Paul lifts a sack weighing 245 newtons vertically from the ground and places it on a platform at a height of 0.7 meters. If he t
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Answer:
Explanation:
The first thing we are asked to find is the Force experienced by the box. That is found in the formula:
F - f = ma where F is the force exerted by the box, f is the friction opposing the box, m is the mass, and a is the acceleration (NOT the same as the pull of gravity). But F can be rewritten in terms of the angle of inclination also:
where w is the weight of the box. We will use this version of the formula because it will help us answer the second question, which is to solve for a. Filling in:
First we need the weight of the box. Having the mass, we find the weight:
w = mg so
w = 25.14(10) so
w = 251.4 N (I am not paying any attention at all to the sig fig's here, since I noticed no one on this site does!) Now we have the weight. Filling that in:
251.4sin(30) - f = ma Before we go on to fill in for f, let's answer the first question. F = 251.4sin(30) so
F = 125.7 And in order to answer what a is equal to, we find f:
f = μ where Fn is the weight of the object.
f = .25(251.4) so
f = 62.85. Filling everything in now altogether to solve for a, the only missing value:
125.7 - 62.85 = 25.14a and
62.85 = 25.14a so
a = 2.5 m/s/s
Now we have to move on to another set of equations to answer the last part. The last part involves the y-dimension. In this dimension, what we know is that
a = -10 m/s/s
v₀ = 0 (it starts from rest)
Δx = -13.30 m (negative because the box falls this fr below the point fro which it started). Putting all that together in the equation for displacement:
Δx = v₀t + and we are solving for time:
and
so
t = 1.6 seconds to reach the bottom of the slope from 13.30 m high.
Answer:
5308.34 N/C
Explanation:
Given:
Surface density of each plate (σ) = 47.0 nC/m² =
Separation between the plates (d) = 2.20 cm
We know, from Gauss law for a thin sheet of plate that, the electric field at a point near the sheet of surface density 'σ' is given as:
Now, as the plates are oppositely charged, so the electric field in the region between the plates will be in same direction and thus their magnitudes gets added up. Therefore,
Now, plug in for 'σ' and
for
and solve for the electric field. This gives,
Therefore, the electric field between the plates has a magnitude of 5308.34 N/C