W^{2}$ ( Where I = $mr^{2}$ )

$\frac{ke}{m}$ = $\frac{r^2}{2}$ $W^{2}$

$\frac{ke}{m}$ = $\frac{r^2}{2}$ $\frac{V^2}{r^2}$ ( $W^{2}$ = $\frac{V^2}{r^2}$ )

$\frac{ke}{m}$ = $\frac{V^2}{2}$

$\frac{ke}{m}$ = $\frac{7000^2}{2}$

$\frac{ke}{m}$ = 24,500,000 J

$\frac{ke}{m}$ = 24,500 KJ

Potential energies/mass

PE = mgh

$\frac{PE}{m}$ = gh

Where

g = 9.807 $\frac{m}{s^2}$

h = 2,210 km x 1,000 = 2,210,000 m

Placing values in the formula

$\frac{PE}{m}$ = 9.807 $\frac{m}{s^2}$ x 2,210,000 m

$\frac{PE}{m}$ = 21,673,470 J

$\frac{PE}{m}$ = 21,673.470 KJ

$\frac{PE}{m}$ = 21,673 KJ

8 0

3 years ago