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2 years ago

What is the ideal mechanical advantage of a lift that allows you raise a 90.0-kg refrigerator with 80.0 N of force?

Physics

1 answer:

Sholpan [36]2 years ago

5 0

Answer:

Explanation:

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Suppose that the resistance between the walls of a biological cell is 6.8 × 109 ω. (a) what is the current when the potential di

GenaCL600 [577]

(a) We can find the current flowing between the walls by using Ohm's law:
$I= \frac{\Delta V}{R}$
where $\Delta V=69 mV=0.069 V$ is the potential difference and $R=6.8\cdot 10^9 \Omega$ is the resistance. Substituting these values, we get
$I=1.01 \cdot 10^{-11} A$

(b) The total charge flowing between the walls is the product between the current and the time interval:
$Q=I \Delta t$
The problem says $\Delta t=0.86 s$ , so the total charge is
$Q=(1.01\cdot 10^{-11} A)(0.86 s)=8.73 \cdot 10^{-12} C$

The current consists of Na+ ions, each of them having a charge of $e=1.6 \cdot 10^{-19} C$ . To find the number of ions flowing, we can simply divide the total charge by the charge of a single ion:
$N= \frac{Q}{e} = \frac{8.73 \cdot 10^{-12}C}{1.6 \cdot 10^{-19}C} = 5.45 \cdot 10^7 ions$

4 0

2 years ago

Can someone please help me?? i’ll give brainilist

Ksivusya [100]

Probably 90 j but im not sure I haven’t done any work like this in a while

7 0

2 years ago

A man is running on the straight road with the uniform velocity of 3 metre per second calculate the acceleration for produced hi

sp2606 [1]

Theoritically

the body moving with uniform velocity has acceleration zero.

Mathmatically,

u=3m/s

v=3m/s (since body is moving with uniform velocity)

a= v-u/t

3-3/t

0/t

0m/s.s

6 0

2 years ago

PLEASE HELP I NEED THIS TO PASS THE EIGHTH GRADE AND I ONLY HAVE A COUPLE HOURS LEFT!!!!!!

zavuch27 [327]

Answer:

the pe at the top of the building: 784 J

the pe halfway through the fall: 392 J

the pe just before hitting the ground: 784 J

Explanation:

Pls brainliest me

I had this question before

7 0

3 years ago

A 0.9 µF capacitor is charged to a potential difference of 10.0 V. The wires connecting the capacitor to the battery are then di

jeyben [28]

Answer:

3.6μF

Explanation:

The charge on the capacitor is defined by the formula

q = CV

because the charge will be conserved

q₁ = C₁V₂

q₂ = C₂V₂ where C₂ V₂ represent the charge on the newly connected capacitor and the voltage drop across the two capacitor will be the same

q = q₁ + q₂ = C₁V₂ + C₂V₂

CV = CV₂ + C₂V₂

CV - CV₂ = C₂V₂

C ( V - V₂) = C₂V₂

C ( V/ V₂ - V₂ /V₂) = C₂

C₂ = 0.9 ( 10 /2) - 1) = 0.9( 5 - 1) = 3.6μF

7 0

3 years ago