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2 years ago

Which of the following is NOT a valid reason that less energy is transferred at each level of the food chain?

Physics

1 answer:

stealth61 [152]2 years ago

5 0

Answer:

Explanation:

I'm pretty sure the answer is B because animals at the top need a lot of energy, so to make up for loss of energy at the higher levels, those organisms at the top have to consume a lot more.

You might be interested in

What are the smallest parts that make up matter?

Alecsey [184]

The smallest parts that make up different types of matter are called atoms.
Atoms are tiny elements that are made up of literally everything on earth.

3 0

3 years ago

A box is pulled with a horizontal force of 500N and moves 5m what is work done

dalvyx [7]

The answer to the question is shown below:

We all know that formula for solving work done is the force multiplied by distance covered:
Work done = Force x distance
Distance = 5m
Force = 500 N
Work done = 500 N * 5m
Work done = 2500 J

4 0

3 years ago

The aurora is caused when electrons and protons, moving in the earth’s magnetic field of ≈5.0×10−5T, collide with molecules of t

ollegr [7]

Answer:

8.79*10^6 rad/s

Explanation:

To find the frequency of the circular orbit for an electron you use the following expression, for the radius of the trajectory of an electron, that travels trough a constant magnetic field:

$r=\frac{mv}{qB}$ (1)

r: radius of the trajectory

m: mass of the electron = 9.1*10^-31 kg

v: speed of the electron = 1.0*10^6 m/s

q: charge of the electron = 1.6*10^-19 C

B: magnitude of the magnetic field = 5.0*10^-5 T

You use the fact that the angular frequency in a circular motion is given by:

$\omega=\frac{v}{r}$

Then, you solve the equation (1) in order to obtain v/r:

$\frac{v}{r}=\omega=\frac{qB}{m}$

Finally, you replace the values of the parameters:

$\omega=\frac{(1.6*10^{-19}C)(5.0*10^{-5}T)}{9.1*10^{-31}kg}\\\\\omega=8.79*10^6\frac{rad}{s}$

hence, the angular frequency is 8.79*10^6 rad/s

The frequency is:

$f=2\pi \omega=5.5*10^7Hz$

5 0

3 years ago

2. a) A disc rotates about its axis at speed 25 revolutions per minute and takes 15 s to stop. Calculate the

ASHA 777 [7]

The statement shows a case of rotational motion, in which the disc decelerates at constant rate.

i) The angular acceleration of the disc ( $\alpha$ ), in revolutions per square second, is found by the following kinematic formula:

$\alpha = \frac{\omega_{f}-\omega_{o}}{t}$ (1)

Where:

$\omega_{o}$ - Initial angular speed, in revolutions per second.
$\omega_{f}$ - Final angular speed, in revolutions per second.
$t$ - Time, in seconds.

If we know that $\omega_{o} = \frac{5}{12}\,\frac{rev}{s}$ , $\omega_{f} = 0\,\frac{rev}{s}$ y $t = 15\,s$ , then the angular acceleration of the disc is:

$\alpha = \frac{0\,\frac{rev}{s}-\frac{5}{12}\,\frac{rev}{s}}{15\,s}$

$\alpha = -\frac{1}{36}\,\frac{rev}{s^{2}}$

The angular acceleration of the disc is $\frac{1}{36}$ radians per square second.

ii) The number of rotations that the disk makes before it stops ( $\Delta \theta$ ), in revolutions, is determined by the following formula:

$\Delta \theta = \frac{\omega_{f}^{2}-\omega_{o}^{2}}{2\cdot \alpha}$ (2)

If we know that $\omega_{o} = \frac{5}{12}\,\frac{rev}{s}$ , $\omega_{f} = 0\,\frac{rev}{s}$ y $\alpha = -\frac{1}{36}\,\frac{rev}{s^{2}}$ , then the number of rotations done by the disc is:

$\Delta \theta = 3.125\,rev$

The disc makes 3.125 revolutions before it stops.

We kindly invite to check this question on rotational motion: brainly.com/question/23933120

6 0

3 years ago

To charge an electroscope negatively by induction you need: a) a positively charged rod and a ground b) a negatively charged rod

Zolol [24]

Answer:

Option a)

Explanation:

In the process of charging anything by the method of induction, a charged body is brought near to the body which is neutral or uncharged without any physical contact and the ground must be provided to the uncharged body.

The charge is induced and the nature of the induced charge is opposite to that of the charge present on the charged body.

So when a positively charged rod is used to charge an electroscope, the rod which is positive attracts the negative charge in the electroscope and the grounding of the electroscope ensures the removal of the positive charge and renders the electroscope negatively charged.

6 0

3 years ago