D lung cancer is not infectious
Answer:
140265.8 C = 1.403 × 10⁵ C
Explanation:
The battery's electric potential energy is used to account for the kinetic and potential work done in moving the car up this hill.
Potential work required to move the 757 kg car up a vertical height of 195 m = mgh
P.E = 757 × 9.8 × 195 = 1446627 J
Kinetic work done = (1/2)(m)(v²)
K.E = (1/2)(757)(25²) = 236562.5 J
Total work done in moving the car up that height = 1446627 + 236562.5 = 1683189.5 J
And this would be equal to the potential of the battery.
For the battery, potential difference = (electric potential energy)/(charges moved)
ΔV = ΔU/q
q = ΔU/ΔV
ΔU = 1683189.5 J
ΔV = 12.0 V
q = 1683189.5/12 = 140265.8 C
Eleven truly inspiring adjectives !
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Do you have a question to ask ?
Answer:
a) x = 4.33 m
, b) w = 2 rad / s
, f = 0.318 Hz
, c) a = - 17.31 cm / s²,
d) T = 3.15 s, e) A = 5.0 cm
Explanation:
In this exercise on simple harmonic motion we are given the expression for motion
x = 5 cos (2t + π / 6)
they ask us for t = 0
a) the position of the particle
x = 5 cos (π / 6)
x = 4.33 m
remember angles are in radians
b) The general form of the equation is
x = A cos (w t + Ф)
when comparing the two equations
w = 2 rad / s
angular velocity and frequency are related
w = 2π f
f = w / 2π
f = 2 / 2pi
f = 0.318 Hz
c) the acceleration is defined by
a == d²x / dt²
a = - A w² cos (wt + Ф)
for t = 0
, we substitute
a = - 5,0 2² cos (π / 6)
a = - 17.31 cm / s²
d) El period is
T = 1/f
T= 1/0.318
T = 3.15 s
e) the amplitude
A = 5.0 cm