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kotegsom [21]
3 years ago
11

A bicycle takes 8.0 seconds to accelerate at a constant rate from rest to a speed of 4.0 m/s. If the mass of the bicycle and rid

er together is 85 kg, what is the net force acting on the bicycle?
Physics
2 answers:
alexira [117]3 years ago
7 0

42.5 or 43 whichever answer you have on your test

Inessa05 [86]3 years ago
4 0

Acceleration = (change in speed) / (time for the change)
Acceleration = (4 m/s) / (8 seconds)
Acceleration = 0.5 m/s²

Force = (mass) x (acceleration)
Force = (85 kg) x (0.5 m/s²)
Force = 42.5 Newtons
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Mamont248 [21]

Answer:

a. True

Explanation:

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5 0
2 years ago
How are chemical changes of matter related to the organization of the periodic table
Charra [1.4K]
As elements in period are arranged according to increasing atomic weight with isotope as protium deuterium and trotium of h2 so chemical property are same but physical property different in period ic table
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3 years ago
.Imagine that you pushed a box, applying a force of 60 newtons, over a distance of 4 meters. how much work have you done? 15 jou
DaniilM [7]
Work=force X distance
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240 joules
5 0
3 years ago
Read 2 more answers
26. The speed of light in a certain medium is
horrorfan [7]

The number we need in order to answer the question belongs in the space between the words "is" and "of".  You left that blank blank, so there really isn't any question here to answer.

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6 0
3 years ago
On a certain planet, which is perfectly spherically symmetric, the free-fall acceleration has magnitude g = go at the north pole
ohaa [14]
The reason why there is a difference between free-fall acceleration is a centrifugal force.
I attached a diagram that shows how this force aligns with the force of gravity.
From the diagram we can see that:
F=F_g-F_{cf}=mg'-mw^2r'cos(\alpha)\\ ma=mg'-mw^2r'cos(\alpha)\\ a=g'-w^2rcos^2(\alpha)\\
Where g' is the free-fall acceleration when there is no centrifugal force, r is the radius of the planet, and w is angular frequency of planet's rotation. \alpha is the latitude.
We can calculate g' and wr^2 from the given conditions in the problem.
g(90)=g_0;\ g_0= g'-w^2rcos^2(90)\\&#10;g_0=g'\\&#10;g(0)=ag_0;\ ag_0=g_0-w^2rcos^2(0)\\&#10;ag_0=g_0-w^2r\\&#10;w^2r=g_0(a-1)&#10;
Our final equation is:
g=g_0-g_0(a-1)cos^2(\alpha)
Colatitude is:
\alpha_c=90^\circ-\alpha
The answer is:
g=g_0-g_0(a-1)cos^2(90-9)=g_0-g_0(a-1)sin^2(9)

5 0
3 years ago
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