Answer:
4. It is the force of the road on the tires (an external force) that stops the car.
Explanation:
If there is no friction between the road and the tires, the car won't stop.
You can see this, for example, when there is ice on the road. You can still apply the brakes (internal force), but since there is no friction (external force) the car won't stop.
The force of the brakes on the wheels is not what makes the car stop, it is the friction of the road against still tires that makes it stop.
Answer:
Area=1.5(1.5)=2.25m^2
Force of gravity=10N
\begin{gathered}\\ \sf\longmapsto Pressure=\dfrac{Force}{Area}\end{gathered}
⟼Pressure=
Area
Force
\begin{gathered}\\ \sf\longmapsto Pressure=\dfrac{10}{2.25}\end{gathered}
⟼Pressure=
2.25
10
\begin{gathered}\\ \sf\longmapsto Pressure=4.4Pa\end{gathered}
⟼Pressure=4.4Pa
The greater the cross-sectional area of an object, the greater the amount of air resistance it encounters since it collides with more air molecules. ... It will have to accelerate for a longer period of time before there is enough upward air resistance to balance the downward force of gravity.
10^9 giga, 10^6 mega, 10^3 kilo, 10^-3 milli, 10^-6 micro, 10^-9 nano, 10^-12 pico
Potentially they might want centi which is 10^-2
Answer:
0.0021576N
Explanation:
F=(k)(q1q2/r^2)
F=(8.99×10^9)(3×10^-6)(2×10^-6)/(5^2)
F=0.0021576N
