John Deere created steel plow
Answer:
yes, though the speed may stay constant, the direction will change. so for example, you're going 70mph on the freeway, but you have to take the exit on your right (the exit continues on to a different freeway), you're not going to speed up or slow down, you'll change your direction which is still accelerating.
Explanation:
Credit goes to @naeAF
Hope this helps :))
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.
The strength of the gravitational field is given by:

where
G is the gravitational constant
M is the Earth's mass
r is the distance measured from the centre of the planet.
In our problem, we are located at 300 km above the surface. Since the Earth radius is R=6370 km, the distance from the Earth's center is:

And now we can use the previous equation to calculate the field strength at that altitude:

And we can see this value is a bit less than the gravitational strength at the surface, which is

.
Answer:
Frequency of oscillation, f = 4 Hz
time period, T = 0.25 s
Angular frequency, 
Given:
Time taken to make one oscillation, T = 0.25 s
Solution:
Frequency, f of oscillation is given as the reciprocal of time taken for one oscillation and is given by:
f = 
f = 
Frequency of oscillation, f = 4 Hz
The period of oscillation can be defined as the time taken by the suspended mass for completion of one oscillation.
Therefore, time period, T = 0.25 s
Angular frequency of oscillation is given by:


