Partially correct answer iconYour answer is partially correct. A proton initially has and then 2.30 s later has (in meters per s
econd).
(a) For that 2.30 s, what is the proton's average acceleration in unit vector notation,
(b) in magnitude, and
(c) the angle between and the positive direction of the x axis
(a) For that 2.30 s, what is the proton's average acceleration in unit vector notation,
(b) in magnitude, and
(c) the angle between and the positive direction of the x axis
1 answer:
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Answer:
(4xy+5ab)(4xy-5ab)
Explanation:
16-25
4^2 is 16 and 5^2 is 25,
Also, (x-a)(x+a) = x^2-a^2
So, this factorized is:
(4xy+5ab)(4xy-5ab)
Hope this helps!
Answer:
0.04225 Nm
Explanation:
N = Force applied = 5 N
= Coefficient of static friction = 0.65
d = Diameter of knob = 1.3 cm
r = Radius of knob =
Force is given by
When we multiply force and radius we get torque
Torque on thumb
Torque on forefinger
The total torque is given by
The most torque that exerted on the knob is 0.04225 Nm