Answer:
Explanation:
Use the one-dimensional equation
Δx = 
where delta x is the displacement of the object, v0 is the velocity of the object, a is the pull of gravity, and t is the time in seconds. That's our unknown.
Δx = -2 (negative because where it ends up is lower than the point at which it started),
, and
a = -9.8
Filling in:
and simplified a bit:
this should look hauntingly familiar (a quadratic, which is parabolic motion...very important in physics!!). We begin by getting everything on one side of the equals sign and solving for t by factoring:
(the 0 is also indicative of the object landing on the ground! Isn't this a beautiful thing, how it all just works so perfectly together?)
When you factor this however your math/physics teacher has you factoring you will get that
t = 1.3 sec and t = -.31 sec
Since we all know that time can NEVER be negative, it takes the ball 1.3 sec to hit the ground from a height of 2 m if it is rolling off the shelf at 5 m/s.
Suppose car A is moving with a velocity Va, and car b with a velocity Vb,
According the principle of conservation of momentum:
Va x Ma + Vb x Mb = (Ma + Mb) V
V = (Va x Ma + Vb x Mb)/(Ma +Mb)
V = speed of cars after coupling
V = (Va x 20 mg + Vb x 15 mg)/(20 mg + 15 mg)
Put in the values of Va and Vb, and get the V
Answer:
More force
Explanation:
Object A has more mass than object B
For object A to accelerate at the same rate as object B, it will need more force.
According to Newton's second law of motion "the net force on a body is the product of its mass and acceleration".
Net force = mass x acceleration
Now, if a body has more mass and needs to accelerate at the same rate as another one with a lower mass, the force on it must be increased.
Answer:
44.4 m/s
Explanation:
d = 80 m, t = 1.8 s, find v
d = v*t
v = d/t = 80/1.8 = 44.4 m/s
Answer:5
Explanation:
Given
speed of object 
radius of circle 
Force towards the center 
Work done is given by the dot product of Force and displacement
and we know know displacement of the object is along the circle which is perpendicular to the force acting therefore Work done will be zero
