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3 years ago

7

Are graded receptor potentials always depolarizing? Do graded receptor potentials always make it easier to induce action potenti

al?

1 answer:

Oksana_A [137]3 years ago

7 0

Answer: Yes,graded receptor potential always depolarize.

Yes,graded receptor potentials must occur to depolarize the neutrons to threshold before action potentials can occur.

Explanation:

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the earth has a radius of 6.38×10^16 meter and turns around once on its axis in 24 hour.what is the radial acceleration of perso

Scrat [10]

337493603.8m/s²

Explanation:

Radius of the earth = 6.38 x 10¹⁶m

time = 24hr (86400s)

Unknown:

Centripetal acceleration = ?

Solution:

The centripetal acceleration is directed inward to keep the body from falling off the surface of the earth.

centripetal acceleration = $\frac{v^{2} }{r}$

where v is the velocity and r is the radius

also;

v = wr

where w is the angular velocity

substituting in the equation for centripetal acceleration gives;

a = w²r

also w = $\frac{2 x pi}{T}$

therefore;

a = $\frac{4 \pi ^{2} r }{T^{2} }$

a = $\frac{ 4 x 3.142^{2} x 6.38 x 10^{16} }{86400^{2} }$

a = 337493603.8m/s²

learn more:

Acceleration brainly.com/question/3820012

#learnwithBrainly

5 0

3 years ago

A girl of mass m1=60 kilograms springs from a trampoline with an initial upward velocity of v1=8.0 meters per second. At height

AleksandrR [38]

a) 5.0 m/s

This first part of the problem can be solved by using the conservation of energy. In fact, the mechanical energy of the girl just after she jumps is equal to her kinetic energy:

$E_i=\frac{1}{2}m_1v_1^2$

where m1 = 60 kg is the girl's mass and v1 = 8.0 m/s is her initial velocity.

When she reaches the height of h = 2.0 m, her mechanical energy is sum of kinetic energy and potential energy:

$E_f = \frac{1}{2}m_1 v_2 ^2 + m_1 gh$

where v2 is the new speed of the girl (before grabbing the box), and h = 2.0m. Equalizing the two equations (because the mechanical energy is conserved), we find

$\frac{1}{2}m_1 v_1^2 = \frac{1}{2}m_1 v_2 ^2 + m_1 gh\\v_1^2 = v_2^2 +2gh\\v_2 = \sqrt{v_1^2 -2gh}=\sqrt{(8.0 m/s)^2-(2)(9.8 m/s^2)(2.0 m)}=5.0 m/s$

b) 4.0 m/s

After the girl grab the box, the total momentum of the system must be conserved. This means that the initial momentum of the girl must be equal to the total momentum of the girl+box after the girl catches the box:

$p_i = p_f\\m_1 v_2 = (m_1 + m_2) v_3$

where m2 = 15 kg is the mass of the box. Solving the equation for v3, the combined velocity of the girl+box, we find

$v_3 = \frac{m_1 v_2}{m_1 + m_2}=\frac{(60 kg)(5.0 m/s)}{60 kg+15 kg}=4 m/s$

c) 2.8 m

We can use again the law of conservation of energy. The total mechanical energy of the girl after she catches the box is sum of kinetic energy and potential energy:

$E_i = \frac{1}{2}(m_1+m_2) v_3^2 + (m_1+m_2)gh=\frac{1}{2}(75 kg)(4 m/s)^2+(75 kg)(9.8 m/s^2)(2.0m)=2070 J$

While at the maximum height, the speed is zero, so all the mechanical energy is just potential energy:

$E_f = (m_1 +m_2)gh_{max}$

where h_max is the maximum height. Equalizing the two expressions (because the mechanical energy must be conserved) and solving for h_max, we find

$E_i = (m_1+m_2)gh_{max}\\h_{max}=\frac{E_i}{(m_1+m_2)g}=\frac{2070 J}{(75 kg)(9.8 m/s^2)}=2.8 m$

4 0

3 years ago

Students measured the mass of 25.0 mL of water and found it be 25.4 g. The accepted mass is 25.0 g. What is the percent error of

Andre45 [30]

Well first of all, I think the students may have been correct.
If they didn't use distilled water, and if it wasn't exactly at
standard temperature, then the mass of 25.0 mL could
very well be 25.4 grams. We don't know that there was
any 'error' in their measurement at all.
But the question says there was, so we'll do the math:

The 'error' was (25.4 - 25.0) = +0.4 gram

As a fraction of the 'real' value, the error was

+0.4 / 25.0 = +0.016 .

To change a decimal to a percent, move the
decimal point two places that way ===> .

+ 0.016 = +1.6 % .

Their measurement was 1.6% too high.

Let's not call it an 'error'. Let's just call it a 'discrepancy'
between the measured value and the 'accepted' value. OK ?

4 0

3 years ago

Read 2 more answers

I Need help with this problem i don’t know what to do

Mazyrski [523]

Answer:

The density of the sample is 36 g/cm³

Explanation:

m= 972g

l=3cm

V = l³ = 3³ = 27 cm³

density = mass/volume

= 972/27

= 36 g/cm³

8 0

3 years ago

The net torce on an object moving with constant speed in circular motion is in which direction?

aleksley [76]

The correct answer is C) towards the center of the circle.

Although the object is moving at a constant speed it is constantly accelerating due to the constant change in direction as it describes the circular path. This causes a constant change in velocity as velocity is a vector quantity.

For the object to maintain the circular path there has to be centripetal force acting on the object and this centripetal force is directed towards the center of the circle.

6 0

3 years ago