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1 year ago

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Q5- Knowing now that your mood might affect your memory, what strategies could you employ to assist your memories and make sure

that you forget less information when you are having a good time?

1 answer:

Tema [17]1 year ago

4 0

You can try recalling and rehearsing information about the memory.

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50 mL of soda in a soda can exerts ____ 50 mL of soda in a 1L bottle.

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More pressure than..

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2 years ago

25. Explain why the speed of sound is faster in solids than in gases. Include two other factors Chapter 17 says the speed of sou

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In solids, particles or atom are very closely arranged compared to gasses. When these particles are arranged in such proximity, vibrations from sound are very easily transmitted from one particle to another in the solid. Hence, the sound vibrations can travel through the solid medium more quickly than through a gas medium.
Speed of sound also depends on its frequency and the wavelength.

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2 years ago

Sharks and related fish can sense the extremely weak electric fields emitted by their prey in the surrounding waters. These dete

Sonja [21]

Answer

2) 1.5×10-2 m

Explanation

The potential difference is related to the electric field by:

$\Delta V=Ed$ (1)

where

$\Delta V$ is the potential difference

E is the electric field

d is the distance

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$E=1 V/cm=100 V/m$

when connected to a battery with potential difference

$\Delta V=1.5 V$

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$d=\frac{\Delta V}{E}=\frac{1.5 V}{100 V/m}=0.015 m=1.5 \cdot 10^{-2} m$

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2 years ago

How would the photos set up change the mechanical advantage of lifting the block as compared with lifting the block without a le

lisabon 2012 [21]

<h3>Answer:</h3>

The mechanical advantage would decrease, making the block more difficult to lift.

<h3>Explanation:</h3>

The mechanical advantage in such a setup is the ratio of distance from A to B to the distance from D to B. In this picture, that ratio is less than 1, meaning the advantage of having this setup is less than the advantage of no setup at all.

While the force required to lift the block is increased by this setup, the distance over which that force is applied will be smaller for raising the block to a given height. (Overall, for the same height, more work is required with the lever setup because you're raising part of the mass of the lever as well as the mass of the block.)

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2 years ago

Read 2 more answers

Chapter 14, Problem 042 A flotation device is in the shape of a right cylinder, with a height of 0.588 m and a face area of 4.19

Alisiya [41]

Answer:

The workdone is $W = 9.28 * 10^{3} J$

Explanation:

From the question we are told that

The height of the cylinder is $h = 0.588\ m$

The face Area is $A = 4.19 \ m^2$

The density of the cylinder is $\rho = 0.346 * \rho_w$

Where $\rho_w$ is the density of freshwater which has a constant value

$\rho_w = 1000 kg/m^3$

Now

Let the final height of the device under the water be $= h_f$

Let the initial volume underwater be $= V_n$

Let the initial height under water be $= h_i$

Let the final volume under water be $= V_f$

According to the rule of floatation

The weight of the cylinder = Upward thrust

This is mathematically represented as

$\rho_c g V_n = \rho_w gV_f$

$\rho_c A h = \rho A h_f$

So $\frac{0.346 \rho_w}{\rho_w} = \frac{h_f}{h}$

=> $\frac{h_f}{h_c} = 0.346$

Now the work done is mathematically represented as

$W = \int\limits^{h_f}_{h} {\rho_w g A (-h)} \, dh$

$= \rho_w g A [\frac{h^2}{2} ] \left | h_f} \atop {h}} \right.$

$= \frac{g A \rho}{2} [h^2 - h_f^2]$

$= \frac{g A \rho}{2} (h^2) [1 - \frac{h_f^2}{h^2} ]$

Substituting values

$W = \frac{(9.8 ) (4.19) (10^3)}{2} (0.588)^2 (1 - 0.346)$

$W = 9.28 * 10^{3} J$

4 0

2 years ago