Answer:
9.39 m/s
Explanation:
Using the y-direction, we can solve for the time t it takes for the cart to reach the ground.
Assume the up direction is positive and the down direction is negative.
- v₀ = 0 m/s
- a = -9.8 m/s²
- Δy = -50 m
- t = ?
Find the constant acceleration equation that contains these four variables.
Substitute known values into this equation.
Multiply and simplify.
Divide both sides of the equation by -4.9.
Square root both sides of the equation.
Now we can use this time t and solve for v₀ in the x-direction. Time is most often our link between vertical and horizontal components of projectile motion.
List out known variables in the x-direction.
- v₀ = ?
- t = 3.194382825 s
- a = 0 m/s²
- Δx = 30 m
Find the constant acceleration equation that contains these four variables.
Substitute known values into the equation.
- 30 = (v₀ · 3.194382825) + 1/2(0)(3.194382825)²
Multiply and simplify.
Divide both sides of the equation by 3.194382825.
The cart was rolling at a velocity of 9.39 m/s (initial velocity) when it left the ledge.
First you need to make a difference between friction while object is stationary and the friction while object is moving. Force required to start moving some object is slightly greater than force required to maintain objects movement. That means that to move a chair you need some force F1 but you can than slightly reduce force and chair will still be moving.
Now to the problem in this question: It can be said that "stationary friction force" is equal to 15 Newtons. Its also good to know that friction force between chair and floor while you are increasing your push is also increasing and is equal to force of your push. Once it reaches 15N which is it "critical value" for that chair, chair starts moving and friction force drops a little bit and now it is called friction force of moving chair.
Answer:
1.59 seconds
12.3 meters
but if you are wise you will read the entire answer.
Explanation:
This is a good question -- if not a bit unusual. You should try and understand the details. It will come in handy.
Time
<u>Given</u>
a = 0 This is the critical point. There is no horizontal acceleration.
d = 20 m
v = 12.6 m/s
<u>Formula</u>
d = vi * t + 1/2at^2
<u>Solution</u>
Since the acceleration is 0, the formula reduces to
d = vi * t
20 = 12.6 * t
t = 20 / 12.6
t = 1.59 seconds.
It takes 1.59 seconds to hit the ground
Height of the building
<u>Givens</u>
t = 1.59 sec
vi = 0 Another critical point. The beginning speed vertically is 0
a = 9.8 m/s^2 The acceleration is vertical.
<u>Formula</u>
d = vi*t + 1/2 a t^2
<u>Solution</u>
d = 1/2 a*t^2
d = 1/2 * 9.8 * 1.59^2
d = 12.3 meters.
The two vi's are not to be confused. The horizontal vi is a number other other 0 (in this case 12.6 m/s horizontally)
The other vi is a vertical speed. It is 0.
The mineral with Mohs hardness would be scratched because the mineral with Mohs 7 hardness is stronger than the Mohs 5 mineral. Eventually, that mineral would turn into dust if you kept rubbing it.
Answer:
Maybe???
Explanation:
but how big is the paper?
but I think even the paper is super big, the ending result will still be too small to observe …
