Answer:
the vertical distance between the two object will increase uniformly when they are dropped after a fixed interval of time
Explanation:
Since airplane is moving horizontally with constant speed v
so when object is dropped from the plane then the speed of the object will be same as that of the speed of the airplane
so we can say that two object when dropped after some interval of time then they always lie in same vertical line
now we know that they both have same acceleration in vertical line so the motion of two objects relative to each other in vertical direction is always uniform motion because they have no acceleration with respect to each other
So the vertical distance between the two object will increase uniformly when they are dropped after a fixed interval of time
Answer:
Distance covered by the sound in air is 800 meter and the time taken by the sound in water for the same distance is 0.5 seconds.
Explanation:
Given:
Speed of sound in air = 320 m/s
Speed of sound in water = 1600 m/s
Time taken to reach certain distance in air = 2.5 sec
a.
We have to find the distance traveled by sound in air.
Distance = Product of speed and time.
⇒ 
⇒ 
⇒
meters.
b.
Now we have to find how much time the sound will take to travel in water.
⇒ Time = Ratio of distance and speed
⇒ 
⇒
<em> ...distance = 800 m and speed = 1600 m/s</em>
⇒ 
⇒
seconds.
Distance covered by the sound in air is 800 meter and the time taken by the sound in water for the same distance is 0.5 seconds.
<em>projectile can only follow the straight line path when it is launched upward straightly so the correct option is <u>90 degree with respect to horizontal x -axis ..:)</u></em>
The second law of thermodynamics states that whenever energy changes occur, DISORDER always increases.
We can first calculate the net force using the given information.
By Newton's second law, F(net) = ma:
F(net) = 25 * 4.3 = 107.5
We can now calculate the frictional force, f, which is working against the applied force, F(app) (this is why the net force is a bit lower):
f = F(net) - F(app) = 150 - 107.5 = 42.5 N
Now we can calculate the coefficient of friction, u, using the normal force, F(N):
f = uF(n) --> u = f/F(N)
u = 42.5/[25(9.8)]
u = 0.17