Answer:
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They circulate blood and oxygen throughout your body
This is possible due to self-discharge. Very small internal currents inevitably occur in these cells over time and they will eventually exhaust the chemistry.
Answer:
Impulse of force = -80 Ns
Explanation:
<u>Given the following data;</u>
Mass = 50kg
Initial velocity = 1.6m/s
Since she glides to a stop, her final velocity equals to zero (0).
Now, we would find the change in velocity.
Substituting into the equation above;
Change in velocity = 0 - 1.6 = 1.6m/s
Substituting into the equation, we have;
<em>Impulse of force = -80 Ns</em>
<em>Therefore, the impulse of the force that stops her is -80 Newton-seconds and it has a negative value because it is working in an opposite direction, thus, bringing her to a stop. </em>
The work done occurs only in the direction the block was moved - horizontally. Work is given by:
W = F(h) * d
Where F(h) is the force applied in that direction (horizontal) and d is the distance in that direction. In this case, F(h) is the horizontal component of the applied force, F(app). However, the question doesn't give us F(app), so we need to find it some other way.
Since the block is moving at a constant speed, we know the horizontal forces must be balanced so that the net force is 0. This means that F(h) must be exactly balanced by the friction force, f. We can express F(h) as a function of F(app):
F(h) = F(app)cos(23)
Friction is a little trickier - since the block is being PUSHED into the ground a bit by the vertical component of the applied force, F(v), the normal force, N, is actually a bit more than mg:
N = mg + F(v) = mg + F(app)sin(23)
Now we can get down to business and solve for F(app) - as mentioned above:
F(h) = f
F(h) = uN
F(h) = u * (mg + F(v))
F(app)cos(23) = 0.20 * (33 * 9.8 + F(app)sin(23))
F(app) = 76.8
Now that we have F(app), we can find the exact value of F(h):
F(h) = F(app)cos(23)
F(h) = 76.8cos(23)
F(h) = 70.7
And now that we have F(h), we can find W:
W = F(h) * d
W = 70.7 * 6.1
W = 431.3
Therefore, the work done by the worker's force is 431.3 J. This also represents the increase in thermal energy of the block-floor system.
