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3 years ago

5

What percentage of energy is available to organisms at each level of an energy pyramid?

1 answer:

Snowcat [4.5K]3 years ago

6 0

The percentage of energy available to each organism is always 10 percent .

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What is the weight of a spring balance when the point is 30

andrezito [222]

Answer:

A 30 lb weight is attached to the end of a spring. The spring is stretched 6 in. Find the equation of motion if the weight is released from rest a point 3 inches above equilibrium position 。x(,) =-2 sin(81) 32 x(t) =-32 cos(80 O x(r) =-icos(81)

Explanation:

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2 years ago

Acceleration of gravity on jupiter

Viefleur [7K]

The acceleration of gravity on Jupiter is listed as <em>24.79 m/s²</em> .

That's roughly 2.53 times its value on Earth. So if you weigh, let's say,
130 pounds on Earth, then you would weigh about 328 pounds on Jupiter.

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3 years ago

Read 2 more answers

A truck is driving over a scale at a weight station. When the front wheels drive over the scale, the scale reads 5800 N. When th

aev [14]

Answer:

$x_2=1.60m$

Explanation:

From the Question We are told that

Initial Force $F_1=5800N$

Final Force $F_2=6500N$

Distance between the front and rear wheels \triangle x=3.20 m

Since

$\triangle x=3.20 m$

Therefore

$x_1+x_2=3.20$

$x_1=3.20-x_2$

Generally the equation for The center of mass is at x_2 is mathematically

given by

$x_2 =\frac{(F_1x_1+F_2x_2)}{(F_1+F_2)}$

$x_2=3.20F_1-\frac{x_2F_1+F_2x_2}{(F_1+F_2)}$

$2*F_1*x_2 =3.20F_1$

$x_2=1.60m$

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2 years ago

Two balls, ball A and ball B, are dropped from the same height onto the same surface. If ball A rebounds to a higher height than

mamaluj [8]

Answer:

its b

Explanation:

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2 years ago

Parallel rays of monochromatic light with wavelength 571nm illuminate two identical slits and produce an interference pattern on

Natasha2012 [34]

Answer:

$8.8\times 10^{-6} W/m^2$

Explanation:

We are given that

Wavelength, $\lambda=571 nm=571\times 10^{-9} m$

$1 nm=10^{-9} m$

R=75 cm= $\frac{75}{100}=0.75 m$

1 m=100 cm

$d=0.640 mm=0.64\times 10^{-3} m$

$1 mm=10^{-3} m$

$a=0.434 mm=0.434\times 10^{-3} m$

$y=0.830 mm=0.830\times 10^{-3} m$

$I_0=5.00\times 10^{-4}W/m^2$

$tan\theta=\frac{y}{R}$

$\theta=\frac{0.830\times 10^{-3}}{0.75\times 10^{-2}}$

$\theta=1.1\times 10^{-3}rad$

$tan\theta\approx \theta$

Because $\theta$ is small.

$\phi=\frac{2\pi dsin\theta}{\lambda}$

$\sin\theta\approx \theta$ ,

Therefore

$\phi=\frac{2\times\pi\times 0.64\times 10^{-3}\times 1.1\times 10^{-3}}{571\times 10^{-9}}$

$\phi=7.74 rad$

$\beta=\frac{2\pi a\theta}{\lambda}$

$\beta=5.3 rad$

$I=I_0cos^2(\frac{\phi}{2})(\frac{sin\frac{\beta}{2}}{\frac{\beta}{2}})^2$

$I=5\times 10^{-4}cos^2(\frac{7.74}{2})(\frac{sin\frac{5.3}{2}}{\frac{5.3}{2}})^2$

$I=8.8\times 10^{-6} W/m^2$

6 0

3 years ago