A. Allow movement.
Muscles connect to your skeleton and they contract and move the skeleton along. <span>They help the process of movement happen in a smoother manner.</span>
A neutral atom is simply an atom that has no charge.
In the given graph, from 4.0 s to 8.0 s, the object is at rest because the speed is zero.
In the given graph we can deduce the following;
- at the time interval, 0 s to 3.5 s, the speed of the object = 1 cm/s
- when the time, t= 4 s, the <em>speed</em> of the object = 0 cm/s
- at the time interval, 4.0 s to 8.0 s, the<em> speed </em>of the object = 0 cm/s
When the <em>speed</em> of an object is zero (0), the object is simply at rest.
Thus, we can conclude that in the given graph, from 4.0 s to 8.0 s, the object is at rest because the speed is zero.
Learn more here:brainly.com/question/10454047
-- We know that the y-component of acceleration is the derivative of the
y-component of velocity.
-- We know that the y-component of velocity is the derivative of the
y-component of position.
-- We're given the y-component of position as a function of time.
So, finding the velocity and acceleration is simply a matter of differentiating
the position function ... twice.
Now, the position function may look big and ugly in the picture. But with the
exception of 't' , everything else in the formula is constants, so we don't even
need any fancy processes of differentiation. The toughest part of this is going
to be trying to write it out, given the text-formatting capabilities of the wonderful
envelope-pushing website we're working on here.
From the picture . . . . . y (t) = (1/2) (a₀ - g) t² - (a₀ / 30t₀⁴ ) t⁶
First derivative . . . y' (t) = (a₀ - g) t - 6 (a₀ / 30t₀⁴ ) t⁵ = (a₀ - g) t - (a₀ / 5t₀⁴ ) t⁵
There's your velocity . . . /\ .
Second derivative . . . y'' (t) = (a₀ - g) - 5 (a₀ / 5t₀⁴ ) t⁴ = (a₀ - g) - (a₀ /t₀⁴ ) t⁴
and there's your acceleration . . . /\ .
That's the one you're supposed to graph.
a₀ is the acceleration due to the model rocket engine thrust
combined with the mass of the model rocket
'g' is the acceleration of gravity ... 9.8 m/s² or 32.2 ft/sec²
t₀ is how long the model rocket engine burns
Pick, or look up, some reasonable figures for a₀ and t₀
and you're in business.
The big name in model rocketry is Estes. Their website will give you
all the real numbers for thrust and burn-time of their engines, if you
want to follow it that far.
Answer:
Wt = 26.84 [N]
Explanation:
In order to solve this problem we must use the definition of work in physics. Which tells us that this is equal to the product of force by distance.
In this case, we must sum the works of the force applied by the box and the friction force that also acts on the box.
The friction force is defined as the product of the normal force by the coefficient of friction.
f = N*μ
where:
N = normal force = m*g [N] (units of Newtons)
m = mass = 72 [kg]
g = gravity acceleration = 9.81 [m/s²]
f = friction force [N]
μ = friction coefficient = 0.21
f = 72*9.81*0.21
f = 148.32 [N]
Now the total work:
Wt = WF - Wf
where:
Wt = total work [J] (units of Joules)
WF = work by the pushing force [J]
Wf = work done by the friction force [J]
Wt = (160*2.3) - (148.32*2.3)
Wt = 26.84 [N]
Note: The friction force exerts a negative work, because this force is acting in opposite direction to the movement, therefore the negative sign.
