If the coefficient of static friction is 0.3, then the minimum force required to get it moving is equal in magnitude to the maximum static friction that can hold the body in place.
By Newton's second law,
• the net vertical force is 0, since the body doesn't move up or down, and in particular
∑ <em>F</em> = <em>n</em> - <em>mg</em> = <em>n</em> - 50 N = 0 ==> <em>n</em> = 50 N
where <em>n</em> is the magnitude of the normal force; and
• the net horizontal force is also 0, since static friction keeps the body from moving, with
∑ <em>F</em> = <em>F'</em> - <em>f</em> = <em>F'</em> - <em>µn</em> = <em>F'</em> - 0.3 (50 N) = 0 ==> <em>F'</em> = 15 N
where <em>F'</em> is the magnitude of the applied force, <em>f</em> is the magnitude of static friction, and <em>µ</em> is the friction coefficient.
The weightiness of the added
water displaced is equivalent to the joined weight of the two extra people who come
to be into the boat:
<span>m water g = 2 x 690 N</span>
<span> =
1,380 N</span>
<span>
</span>
The mass of the water displace
is then
<span>m water g = 1,380 N</span>
<span> = 1,380 N / 9.8 m/s^2</span>
<span> = 141 kg</span>
<span>
</span>
Compute the calculation for
density for the volume of water displace and practice this outcome for the mass
of the water displace to get the answer:
<span>p water = mass of water / volume of water</span>
<span>
</span>
<span>volume of water = mass of water / p water</span>
<span> = 141 kg / 1000 kg /m^3 eliminate
kilogram</span>
<span> = 0.14 m^3 the additional volume
of water that is displaced</span>
Answer:
The first is that the object moves
Answer:
s = 1.7 m
Explanation:
from the question we are given the following:
Mass of package (m) = 5 kg
mass of the asteriod (M) = 7.6 x 10^{20} kg
radius = 8 x 10^5 m
velocity of package (v) = 170 m/s
spring constant (k) = 2.8 N/m
compression (s) = ?
Assuming that no non conservative force is acting on the system here, the initial and final energies of the system will be the same. Therefore
• Ei = Ef
• Ei = energy in the spring + gravitational potential energy of the system
• Ei = \frac{1}{2}ks^{2} + \frac{GMm}{r}
• Ef = kinetic energy of the object
• Ef = \frac{1}{2}mv^{2}
• \frac{1}{2}ks^{2} + (-\frac{GMm}{r}) = \frac{1}{2}mv^{2}
• s = 
s = 
s = 1.7 m
