The maximum force that the athlete exerts on the bag is equal to 1,500 N and in the opposite direction as the force that the bag exerts on the athlete.
<h3>
Newton's third law of motion</h3>
Newton's third law of motion states that action and reaction are equal and opposite.
Fa = -Fb
The force exerted by the athlete on the bag is equal to the force the bag exerted on the athlete but in opposite direction.
Thus, the maximum force that the athlete exerts on the bag is equal to 1,500 newtons and in the opposite direction as the force that the bag exerts on the athlete.
Learn more about force here: brainly.com/question/12970081
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Answer:
A. El volumen
B. La densidad.
Explanation:
A derived quantity is defined as one that has to be calculated by using two or more other measurements.
Volume is a derived quantity because it requires one to use different measurements to determine it. For instance, in the case of a cube, the length, width and height of the cube are all needed to calculate volume.
Density is also a derived quantity because it needs both volume and mass for it to be calculated.
Answer:
7m/s^2
Explanation:
using v=u+at
since the car started from rest, u=0 , v=14m/s t=2s
a =acceleration.
14=0+a×2
14=0+2a
14=2a
a= 14/2 =7
a=7m/s^2
So the area under a velocity time graph is distance or displacement, if you have done calculus yet you will understand that if you take the integral of a velocity function then you end up with displacement. Thats for later understanding however.
So this appears to be a right triangle so we can find the area of a triangle as:
0.5bh = A
Since our area is 10 meters lets alter our formula a bit to fit the situation:
Our base here is time and our height is velocity so:
0.5tv = Δx
So we can read off the graph that our velocity at the end, or our final velocity appears to be near 2.0 m/s
So we have v, and Δx so lets isolate for time by dividing by v and 0.5
t = Δx / 0.5v
Now lets plug all that in:
t = 10 / 0.5(2)
t = 10 seconds
Hope this helped!
Answer:
makinig ka kasi sa titser mo tas intindihin mo bobo
