Answer:
0 kg m/s before and after collision
Explanation:
Let m, v be the mass and speed of the 2 balls, respectively, before the collision. Since they have the same mass and same speed but in opposite direction, the total momentum of the system would be:
P = mv - mv = 0 kg m/s
As the collision is elastic. The total momentum after the collision is the same as the total momentum before the collision, which is 0.
<span>Answer:
KE = (11/2)mω²r²,
particle B must have mass of 2m, while A has mass m.
Then the moment of inertia of the system is
I = Σ md² = m*(3r)² + 2m*r² = 11mr²
and then
KE = ½Iω² = ½ * 11mr² * ω² = 11mr²ω² / 2
So I'll proceed under that assumption.
For particle A, translational KEa = ½mv²
but v = ω*d = ω*3r, so KEa = ½m(3ωr)² = (9/2)mω²r²
For particld B, translational KEb = ½(2m)v²
but v = ω*r, so KEb = ½(2m)ω²r²
so total translational KE = (9/2 + 2/2)mω²r² = 11mω²r² / 2
which is equal to our rotational KE.</span>
Answer:
(a) F= 6.68*10¹¹⁴ N (-k)
(b) F =( 6.68*10¹¹⁴ i + 7.27*10¹¹⁴ j ) N
Explanation
To find the magnetic force in terms of a fixed amount of charge q that moves at a constant speed v in a uniform magnetic field B we apply the following formula:
F=q* v X B Formula (1 )
q: charge (C)
v: velocity (m/s)
B: magnetic field (T)
vXB : cross product between the velocity vector and the magnetic field vector
Data
q= -1.24 * 10¹¹⁰ C
v= (4.19 * 10⁴ m/s)î + (-3.85 * 10⁴m/s)j
B =(1.40 T)i
B =(1.40 T)k
Problem development
a) vXB = (4.19 * 10⁴ m/s)î + (-3.85* 10⁴m/s)j X (1.40 T)i =
= - (-3.85*1.4) k = 5.39* 10⁴ m/s*T (k)
1T= 1 N/ C*m/s
We apply the formula (1)
F= 1.24 * 10¹¹⁰ C* 5.39* 10⁴ m/s* N/ C*m/s (-k)
F= 6.68*10¹¹⁴ N (-k)
a) vXB = (4.19 * 10⁴ m/s)î + (-3.85* 10⁴m/s)j X (1.40 T)k =
=( - 5.39* 10⁴i - 5.87* 10⁴j)m/s*T
1T= 1 N/ C*m/s
We apply the formula (1)
F= 1.24 * 10¹¹⁰ C* ( 5.39* 10⁴i + 5.87* 10⁴j) m/s* N/ C*m/s
F =( 6.68*10¹¹⁴ i + 7.27*10¹¹⁴ j ) N
Answer:
7 deg
Explanation:
= mass of the rod = 
= weight of the rod = 
= spring constant for left spring = 
= spring constant for right spring = 
= stretch in the left spring
= stretch in the right spring
= length of the rod = 0.75 m
= Angle the rod makes with the horizontal
Using equilibrium of force in vertical direction for left spring
Using equilibrium of force in vertical direction for right spring
Angle made with the horizontal is given as
Answer:
The wire meet the ground at an angle of 56.4 degrees
Explanation:
It is given that,
To support a tree damaged in a storm, a 12-foot wire is secured from the ground to the tree at a point 10 feet off the ground.
The hypotenuse is, H = 12 foot
The perpendicular distance is, P = 10 feet
The angle between the tree and the ground is 90 degrees
Using Pythagoras theorem as :
So, the wire meet the ground at an angle of 56.4 degrees. Hence, the correct option is (d).
