Acceleration a=3m/s^2
time t= 4.1seconds
Final velocity V= 55km/h
initial velocity U= ?
First convert V to m/s
36km/h=10m/s
55km/h= 55*10/36=15.28m/s
Using the formula V= U+at
U= V-at
U= 15.28-3*4.1=15.28-12.3=2.98m/s
Initial velocity U= 2.98m/s or 10.73km/h (Using the conversion rate 36km/h=10m/s)
Answer:
-2.26×10^-4 radians
Explanation:
The solution involves a right angle triangle
Length is z while the horizontal is the height x
X^2+ 100^2=z^2
Taking the derivatives
2x(dx/dt)=Z^2(dz/dt)
Specific moments = Z= 200 ,X= 100sqrt3 and dx/dt= 11
dz/dt= 1100sqrt3/200 = 9.53
Sin a= 100/a
Taking derivatives in terms of t
Cos a(da/dt)=100/z^2 dz/dt
a= 30°
Cos (30°)da/dt= (-100/40000×9.5)
a= -2.26×10^-4radians
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
