Answer:
Horizontal component of velocity = 241.61 m/s
Explanation:
Given that,
Velocity of a jetliner is 245 m/s
The vertical component of the plane's velocity is 40.6 m/s, 
We need to find the magnitude of the horizontal component of the plane velocity.
The resultant velocity is given by :

Where,
is the horizontal component of the plane velocity

So, the magnitude of the horizontal component of the plane velocity is 241.61 m/s.
The complete question is missing, so i have attached the complete question.
Answer:
A) FBD is attached.
B) The condition that must be satisfied is for ω_min = √(g/r)
C) The tension in the string would be zero. This is because at the smallest frequency, the only radially inward force at that point is the weight(force of gravity).
Explanation:
A) I've attached the image of the free body diagram.
B) The formula for the net force is given as;
F_net = mv²/r
We know that angular velocity;ω = v/r
Thus;
F_net = mω²r
Now, the minimum downward force is the weight and so;
mg = m(ω_min)²r
m will cancel out to give;
g = (ω_min)²r
(ω_min)² = g/r
ω_min = √(g/r)
The condition that must be satisfied is for ω_min = √(g/r)
C) The tension in the string would be zero. This is because at the smallest frequency, the only radially inward force at that point is the weight(force of gravity).
Answer:
Explanation:
Consider another special case in which the inclined plane is vertical (θ=π/2). In this case, for what value of m1 would the acceleration of the two blocks be equal to zero
F - Force
T = Tension
m = mass
a = acceleration
g = gravitational force
Let the given Normal on block 2 = N
and 
and the tension in the given string is said to be 
When the acceleration 
for the said block 1.
It will definite be zero only when Force is zero , F=0.
Here by Force, F
I refer net force on block 1.
Now we know

It is known that if the said
,
then Tension
,
Now making 
So If we are to make Force equal to zero

Answer:
Explanation:
Given
mass of marble 
velocity of marble 
mass of second marble 
Velocity of second marble 
After collision 13.5 gm marble moves to the right at i.e. 
Conserving momentum




