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3 years ago

15

Suppose you are investigating how the day ends hamsters affect how much they run on the hamster wheel. In order to control the e

xperiment which of these should you do?
A. Change only the diet of each hampster.

B. Give all the hamsters the exact same diet.

C. Only allow some hampsters to run on the wheel.

D. Compare the behavior of hamsters and gerbils

2 answers:

Rzqust [24]3 years ago

7 0

Answer:

B

Explanation:

If you want to control the experiment, you want to make sure that no outside factor could potentially influence your results.

Hence, changing the diet of a hamster would significantly alter any results you may collect.

For example, if you allow hamster A to eat a nutritious diet, while not giving a proper diet to hamster B. Hamster A will perform better, and run further and for longer, because he has more energy.

The answer is B, because you want to make sure no hamster is at an advantage.

aksik [14]3 years ago

5 0

The answer is C because you need non running hamsters to compare running hamsters with.

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Masteriza [31]

The net current in the conductors and the value of the line integral

$I=\frac{3.2\cdot 10^{-4}}{4\pi \cdot 10^{-7}}=254.77\, A$
The resultant remains same 3.2 *10^4 Tm

This is further explained below.

<h3>What is the net current in the conductors?</h3>

Generally,

To put it another way, the total current In flowing across a surface S (contained by C) is proportional to the line integral of the magnetic B-field (in tesla, T).

$\oint_C \mathbf{B} \cdot \mathrm{d}\boldsymbol{\ell} = \mu_0 \iint_S \mathbf{J} \cdot \mathrm{d}\mathbf{S} = \mu_0I_\mathrm{enc}$

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B)

In conclusion, It is possible for the line integral to go around the loop in either direction (clockwise or counterclockwise), the vector area dS to point in either of the two normal directions and Ienc, which is the net current passing through the surface S, to be positive in either direction—but both directions can be chosen as positive in this example. The right-hand rule solves these ambiguities.

The resultant remains the same at 3.2 *10^4 Tm

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2 years ago

From what three regions do most of the small objects in the universe come?

Ivenika [448]

Answer:

Scientists who study the Sun usually divide it up into three main regions: the Sun's interior, the solar atmosphere, and the visible "surface" of the Sun which lies between the interior and the atmosphere. There are three main parts to the Sun's interior: the core, the radiative zone, and the convective zone.

Explanation:

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Dr. John Paul Stapp was U.S. Air Force officer who studied the effects of extreme deceleration on the human body. On December 10

Mars2501 [29]

Answer:

a) $a=5.7551 \times g$

b) $d=20.5539\times g$

Explanation:

Given:

speed of rocket initially, $v_i=0\ m.s^{-1}$

top speed of rocket after acceleration, $v=282\ m.s^{-1}$

time taken to get to the top speed, $t_i=5\ m.s^{-1}$

final speed of the rocket, $v_f=0\ m.s^{-1}$

time taken to get to the final speed after reaching the top speed, $t_f=1.4\ s$

Now the acceleration:

$a=\frac{v-v_i}{t_i}$

$a=\frac{282-0}{5}$

$a=56.4\ m.s^{-2}$

Now as a fraction of gravity:

$a=\frac{56.4}{9.8}\times g$

$a=5.7551 \times g$

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$d=\frac{0-282}{1.4}$

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3 years ago

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A 4.0-kg object is moving with speed 2.0 m/s. a 1.0-kg object is moving with speed 4.0 m/s. both objects encounter the same cons

LenKa [72]

Newton's second law states that the product between the mass and the acceleration of an object is equal to the force applied:
$F=ma$
from which we find an expression for the acceleration:
$a= \frac{F}{m}$ (1)

Both objects are moving by uniformly accelerated motion (because the force applied is constant), so we can also using the following relationship
$v_f^2 - v_i^2 = 2 a S$ (2)
where
$v_f$ is the final speed of the object
$v_i$ is the initial speed
S is the distance covered
By substituting (1) into (2), and by removing $v_f$ (since the final velocity of the two objects is zero), we find
$-v_i^2 = 2 \frac{F}{m}S$
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$S= \frac{(2.0 m/s)^2 (4.0 kg)}{2F} = \frac{8}{F} [m]$
And for the second object we have
$S= \frac{(4.0 m/s)^2 (1.0 kg)}{2F} = \frac{8}{F} [m]$

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