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3 years ago

When cell A sends a message to

Physics

1 answer:

IRISSAK [1]3 years ago

3 0

Answer:

C synaptic transmission

Explanation:

Hope this helped :)

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A time-dependent but otherwise uniform magnetic field of magnitude B0(t) is confined in a cylindrical region of radius 6.5 cm. I

vodka [1.7K]

Answer:

a = 603.59 m/s^2

Explanation:

from the data given . the rate of change in magnetic field is as follow

$\frac{dB}{dt} = 280 G/s = 280 \times 10^{-4} T/s$

from the faraday's law of induction , the expression for the induced emf in region of radius r as follow

$\epsilon = \frac{d \phi}{dt}$

$\int E.dl = \frac{d(BA)}{dt}$

$E(2\pi r)= \pi r^2 \frac{dB}{dt}$

$E = \frac{r}{2} \frac{dB}{dt}$

electric field at point P_1 as follow

$E = \frac{r}{2} \frac{dB}{dt}$

$E = \frac{1.5\times 10^{-2}}{2} 280 \times 10^{-4}$

$E = 6.3\times 10^{-6} V/m$

from newton 2nd law of motion, the acceleration of proton is

F = ma

qE = ma

$a = \frac{qE}{m}$

$a = \frac{1.6 \times 10^{-19} (6.3\times 10^{-6})}{1.67\times 10^{-27}}$

a = 603.59 m/s^2

5 0

3 years ago

Choose the correct association for: dense bushes<br><br> savanna<br> rain forest or jungle

Gemiola [76]

Answer:

Climate is determined by averaging the seasonal weather conditions for a region over a period of many ______ years

Choose the correct association for: dense bushes rain forest or jungle

Choose the correct association for: plains savanna

3 0

2 years ago

Hey could i have some Help??

bagirrra123 [75]

Answer:

it is biology i think

4 0

3 years ago

Read 2 more answers

Imagine a particle that has three times the mass of the electron. All other quantities given above remain the same. What is the

melamori03 [73]

Answer:

The only parameter that changes is mass m

It is only necessary to calculate the ratio Eh/Ee

$m_{h}=3m_{e}\\E_{h}=\frac{3m_{e}v^{2}}{2}\\E_{e}=\frac{m_{e}v^{2}}{2}\\\frac{E{h}}{E{e}}=3$

The kinetic energy of the heavy paricle is three times the kinetic energy of an electron

5 0

3 years ago

Three point charges have equal magnitudes, two being positive and one negative. These charges are fixed to the corners of an equ

Irina-Kira [14]

Answer:

Magnitude of the net force on q₁-

Fn₁=1403 N

Magnitude of the net force on q₂+

Fn₂= 810 N

Magnitude of the net force on q₃+

Fn₃= 810 N

Explanation:

Look at the attached graphic:

The charges of the same sign exert forces of repulsion and the charges of opposite sign exert forces of attraction.

Each of the charges experiences 2 forces and these forces are equal and we calculate them with Coulomb's law:

F= (k*q*q)/(d)²

F= (9*10⁹*3*10⁻⁶*3*10⁻⁶)(0.01)² =810N

Magnitude of the net force on q₁-

Fn₁x= 0

Fn₁y= 2*F*sin60 = 2*810*sin60° = 1403 N

Fn₁=1403 N

Magnitude of the net force on q₃+

Fn₃x= 810- 810 cos 60° = 405 N

Fn₃y= 810*sin 60° = 701.5 N

$Fn_{3} = \sqrt{405^{2}+701.5^{2} }$

Fn₃ = 810 N

Magnitude of the net force on q₂+

Fn₂ = Fn₃ = 810 N

6 0

3 years ago