Atomic Number
or
Number of Protons
ΩΩΩΩΩΩΩΩΩΩ
Answer:
For example, when a car travels at a constant speed, the driving force from the engine is balanced by resistive forces such as air resistance and friction in the car's moving parts. The resultant force on the car is zero.
Explanation:
hope this helps
Answer:
6.003×10¯⁶ N
Explanation:
We'll begin by converting 1 cm to m. This can be obtained as follow:
100 cm = 1 m
Therefore,
1 cm = 1 cm × 1 m / 100 cm
1 cm = 0.01 m
Finally, we shall determine the gravitational attraction. This can be obtained as follow:
Mass 1 (M₁) = 3 Kg
Mass 2 (M₂) = 3 Kg
Distance apart (r) = 0.01 m
Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²
Force of attraction (F) =?
F = GM₁M₂ / r²
F = 6.67×10¯¹¹ × 3 × / 0.01²
F = 6.003×10¯¹⁰ / 1×10¯⁴
F = 6.003×10¯⁶ N
Thus the gravitational attraction is 6.003×10¯⁶ N
Answer:
E = 1/2 M V^2 = 1/2 P V since P = M V
E2 / E1 = P2 V2 / (P1 V1)
P2 / P1 = E2 V1 / (E1 V2) = V2^2 V1 / (V1^2 V2) = V2 / V1
E2 / E1 = V2^2 / V1^2
V2 / V1 = (E2 / E1)^1/2
V2 / V1 = (.9)1/2 = .95
The linear momentum would have to decrease by 5%
Answer:
F= 0
Explanation:
This exercise we use Newton's second law,
F = m a
in this case as the speed is constant the acceleration is zero therefore the force is zero.
Change we can solve it using Newton's first law, which states that every vehicle remains still or with constant speed if there is no extensive outside acting on it
We see that with any of the two forms the sum of the applied forces is zero
∑ F = 0
