The car's speed was zero at the beginning of the 12 seconds,
and 18 m/s at the end of it. Since the acceleration was 'uniform'
during that time, the car's average speed was (1/2)(0 + 18) = 9 m/s.
12 seconds at an average speed of 9 m/s ==> (12 x 9) = 108 meters .
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That's the way I like to brain it out. If you prefer to use the formula,
the first problem you run into is: You need to remember the formula !
The formula is D = 1/2 a T²
Distance = (1/2 acceleration) x (time in seconds)²
Acceleration = (change in speed) / (time for the change)
= (18 m/s) / (12 sec)
= 1.5 m/s² .
Distance = (1/2 x 1.5 m/s²) x (12 sec)²
= (0.75 m/s²) x (144 sec²) = 108 meters .
Answer:
W= 4.89 KJ
Explanation:
Lets take
temperature of hot water T₁ = 100⁰C
T₁ = 373 K
Temperature of cold ice T₂= 0⁰C
T₂ = 273 K
The latent heat of ice LH= 334 KJ
The heat rejected by the engine Q= m .LH
Q₂= 0.04 x 334
Q₂= 13.36 KJ
Heat gain by engine = Q₁
For Carnot engine
Q₁ = 18.25 KJ
The work W= Q₁ - Q₂
W= 18.25 - 13.36 KJ
W= 4.89 KJ
Answer:
0.015m^3
Explanation:
1 m^3 = 1000 liters
x m^3 = 15 liters
Cross multiply
xm^3 x 1000 l = 15 l
Divide both sides by 1000
xm^3 x1000/1000 = 15/1000
xm^3 = 0.015m^3
Therefore 15 liter = 0.015m^3
The charge on the electron is 1.6x10^-19C. So, 10^24 of them will be a charge of 1.6x10^5C, F = q1xq2/[(4pi epsilon nought)r^2]
Answer:
Explanation:
Electron information needed to solve the question:
![E=\frac{9.11x10{-31}kg*3.0x10^{12}m/s^2}{-1.6x10{-19}C}-[(19.0x10^3mj+18.0x10^3m)xi(400x10^{-6}T)]](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B9.11x10%7B-31%7Dkg%2A3.0x10%5E%7B12%7Dm%2Fs%5E2%7D%7B-1.6x10%7B-19%7DC%7D-%5B%2819.0x10%5E3mj%2B18.0x10%5E3m%29xi%28400x10%5E%7B-6%7DT%29%5D)
![E=-i17.08N/C-[7.6(-k)+7.2(j)]N/C](https://tex.z-dn.net/?f=E%3D-i17.08N%2FC-%5B7.6%28-k%29%2B7.2%28j%29%5DN%2FC)
