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masya89 [10]
2 years ago
6

A 380-N girl walks down a flight of stairs so that she is 2.5 m below her starting level. What is the change in the girl's gravi

tational potential energy?
Physics
2 answers:
Svetach [21]2 years ago
8 0
PEg = Mass x Height x Gravity. So it doesn't matter how much PEg she started with, since you're finding how much it changed when she walked down 2.5 meters of stairs. So by plugging into the equation, you'll find how much potential energy was lost by walking down. (I'll leave it up to you since idk if you use 10 or 9.8 as gravity)
White raven [17]2 years ago
6 0

To find gravitational potential energy you need to multiply weight * height. The weight in this situation would be 380-N. Our height is 2.5 m. 380 *2.5= 950, or 950 J of gravitational potential energy.

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Reptile [31]

Answer:

B = 7.9012*10^{-5}T

Explanation:

To solve the problem, the concepts related to the magnetic field and the current produced in a lightning bolt are necessary.

The current is defined by the load due to time, that is to say

I= \frac{q}{t}

Where,

q= Charge

t = time

So the current can be expressed as:

I = \frac{16}{1.5*10^{-3}}

I = 10666.67A

Once the current is found it is now possible to find the magnetic field, as this is given by the equation,

B =\frac{\mu_0}{2\pi}\frac{I}{r}

Where,

\mu_0 =Permeability Constant

I= Current

r= radius

Replacing the values we have

B=\frac{4\pi*10^{-7}}{2\pi}(\frac{10666.67}{27})

B = 7.9012*10^{-5}T

7 0
3 years ago
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Maru [420]

Answer:

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3 0
2 years ago
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Suppose that you have a 680 Ω, a 720 Ω and a 1.20 kΩ resistor. (a) What is the maximum resistance you can obtain by combining th
Delvig [45]

Explanation:

As the given data is as follows.

    R_{1} = 680 \ohm ohm\ohm,    R_{2} = 720 \ohm ohm,

   R_{3} = 1.2 k\ohm = 1200 \ohm   (as 1 k ohm = 1000 m)

(a)   We will calculate the maximum resistance by combining the given resistances as follows.

      Max. Resistance = R_{1} + R_{2} + R_{3}

                                  = (680 + 720 + 1200) \ohm ohm

                                  = 2600 ohm

or,                               = 2.6 k\ohm ohm

Therefore, the maximum resistance you can obtain by combining these is 2.6 k\ohm ohm.

(b)   Now, the minimum resistance is calculated as follows.

      Min. Resistance = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}

                                 = \frac{1}{680} + \frac{1}{720} + \frac{1}{1200}

                                 = 3.683 \times 10^{-3} ohm

Hence, we can conclude that minimum resistance you can obtain by combining these is 3.683 \times 10^{-3} ohm.

3 0
2 years ago
A potential difference of 107 mV exists between the inner and outer surfaces of a cell membrane. The inner surface is negative r
sergij07 [2.7K]

Answer:

The workdone is  W = 1.712 *10^{-20 } \  J  

Explanation:

From the question we are told that

    The potential difference is  V  =  107 mV =  107 *10^{-3} \  V

Generally the charge on  Na^{+} is  Q_{Na^{+}} = 1.60 *10^{-19 } \  C

 Generally the workdone is mathematically represented as

         W =  Q_{Na^{+}}V

=>     W = 1.60 *10^{-19 } *  107 *10^{-3}    

=>     W = 1.712 *10^{-20 } \  J    

8 0
2 years ago
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