Explanation:
Given:
v₀ = 22 m/s
v = 0 m/s
t = 17.32 s
Find: a
v = at + v₀
(0 m/s) = a (17.32 s) + (22 m/s)
a = -1.270 m/s²
Round as needed.
Answer:
The resultant force would (still) be zero.
Explanation:
Before the 600-N force is removed, the crate is not moving (relative to the surface.) Its velocity would be zero. Since its velocity isn't changing, its acceleration would also be zero.
In effect, the 600-N force to the left and 200-N force to the right combines and acts like a 400-N force to the left.
By Newton's Second Law, the resultant force on the crate would be zero. As a result, friction (the only other horizontal force on the crate) should balance that 400-N force. In this case, the friction should act in the opposite direction with a size of 400 N.
When the 600-N force is removed, there would only be two horizontal forces on the crate: the 200-N force to the right, and friction. The maximum friction possible must be at least 200 N such that the resultant force would still be zero. In this case, the static friction coefficient isn't known. As a result, it won't be possible to find the exact value of the maximum friction on the crate.
However, recall that before the 600-N force is removed, the friction on the crate is 400 N. The normal force on the crate (which is in the vertical direction) did not change. As a result, one can hence be assured that the maximum friction would be at least 400 N. That's sufficient for balancing the 200-N force to the right. Hence, the resultant force on the crate would still be zero, and the crate won't move.
Answer:
By Applying pressure to the brakes
Explanation:
Driving cars through deep water that is more than 10cm can make the cars to float. Most modern cars are usually water- tight so they can start to float through water that is about 30cm deep, fast moving water is very powerful so one needs to be very careful when driving.
If the brakes are wet test them by pressing or tapping on them gently.
You can as well dry brakes by driving in low gear and applying pressure to the brakes.
The y-component of the acceleration is 
Explanation:
The y-component of the ice skater acceleration can be calculated with the equation

where
is the y-component of the final velocity
is the y-component of the initial velocity
t is the time elapsed
Here we have:
- Initial velocity is
at
, so its y-component is 
- Final velocity is
at
, so its y-component is 
The time elapsed is
t = 8.33 s
Therefore, the y-component of the acceleration is

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