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alexira [117]
3 years ago
7

You want to use a rope to pull a 10-kg box of books up a plane inclined 30∘ above the horizontal. The coefficient of kinetic fri

ction is 0.29. The rope pulls parallel to the incline.
What force do you need to exert on the rope if you want to pull the box with a constant acceleration of 0.50 m/s2 up the plane?

Physics
1 answer:
fgiga [73]3 years ago
5 0

Answer:   78.61N

Explanation

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List several examples of applied force, normal force, and friction that you’ve observed in your life.
grandymaker [24]

applied forces would be push for example.

normal forces would seem to be a force such as gravity.

friction for example when you try to slide on carpet but the fabric or whatever its made of stops you.

3 0
3 years ago
What is the speed of a car that travels 48 km in 2 hours?
ankoles [38]
Speed=distance/time

distance =48km
time=2 hours

speed=48/2
=24km/h


speed=24km/h

FORMULAS

speed = distance/time
time = distance/speed
distance = speed×time

3 0
3 years ago
Which of the following statements is an accurate description of vibrations?
dexar [7]
<span>So we want to know what statement is an accurate description of vibrations. So humans can hear sound frequencies from 20-20000 Hz. Below 20 Hz is infra sound and above 20000 Hz is ultra sound. Humans cant hear both infra sound and ultra sound so the correct answer is A.</span>
6 0
3 years ago
Read 2 more answers
The center of mass is
PolarNik [594]
D is the best answer. In many physics problems we treat an extended object as if it were a point with the same mass located at the center of mass.
5 0
3 years ago
Find the potential inside and outside a uniformly charged solid sphere whose radius is R and whose total charge is q. Use infini
amid [387]

Answer:

Recall that the electric field outside  a uniformly charged solid sphere  is exactly the same as if the charge were all at a point in the centre of the  sphere:

E_{outside} =\frac{1}{4\pi(e_{0})}\frac{Q}{r^{2} } r^{'}

lnside the sphere, the electric field also acts like a point charge, but only for the proportion of the charge further inside than the point r:

E_{inside} =\frac{1}{4\pi(e_{0})}\frac{Q}{R^{2} } \frac{r}{R} r^{'}

To find the potential, we integrate the electric field on a path from infinity (where of course, we take the direct path so that we can write the it as a 1 D integral):

V(r>R)=\int\limits^r_\infty {\frac{1}{4\pi(e_{0)} }\frac{Q}{r^2}  } \, dr=\frac{q}{4\pi(e_{0)} } \frac{1}{r} \\V(r

=\frac{q}{4\pi e_{0} } [\frac{1}{R} -\frac{r^{2}-R^{2}  }{2R^{3} } ]

∴NOTE: Graph is attached

8 0
3 years ago
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