Hi there!
Impulse = Change in momentum
I = Δp = mΔv = m(vf - vi)
Where:
m = mass of object (kg)
vf = final velocity (m/s)
vi = initial velocity (m/s)
Begin by converting grams to kilograms:
1 kg = 1000g ⇒ 145g = .145kg
Now, plug in the given values. Remember to assign directions since velocity is a vector. Let the initial direction be positive and the opposite be negative.
I = (.145)(-20 - 17) = -5.365 Ns
The magnitude is the absolute value, so:
|-5.365| = 5.365 Ns
Answer:
Explanation:
The unknown charge can not remain in between the charge given because force on the middle charge will act in the same direction due to both the remaining charges.
So the unknown charge is somewhere on negative side of x axis . Its charge will be negative . Let it be - Q and let it be at distance - x on x axis.
force on it due to rest of the charges will be equal and opposite so
k3q Q / x² =k 8q Q / (L+x)²
8x² = 3 (L+x)²
2√2 x = √3 (L+x)
2√2 x - √3 x = √3 L
x(2√2 - √3 ) = √3 L
x = √3 L / (2√2 - √3 )
Let us consider the balancing force on 3q
force on it due to -Q and -8q will be equal
kQ . 3q / x² = k3q 8q / L²
Q = 8q (x² / L²)
so charge required = - 8q (x² / L²)
and its distance from x on negative x side = √3 L / (2√2 - √3 )
Answer:
sure, do you have more detailed of the assignments...?
Explanation:
The bearing could be the below:
oppositely charged, same initial direction
same charge, opposite initial direction
You can decide by utilizing your correct hand and put your fingers toward the attractive field (North to South). Thumb toward present or charged molecule. The course of your palm will demonstrate the heading of compelling set on a decidedly charged molecule and the bearing of the back of your hand will demonstrate the bearing of a contrarily charged molecule.
