The horizontal force applied is 160 N while the velocity is 2.03 m/s.
<h3>What is the speed of the car?</h3>
The work done by the car is obtained as the product of the force and the distance;
W = F x
F = ?
x = 30.0 m
W = 4,800 J
F = 4,800 J/30.0 m
F = 160 N
But F = ma
a = F/m
a = 160 N/2.30 ✕ 10^3-kg
a= 0.069 m/s
Now;
v^2 = u^2 + 2as
u = 0/ms because the car started from rest
v = √2as
v = √2 * 0.069 * 30
v = 2.03 m/s
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Expansion work against constant external pressure: w=-pex Δ Δ V 3. The attempt at a solution . I tried following that. Because Vf>>Vi, and Vf=nRT/pex, then w=-pex x nRT/pex=-nRT (im assuming n is number of moles of CO2?). 1 mole of CaCO3 makes 1 mole of CO2, so plugging in numbers, I get 8.9kJ, although I dont use the 1 atm pressure at all
The Voyager and Pioneer flybys of the 1970s and 1980s provided rough sketches of Saturn’s moons. But during its many years in Saturn orbit, Cassini discovered previously unknown moons, solved mysteries about known ones, studied their interactions with the rings and revealed how sharply different the moons are from one another.
This question is incomplete, the complete question is;
An air-filled capacitor stores a potential energy of 6.00 mJ due to its charge. It is accidentally filled with water in such a way as not to discharge its plates. How much energy does it continue to store after it is filled?
(The dielectric constant for water is 78 and for air it is 1.0006.)
Answer: it continue to store 0.07692 mJ after it was filled
Explanation:
Given that;
stored potential energy = 6.00 mJ = 0.006 J
dielectric constant for water K = 78
Energy stored U = Q² / 2C = 0.006 J
C = ∈₀A/d { Air}
C = K∈₀A/d { Water, k = 78 }
so
U = 0.006 / 78
U = 7.6723 × 10⁻⁵J
U = 0.07692 mJ
Therefore it continue to store 0.07692 mJ after it was filled
