Answer;
B. at your original position.
Explanation.
It the train does not slow down the to the force of gravity you would land far behind the train. Here, it slows down and you already jumped vertically up, you cover some certain distance from the train up in the air and it is also slowing down you will likely come to your original position.
Vertically, the object is in equilibrium, so that the net force in this direction is
∑ <em>F</em> (vertical) = <em>n</em> - <em>mg</em> = 0
where <em>n</em> is the magnitude of the normal force due to the contact between the object and surface. You're given that the object's weight is <em>mg</em> = 550 N, so <em>n</em> = 550 N as well.
Horizontally, the net force would be
∑ <em>F</em> (horizontal) = <em>p</em> - <em>f</em> = 0
where <em>p</em> is the magnitude of the applied force and <em>f</em> is the magnitude of (kinetic) friction opposing <em>p</em>. Now,
<em>f</em> = 0.012<em>n</em> = 0.012 (550 N) = 6.6 N
so that you need to apply a force of <em>p</em> = 6.6 N to keep the object sliding at a steady pace.
To solve this problem we will apply the concepts of acceleration according to the laws of kinetics. Acceleration can be defined as the change in velocity per unit time, that is,
Here,
v = Final velocity
u = Initial velocity
t = Time
Rearranging to find the initial velocity,
Now the acceleration is equal to the gravity, then,
The velocity of the shell at 5.5s after the launch is,
Therefore the magnitude of the velocity of the shell at 5.5s is -19.26m/s
Answer:
286
Explanation:
p1v1/T1=p2v2/T2
then subtitude your values
T2=760*0.65*273/210*0.040
T2=135/8.4=16+273=289
Answer:
Zero
Explanation:
here, the inductive reactance and the capacitive reactance is same, so this is the condition for resonance.
In the condition for resonance,
the circuit and the voltage in the circuit is in the same phase and the impedance of the circuit is minimum which is equal to the resistance of the circuit.
The phase angle is given by
Ф = 0
