Answer:
Atom, smallest unit into which matter can be divided without the release of electrically charged particles.
an element is a pure substance which cannot be broken down by chemical means
Answer:
a)0.22 m/s².
Explanation:
Given that
Net force ,F= 6.8 N
mass ,m = 31 kg
From the second law of Newton's
F = m a ---------------1
Where
F=Net force ,m=mass
a=Acceleration
Now by putting the values in the equation 1
F = m a
6.8 = 31 x a



Therefore the acceleration of the scooter will be 0.22 m/s².
The answer will be "a".
a)0.22 m/s².
Answer:
R: 27 u
ou 27g/mol
Explanation:
A massa atômica de um elemento é dada pela soma do número atômico (número de protons) com o número de neutrons. Assim:
A= 13+14 = 27
The function y must be equal to 0 on any interval on which it is defined. The function y must be increasing (or equal to 0) on any interval on which it is defined.
Analysis of solution by seeing differential equation:
Given differential equation is: y' = (1/2)y2
How do deduce the results just by seeing them?
The equation tells us that:
rate = positive of ( y^2 )
rate = positive of (positive or zero) = positive or zero
Thus, the rate is positive or zero no matter what value we put in the place of y from its valid domain, since.
When the rate is positive or zero, that means the function will never grow upwards. Thus, either increasing or staying at the same level.
Learn more about differential equations here:
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The average force on the ball is 287.3 N.
Explanation:
The impulse exerted on an object, which is equal to the product between the force exerted and the duration of the collision, is equal to the change in momentum of the object.
If we apply this to the ball, we can write:

where
F is the force exerted on the ball
is the duration of the collision
m = 513 g = 0.513 kg is the mass of the ball
u = 12.1 m/s is the initial velocity of the ball
v = -13.1 m/s is the final velocity (negative since the ball rebounds in the opposite direction)
And solving for F, we find:

So, the magnitude of the average force is 287.3 N.
Learn more about impulse and change in momentum:
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