a rifle is fired and recoils when the bullet leaves the gun. this is an example of newton's 3rd law. the force on the bullet is.
2 answers:
You might be interested in
Answers:
A. A car driving at steady speed on a straight and level road.
B. A car driving at steady speed up a 10∘ incline.
Explanation:
An object is said to be in an inertial reference frame if the net force acting on the object is zero. According to Newton's second law, this also means that the acceleration of the object is also zero:
Since F=0, a=0 as well.
Let's now analyze each case.
A. A car driving at steady speed on a straight and level road. --> YES: this is an inertial reference frame, because the car is keeping a constant speed and a constant direction, so its velocity is not changing, and its acceleration is zero.
B. A car driving at steady speed up a 10∘ incline. --> YES: this is an inertial reference frame, because the car is keeping a constant speed and a constant direction, so its velocity is not changing, and its acceleration is zero.
C. A car speeding up after leaving a stop sign. --> NO: this is not an intertial reference frame, because the car is speeding up, so it is accelerating.
D. A car driving at steady speed around a curve. --> NO: this is not an inertial reference frame, because the car is changing direction, therefore its velocity is changing and so the car is accelerating.
So the only two choices which are correct are A and B.
Answer:
Explanation:
Given that,
Mass of the heavier car m_1 = 1750 kg
Mass of the lighter car m_2 = 1350 kg
The speed of the lighter car just after collision can be represented as follows
b) the change in the combined kinetic energy of the two-car system during this collision
substitute the value in the equation above
Hence, the change in combine kinetic energy is -2534.78J
Answer: 39.8 μC
Explanation:
The magnitude of the electric field generated by a capacitor is given by:
d is the distance between the plates.
For a capacitor, charge Q = CV where C is the capacitance and V is the voltage.
where A is the area of the plate and ε₀ is the absolute permittivity.
substituting, we get
It is given that the magnitude of the electric field that can exist in the capacitor before air breaks down is, E = 3 × 10⁶ N/C.
radius of the plates of the capacitor, r = 69 cm = 0.69 m
Area of the plates, A = πr² = 1.5 m²
Thus, the maximum charge that can be placed on disks without a spark is:
Q = E×ε₀×A
⇒ Q = 3 × 10⁶ N/C × 8.85 × 10⁻¹² F/m × 1.5 m² = 39.8 × 10⁻⁶ C = 39.8 μC.