Answer:
When an object moves in a straight line, it is said to be in linear motion. By Newton's first law of motion, a body tends to be in rest or motion in a straight line until a net non-zero force acts on it.
Rate of change of position with respect to time is known as velocity. Uniformly accelerated motion refers to the motion where the rate of change of velocity with respect to time is constant.
Kinematic equations can be used to measure different aspects of a linear motion:
v = u + a t
s = u t + 0.5 a t²
v²= u² + 2 a s
where, u is initial velocity, v is final velocity, a is acceleration, t is time and s is displacement.
Answer:
So, for noble gases, they already have their outermost shells filled completely so they need not donate or accept any electrons. For example: Argon has atomic no 18 ,i.e E.C = 2,8,8. They are already stable. So, they don't need to form ions.
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Answer:
Acceleration(a) = 2.588 m/s²
TL = 230.784 N
TR = 220.5 N
Explanation:
Given:
M = 7.95 kg
mL = 32 kg
mR = 17.8 kg
g = 9.8 m/s²
Find:
Acceleration(a)
TL
TR
Computation:
Acceleration(a) = [(mL - mR)g] / [mL + mR + M/2]
Acceleration(a) = [(32 - 17.8)9.8] / [32 + 17.8 + 7.95/2]
Acceleration(a) = [139.16] / [53.775]
Acceleration(a) = 2.588 m/s²
TL = mL(g-a)
TL = 32(9.8-2.588)
TL = 230.784 N
TR = mR(g+a)
TR = 17.8(9.8+2.588)
TR = 220.5 N
Answer:
F = F₀ 0.2
Explanation:
For this exercise we apply Coulomb's law with the initial data
F₀ = k q_A q_B / d²
indicate several changes
q_A ’= ½ q_A
q_B ’= 1/10 q_B
d ’= ½ d
let's substitute these new values in the Coulomb equation
F = k q_A ’q_B’ / d’²
F = k ½ q_A 1/10 q_B / (1/2 d)²
F = (k q_A q_B / d2) ½ 1/10 2²
F = F₀ 0.2
B- the acceleration is greater for the more massive rock