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Mumz [18]
4 years ago
10

Which describes how the same force affects a small mass and a large mass

Physics
1 answer:
goldenfox [79]4 years ago
5 0
You shall use the equation for force given by the second Law of Newton, this is F = m*a, where F is the net force that acts over the object, m is the mass of the object and a is the acceleration that the object will acquire. From that equation you can find a = F/m, which means that a is direct proportional to F and invsersely related to m. So, small masses accelerate faster than large masses, and <span>the answer is the option B. the small mass accelerates faster.</span>
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Ions are unequally distributed across the plasma membrane of all cells. This ion distribution creates an electrical potential di
nekit [7.7K]

Answer: Resting Membrane Potential

Explanation:

The <u>resting membrane potential</u> refers to the difference in voltage between the inside and outside of the cell membrane when the cell is at physiological rest. It should be noted that <u>the cell membrane is a selective semipermeable barrier, which only allows the transit through it of certain molecules and prevents the transit of others. </u>

This selectivity causes an uneven distribution of charged particles (ions), as the membrane only accepts some types of ions.

Now, in the case of neurons, which are electrically excitable nerve cells; the transport of electrical signals is due to these changes in the permeability and asymmetric distribution of ions (mainly sodium and potassium) when the neuron is not excited (at rest).

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3 years ago
What exercise will strengthen your legs?
Olin [163]

Answer: Squats and leg lifts

Explanation:

They strengthen your legs!

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3 years ago
A.) Review the graphical data above and create a data table that reflects the graph (10<br> pts.)
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3 years ago
____________ are used to calculate the distance a continent has moved in a year.
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6 0
3 years ago
A 1.2 W point source emits sound waves isotropically. Assuming that the energy of the waves is conserved, find the intensity (a)
pshichka [43]

Answer:

(a) I = 3.7\times 10^{-2}\ W/m^{2}

(b) I = 1.9\times 10^{-2}\ W/m^{2}

Explanation:

Given:

Point source P_{b}=1.2\ W

Distance from source Part a L = 1.6\ m

Distance from source Part b L = 2.2\ m

Solution:

Part (a)

Using intensity formula at distance L from an isotropic point.

I = \frac{P_{b}}{4\pi(L)^{2}}  -------------------(1)

Substitute L = 1.6\ m and P_{b}=1.2\ W in equation 1..

I = \frac{1.2}{4\times 3.14(1.6)^{2}}

I = \frac{1.2}{32.15}

I = 0.037\ W/m^{2}

I = 3.7\times 10^{-2}\ W/m^{2}

Part (b)

Substitute L = 2.2\ m and P_{b}=1.2\ W in equation 1.

I = \frac{1.2}{4\times 3.14(2.2)^{2}}

I = \frac{1.2}{60.79}

I = 0.019\ W/m^{2}

I = 1.9\times 10^{-2}\ W/m^{2}

Therefore, the intensity at distance 1.6 m from the source:

I = 1.9\times 10^{-2}\ W/m^{2}.

And, the intensity at distance 2.2 m from the source:

I = 1.9\times 10^{-2}\ W/m^{2}.

7 0
3 years ago
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