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ehidna [41]
2 years ago
6

The electric field inside a cell membrane is 8.0 MN/C. Part A What's the force on a singly charged ion in this field?

Physics
1 answer:
dlinn [17]2 years ago
7 0

Answer:

Force on the singly charged ion will be 12.6\times 10^{-13}N

Explanation:

We have given electric field E = 8 MN/C

So electric field in N/C will be E=8MN/C=8\times 10^6N/C

It is given that ion so charge on ion will be equal to e=1.6\times 10^{-19}C

We have to find the electric force on the ion

Electric force is equal to F=qE, here q is charge and E is electric field

So force on the charge will be equal to F=8\times 10^6\times 1.6\times 10^{-19}=12.6\times 10^{-13}N

So force on the singly charged ion will be 12.6\times 10^{-13}N

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A toy car of mass 0.15kg accelerates from a speed of 10 cm/s to a speed of 15 cm/s. What is the impulse acting on the car?
OLga [1]

v=v2-v1=15-10=5 cm/s

p=mv=0.15•5=0.75 kg•cm/s

7 0
2 years ago
Read 2 more answers
Two flat 4.0 cm × 4.0 cm electrodes carrying equal but opposite charges are spaced 2.0 mm apart with their midpoints opposite ea
serious [3.7K]

Answer:

1.77 x 10^-8 C

Explanation:

Let the surface charge density of each of the plate is σ.

A = 4 x 4 = 16 cm^2 = 16 x 10^-4 m^2

d = 2 mm

E = 2.5 x 10^6 N/C

ε0 = 8.85 × 10-12 C2/N ∙ m2

Electric filed between the plates (two oppositively charged)

E = σ / ε0

σ = ε0 x E

σ = 8.85 x 10^-12 x 2.5 x 10^6 = 22.125 x 10^-6 C/m^2

The surface charge density of each plate is ± σ / 2

So, the surface charge density on each = ± 22.125 x 10^-6 / 2

                                                                 = ± 11.0625 x 10^-6 C/m^2  

Charge on each plate = Surface charge density on each plate x area of each plate

Charge on each plate = ± 11.0625 x 10^-6  x 16 x 10^-4 = ± 1.77 x 10^-8 C

7 0
2 years ago
A 3kW oven supplied with 9mJ of energy.How many minutes can it run for?
andreev551 [17]

<span>Hello!
 
We have the following data:
</span>
Time (T) = ? (in minutes)
Power (P) = 3 kW → 3000 W
Energy (E) = 9 MJ → 9000000 J or (W/s)

Formula of the consumption of electric energy:

P =  \frac{E}{T}

Solving:

P = \frac{E}{T}

P = \frac{E}{T} \to T =  \frac{E}{P}

T =  \frac{9000\diagup\!\!\!\!0\diagup\!\!\!\!0\diagup\!\!\!\!0\:\diagup\!\!\!\!W/s}{3\diagup\!\!\!\!0\diagup\!\!\!\!0\diagup\!\!\!\!0\:\diagup\!\!\!\!W}

\boxed{T = 3000\:seconds}

How many minutes can it run for? (<span>Let's convert in minutes)
</span>
1 minute --------- 60 seconds
y minute --------- 3000 seconds

\frac{1}{y} = \frac{60}{3000}

<span>Product of extremes equals product of means
</span>
60*y = 1*3000

60y = 3000

y =  \frac{3000}{60}

\boxed{\boxed{y = 50\:minutes}}\end{array}}\qquad\quad\checkmark


I hope this helps! =)
<span>

</span>
7 0
2 years ago
Read 2 more answers
Pions have a half-life of 1.8 x 10^-8 s. A pion beam leaves an accelerator at a speed of 0.8c. What is the expected distance ove
Nuetrik [128]

Answer:

the expected distance is 4.32 m

Explanation:

given data

half life time = 1.8 × 10^{-8} s

speed = 0.8 c = 0.8 × 3 × 10^{8}

to find out

expected distance over

solution

we know c is speed of light in air is 3 × 10^{8} m/s

we calculate expected distance by given formula that is

expected distance = half life time × speed   .........1

put here all these value

expected distance = half life time × speed

expected distance = 1.8 × 10^{-8} ×  0.8 × 3 × 10^{8}

expected distance = 4.32

so the expected distance is 4.32 m

5 0
2 years ago
If Emily throws the ball at an angle of 30∘ below the horizontal with a speed of 14m/s, how far from the base of the dorm should
liubo4ka [24]

Emily throws the ball at 30 degree below the horizontal

so here the speed is 14 m/s and hence we will find its horizontal and vertical components

v_x = 14 cos30 = 12.12 m/s

v_y = 14 sin30 = 7 m/s

vertical distance between them

\delta y = 4 m

now we will use kinematics in order to find the time taken by the ball to reach at Allison

\delat y = v_y *t + \frac{1}{2} at^2

here acceleration is due to gravity

a = 9.8 m/s^2

now we will have

4 = 7 * t + \frac{1}{2}*9.8 * t^2

now solving above quadratic equation we have

t = 0.44 s

now in order to find the horizontal distance where ball will fall is given as

d = v_x * t

here it shows that horizontal motion is uniform motion and it is not accelerated so we can use distance = speed * time

d = 12.12 * 0.44 = 5.33 m

so the distance at which Allison is standing to catch the ball will be 5.33 m

8 0
2 years ago
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