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Kipish [7]
2 years ago
11

Two point charge

Physics
1 answer:
timama [110]2 years ago
4 0

Answer:

Showing results for Two point charge q, separated by 1.5cm have change value of +2.0 and -4.0AND/C respectively what is the magnitude of the Electric force midway between them?

Search instead for Two point charge q, seperated by 1.5cm have change value of +2.0 and -4.0N/C respectively what is the magnitude of the Electric force midway between them?

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A motorcycle accelerated from 10 metre per second to 30 metre per second in 6 seconds. How far did it travel in this time ?
Ber [7]
U=10 m/s
v=30 m/s
t=6 sec

therefore, a=(v-u)/t
                   =(30-10)/6
                   =(10/3) ms^-2

now, displacement=ut+0.5*a*t^2
                              =60+ 0.5*(10/3)*36
                              =120 m
And you can solve it in another way:

v^2=u^2+2as
or, s=(v^2-u^2)/2a
       =(900-100)/6.6666666.......
       =120 m

7 0
4 years ago
Caroline, a piano tuner, suspects that a piano's G4 key is out of tune. Normally, she would play the key along with her G4 tunin
alexira [117]

Answer:

e. 425.9 Hz

Explanation:

The computation of the frequency is being played by the out-of-tune key is shown below;

Given that

Δf1  = x - 349.2 = 76.7.........(1)

Δf2 = 440 - x  = 14.1......(2)

Now solve (1) and (2)

(440 - x) - x + 349.2 = 14.1 - 76.7

789.2 + (-2x) = -62.6

x = 425.9 Hz

Hence, the frequency is being played by the out-of-tune key is 425.9 Hz

Therefore the option e is correct

6 0
3 years ago
How to calculate the slope of position time graph?
Mashcka [7]

Answer:

The slope of a position-time graph can be calculated as:

m=\frac{\Delta y}{\Delta x}

where

\Delta y is the increment in the y-variable

\Delta x is the increment in the x-variable

We can verify that the slope of this graph is actually equal to the velocity. In fact:

\Delta y corresponds to the change in position, so it is the displacement, \Delta s

\Delta x corresponds to the change in time \Delta t, so the time interval

Therefore the slope of the graph is equal to

m=\frac{\Delta s}{\Delta t}

which corresponds to the definition of velocity.

3 0
3 years ago
The modulus of elasticity for a ceramic material having 4.7 vol% porosity is 317 GPa. (a) Calculate the modulus of elasticity (i
elena-s [515]

Answer:

The answer is below

Explanation:

a) Given that the modulus of elasticity (E) = 317 GPa, to find the modulus of elasticity (in GPa) for the nonporous material (E_o), we use the formula:

E_0=\frac{E}{1-1.9P+0.9P^2}\\\\where\ P=4.7\%=0.047,hence:\\\\ E_0=\frac{317}{1-1.9(0.047)+0.9(0.047)^2}\\\\E_0=347.3\ GPa

b) If the porosity P = 11.1%, then the modulus of elasticity is:

E=E_0(1-1.9P+0.9P^2)=347.3(1-1.9(0.111)+0.9(0.111)^2)=278\ GPa

5 0
3 years ago
16 grams of ice at –32°C is to be changed to steam at 182°C. The entire process requires _____ cal. Round your answer to the nea
ICE Princess25 [194]

Answer:

12432 cal.

Explanation:

The process to change ice at -32 ºC to steam at 182 ºC can be divided into 5 steps:

1. Heat the ice to 0 ºC, which is the fusion temperature.

2. Melt the ice (obtaining liquid water), which is a process at constant pressure and temperature, so the liquid obtained is also at 0ºC.

3. Heat the liquid water from 0 ºC to 100 ºC, which is the vaporization normal temperature of the water.

4. Vaporization of all the water; this is also a process that occurs at constant pressure and temperature, so the produced steam will be at 100ºC.

5. Heat the steam from 100 ºC to 182 ºC.

Each process has a required energy, and the sum of the energy required for each and all of the steps is the total amount of energy required for the whole process:

E_T=E_1+E_2+E_3+E_4+E_5

E_1 is a heating process for the ice, so we know that the energy required is proportional to the temperature difference through the specific heat:

E_1=m*Cp_{sol}*(T_2-T_1)\\E_1=16g*0.5\frac{cal}{gC}*(0-(-32))=256cal

E_2 is a phase change process, so we do not use the specific heat (sensible heat), but the fusion heat (latent heat), so:

E_{2}=m*dh_{f}={16g*80\frac{cal}{g}}=1280cal

Analogously,

E_3=m*Cp_{liq}*(T_3-T_2)=16g*1.00\frac{cal}{gC}*(100-0)C = 1600 cal

E_{4}=m*{dh_{vap}}\\\\E_4=16g*540\frac{cal}{g} =8640cal

E_{5}=m*Cp_{vap}*(T_{5}-T_{4})\\E_{5}={16g*0.5\frac{cal}{gK}*(182-100)K}=656cal

Finally, the total energy required is:

E_T=256cal+1280cal+1600cal+8640cal+656cal\\E_T=12432cal

8 0
3 years ago
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