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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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A dog runs down his driveway with an initial speed of 5 m/s for 8 s, and then uniformly increases his speed to 10 m/s in 5 s.

MakcuM [25]

Acceleration is given by change in velocity divided by change in time, so his acceleration should just be (10-5)/5 which is [tex] \frac{5}{2} \frac{m}{s^{2}} [tex]

The driveway is 40 meters plus 225/4. You can do the math.

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3 years ago

Is there supposed to be a slit at the top of your adams apple?

Allushta [10]

No, there isn't. Please consult your doctor if this is the case with yours or someone you know.

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2 years ago

A car traveling at 60 mph has how much more energy than a car going at 15 mph?

VikaD [51]

Answer:

No, if a car is going faster. The RPM is obviously higher. If that is higher, you can burn through gas and energy much faster. A car going at 15mph would be cruising and wouldn't have to worry too much about burning our your vehicle.

Explanation:

7 0

2 years ago

(II) You buy a 75-W lightbulb in Europe, where electricity is delivered at 240 V. If you use the bulb in the United States at 12

Elodia [21]

Answer:

Explanation:

You are looking for the resistance to start with

W = E * E/R

75 = 240 * 240 / R

75 * R = 240 * 240

R = 240 * 240 / 75

R = 57600 / 75

R = 768

Now let's see what happens when you try putting this into 110

W = E^2 / R

W = 120^2 / 768

W = 18.75

So the wattage is rated at 75. 18.75 is a far cry from that. I think they intend you to set up a ratio of

18.75 / 75 = 0.25

This is the long sure way of solving it. The quick way is to realize that the voltage is the only thing that is going to change. 120 * 120 / (240 * 240) = 1/2*1/2 = 1/4 = 0.25

4 0

2 years ago

Consider a transformer. used to recharge rechargeable flashlight batteries, that has 500 turns in its primary coil, 3 turns in i

Rashid [163]

Answer:

<em>a) 0.72 V</em>

<em>b) 19.2 mA</em>

<em>c) 2.304 Watts</em>

Explanation:

A transformer is used to step-up or step-down voltage and current. It uses the principle of electromagnetic induction. When the primary coil is greater than the secondary coil, the it is a step-down transformer, and when the primary coil is less than the secondary coil, the it is a step-up transformer.

number of primary turns = $N_{p}$ = 500 turns

input voltage = $V_{p}$ = 120 V

number of secondary turns = $N_{s}$ = 3 turns

output voltage = $V_{s}$ = ?

using the equation for a transformer

$\frac{V_{s} }{V_{p} } = \frac{N_{s} }{N_{p} }$

substituting values, we have

$\frac{V_{s} }{120 } = \frac{3 }{500} }$

$500V_{p} = 120*3\\500V_{p} = 360$

$V_{p}$ = 360/500 =<em> 0.72 V</em>

<em></em>

b) by law of energy conservation,

$I_{P}V_{p} = I_{s}V_{s}$

where

$I_{p}$ = input current = ?

$I_{s}$ = output voltage = 3.2 A

$V_{s}$ = output voltage = 0.72 V

$V_{p}$ = input voltage = 120 V

substituting values, we have

120 $I_{p}$ = 3.2 x 0.72

120 $I_{p}$ = 2.304

$I_{p}$ = 2.304/120 = 0.0192 A

= <em>19.2 mA</em>

<em></em>

c) power input = $I_{p} V_{p}$

==> 0.0192 x 120 = <em>2.304 Watts</em>

7 0

2 years ago