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alexandr402 [8]
3 years ago
13

If the velocity of a runner changes from -2 m/s to -4 m/s over a period of time, the

Physics
1 answer:
Mnenie [13.5K]3 years ago
3 0

Answer:

It will be A. So since its 2 times more the kinetic energy. But then you have to square it 2^2 = 4

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Homework help asap!!
ozzi
Where is the picture cant6 see it

7 0
3 years ago
Enny and Anne wanted to see who could throw a ball the hardest. They decided to each throw a ball against a wall as hard as they
mihalych1998 [28]

Answer:

wha t kind of variables

independent variables

dependent variables

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4 0
2 years ago
In a simple RC circuit, at t=0 the switch is closed with the capacitor uncharged. If C=30µF, =50V and R=10k, what is the poten
sergij07 [2.7K]

Answer:

Voltage across the capacitor is 30 V and rate of energy across the capacitor is 0.06 W

Explanation:

As we know that the current in the circuit at given instant of time is

i = 2.0 mA

R = 10 k ohm

now we know by ohm's law

V = iR

V = (2 mA)(10 kohm)

V = 20 volts

so voltage across the capacitor + voltage across resistor = V

V_c + 20 = 50

V_c = 30 V

Now we know that

U = \frac{q^2}{2C}

here rate of change in energy of the capacitor is given as

\frac{dU}{dt} = \frac{q}{C} \frac{dq}{dt}

\frac{dU}{dt} = (30)(2 mA)

\frac{dU}{dt} = 0.06 W

3 0
3 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
3 years ago
The latent heat of fusion for Aluminium is 3.97 x 105. How much energy would be required to melt 0.75 kg of it?
RoseWind [281]

Answer:

E = 2.9775\times10^5 J

Explanation:

Given:  The latent heat of fusion for Aluminum is L = 3.97\times10^5  J/Kg

mass to be malted m = 0.75 Kg

Energy require to melt E = mL

E = 3.97\times10^5\times0.75 = 2.9775\times10^5 J

Therefore, energy required to melt 0.75 Kg aluminum

E = 2.9775\times10^5 J

5 0
3 years ago
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