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3 years ago

11

Which level of organisms get the most energy out of the food they make or consume?

1 answer:

boyakko [2]3 years ago

6 0

Higher-level consumers get the most energy from food they eat.
An example is a lion eating a zebra. The zebra has more energy than a handful of grass, which has almost none to offer.

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.. Which location shows AUTUMN (fall) in the NORTHERN hemisphere? *

Lena [83]

Answer:

Location 2

Explanation:

This shows fall because the Earth is a little tilted not giving as much light

8 0

3 years ago

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A ship anchored at sea is rocked by waves that have crests Lim apart the waves travel at 70m/S, at what frequency do the waves r

inessss [21]

Question: A ship anchored at sea is rocked by waves that have crests 100 m apart the waves travel at 70m/S, at what frequency do the waves reach the ship?

Answer:

0.7 Hz

Explanation:

Applying,

v = λf............... Equation 1

Where v = velocity of the wave, f = frequency fo the wave, λ = wavelength of the wave

make f the subject of the equation

f = v/λ................. Equation 2

From the question,

Given: v = 70 m/s, λ = 100 m ( distance between successive crest)

Substitute these values into equation 2

f = 70/100

f = 0.7 Hz

Hence the frequency at which the wave reach the ship is 0.7 Hz

3 0

3 years ago

The parasailing system shown uses a winch to pull the rider in towards the boat, which is traveling with a constant velocity. Du

leonid [27]

Answer:

The magnitude of the force is $F_{net}= 1.837 *10^4N$

the direction is 57.98° from the horizontal plane in a counter clockwise direction

Explanation:

From the question we are told that

At t = 0 , $\theta = 20^o$

The rate at which the angle increases is $w = 2 \ ^o/s$

Converting this to revolution per second $\theta ' = 2 \ ^o/s * \frac{\pi}{180} =0.0349\ rps$

The length of the rope is defined by

$r = 125- \frac{1}{3}t^{\frac{3}{2} }$

At $\theta =30^o$ , The tension on the rope T = 18 kN

Mass of the para-sailor is $M_p = 75kg$

Looking at the question we see that we can also denote the equation by which the length is defied as an an equation that define the linear displacement

Now the derivative of displacement is velocity

So

$r' = -\frac{1}{3} [\frac{3}{2} ] t^{\frac{1}{2} }$

represents the velocity, again the derivative of velocity gives us acceleration

So

$r'' = -\frac{1}{4} t^{-\frac{1}{2} }$

Now to the time when the rope made angle of 30° with the water

generally angular velocity is mathematically represented as

$w = \frac{\Delta \theta}{\Delta t}$

Where $\theta$ is the angular displacement

Now considering the interval between $20^o \ to \ 30^o$ we have

$2 = \frac{30 -20 }{t -0}$

making t the subject

$t = \frac{10}{2}$

$= 5s$

Now at this time the displacement is

$r = 125- \frac{1}{3}(5)^{\frac{3}{2} }$

$= 121.273 m$

The linear velocity is

$r' = -\frac{1}{3} [\frac{3}{2} ] (5)^{\frac{1}{2} }$

$= -1.118 m/s$

The linear acceleration is

$r'' = -\frac{1}{4} (5)^{-\frac{1}{2} }$

$= -0.112m/s^2$

Generally radial acceleration is mathematically represented by

$\alpha _R = r'' -r \theta'^2$

$= -0.112 - (121.273)[0.0349]^2$

$= 0.271 m/s^2$

Generally angular acceleration is mathematically represented by

$\alpha_t = r \theta'' + 2 r' \theta '$

Now $\theta '' = \frac{d (0.0349)}{dt} = 0$

So

$\alpha _t = 121.273 * 0 + 2 * (-1.118)(0.0349)$

$= -0.07805 m/s^2$

The net resultant acceleration is mathematically represented as

$a = \sqrt{\alpha_R^2 + \alpha_t^2 }$

$= \sqrt{(-0.07805)^2 +(-0.027)^2}$

$= 0.272 m/s^2$

Now the direction of the is acceleration is mathematically represented as

$tan \theta_a = \frac{\alpha_R }{\alpha_t }$

$\theta_a = tan^{-1} \frac{-0.271}{-0.07805}$

$= 73.26^o$

The force on the para-sailor along y-axis is mathematically represented as

$F_y = mg + Tsin 30^o + ma sin(90- \theta )$

$= (75 * 9.8) + (18 *10^3) sin 30 + (75 * 0.272)sin(90-73.26)$

$= 9.74*10^3 N$

The force on the para-sailor along x-axis is mathematically represented as

$F_x = mg + Tcos 30^o + ma cos(90- \theta )$

$= (75 * 9.8) + (18 *10^3) cos 30 + (75 * 0.272)cos(90-73.26)$

$= 1.557 *10^4 N$

The net resultant force is mathematically evaluated as

$F_{net} = \sqrt{F_x^2 + F_y^2}$

$=\sqrt{(1.557 *10^4)^2 + (9.74*10^3)^2}$

$F_{net}= 1.837 *10^4N$

The direction of the force is

$tan \theta_f = \frac{F_y}{F_x}$

$\theta_f = tan^{-1} [\frac{1.557*10^4}{9.74*10^3} ]$

$= tan^{-1} (1.599)$

$= 57.98^o$

7 0

3 years ago

A grindstone increases in angular speed from 4.00 rad/s to 12.00 rad/s in 4.00 s. Through what angle does it turn during that ti

Sophie [7]

Answer:

d. 32.0 rad

Explanation:

The angular speed, denoted by ω can be calculated thus;

ω = θ /t

Where;

ω = angular speed in radians/sec

θ = angle in radians

t = time in seconds

∆ω = final ω - initial ω

ω = 12.00 rad/s - 4.00 rad/s

ω = 8.00 rad/s

Hence, using ω = θ/t

8.00 = θ/4

θ = 32.00rad

4 0

3 years ago

What are the forces on a long jumper when he runs jumps and lands xxx

Makovka662 [10]

Gravity and friction would affect the long jumper.

4 0

3 years ago

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