Answer:
The linear momentum of a particle with mass m moving with velocity v is defined as
p = mv (7.1)
Linear momentum is a vector . When giving the linear momentum of a particle you must
specify its magnitude and direction. We can see from the definition that its units must be
kg·m
s
. Oddly enough, this combination of SI units does not have a commonly–used named so
we leave it as kg·m
s
!
The momentum of a particle is related to the net force on that particle in a simple way;
since the mass of a particle remains constant, if we take the time derivative of a particle’s
momentum we find
dp
dt = m
dv
dt = ma = Fnet
so that
Fnet =
dp
dt (7.2)
Answer: An exercise, activity, or workout that targets a specific training need Progression - Gradually increasing your skill level and ability over time Overload -Working your body harder than usual to improve how it functions Reversibility - Losing what you have gained by taking time off Tedium - getting bored by doing the same thing
Explanation:
Answer:
The charge on the ball bearing 4.507 × 10^-8 C
Explanation:
From Coulomb's law
F = kq1q2/r²
make q2 the subject
q2 = Fr²/kq1
q2 = (1.8×10^-2 × 0.026²) ÷ (9×10^9 × 30×10^-9)
q2 = 4.507 × 10^-8 C
"v0" means that there are no friction forces at that speed
<span>mgsinΘ = (mv0²/r)cosΘ → the variable m cancels </span>
<span>sinΘ/cosΘ = tanΘ = v0² / gr
</span><span>Θ = arctan(v0² / gr) </span>
<span>When v > v0, friction points downslope: </span>
<span>mgsinΘ + µ(mgcosΘ + (mv²/r)sinΘ) = (mv²/r)cosΘ → m cancels: </span>
<span>gsinΘ + µ(gcosΘ + (v²/r)sinΘ) = (v²/r)cosΘ </span>
<span>µ = ((v²/r)cosΘ - gsinΘ) / (gcosΘ + (v²/r)sinΘ) </span>
<span>where Θ is defined above. </span>
<span>When v > v0, friction points upslope: </span>
<span>mgsinΘ - µ(mgcosΘ + (mv²/r)sinΘ) = (mv²/r)cosΘ → m cancels: </span>
<span>gsinΘ - µ(gcosΘ + (v²/r)sinΘ) = (v²/r)cosΘ </span>
<span>µ = (gsinΘ - (v²/r)cosΘ) / (gcosΘ + (v²/r)sinΘ) </span>
<span>where Θ is defined above. </span>
Newton's second law tells you:
Sum of forces on an object = ma
Here, the forces acting on the bundle are the tension in the string and the force of gravity, these two must combine to yield the acceleration of the bundle.
So we have:
T-mg = ma
or T=m(g+a)
We know m=8.7kg, we need to find a from the information
starting from rest, an accelerating object covers distance according to:
<span>dist = 1/2 at^2 </span>
to cover 1m in 1.8s, we have:
a=2d/t^2 = 2x1/1.8^2 = 0.62 m/s/s
Thus, the tension in the string is:
<span>T = m(g+a)
= 8.7</span>kg(9.8m/s/s+0.62m/s/s)
<span>
<span>T = 90.654 N
</span>
I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
</span>
