The distance in meters she would have moved before she begins to slow down is 11.25 m
<h3>
LINEAR MOTION</h3>
A straight line movement is known as linear motion
Given that Ann is driving down a street at 15 m/s. Suddenly a child runs into the street. It takes Ann 0.75 seconds to react and apply the brakes.
To know how many meters will she have moved before she begins to slow down, we need to first list all the given parameters.
From definition of speed,
speed = distance / time
Make distance the subject of the formula
distance = speed x time
distance = 15 x 0.75
distance = 11.25m
Therefore, the distance in meters she would have moved before she begins to slow down is 11.25 m
Learn more about Linear motion here: brainly.com/question/13665920
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:
The value of the spring constant of this spring is 1000 N/m
Explanation:
Given;
equilibrium length of the spring, L = 10.0 cm
new length of the spring, L₀ = 14 cm
applied force on the spring, F = 40 N
extension of the spring due to applied force, e = L₀ - L = 14 cm - 10 cm = 4 cm
From Hook's law
Force applied to a spring is directly proportional to the extension produced, provided the elastic limit is not exceeded.
F ∝ e
F = ke
where;
k is the spring constant
k = F / e
k = 40 / 0.04
k = 1000 N/m
Therefore, the value of the spring constant of this spring is 1000 N/m
Answer:
the middle
Explanation:
the left one bulb gets power from the outher bulb
the one on right has more bulbs
Answer:
The coefficient of performance for the cycle is 2.33.
Explanation:
Given that,
Output energy 
Work done 
We need to calculate the coefficient of performance
Using formula of the coefficient of performance
We need to calculate the 
Put the value into the formula
Now put the value of into the formula of COP
Hence, The coefficient of performance for the cycle is 2.33.
