A 10n force is applied to a 25kg mass to slide it across a frictional surface. What is the acceleration of the mass?
2 answers:
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Answer:
a) -1.27*10³ N/C b) 0 c) -0.21*10³ N/C d) 0.1*10³ N/C
Explanation:
a) r = 0.76R
As this distance is inside the sphere, we need to know how much charge is enclosed within this distance for the center, as follows:
Q = ρ*V(r) = ρ*
where r = 0.760* R = 0.760* 0.295 m = 0.224 m, and ρ = -151 nC/m³
Applying Gauss' Law to a spherical gaussian surface of r= 0.76R, as the electric field is radial, and directed inward, we can write the following equation:
E*A = Q/ε₀, where Q= -7.11 nC, A= 4*π*(0.76R)² and ε₀ =8.85*10⁻¹² C²/N*m²
We can solve for E, as follows:
⇒ E = -1.27*10³ N/C
b) r= 3.90 R
As this distance falls inside the conducting shell, and no electric field can exist within a conductor in electrostatic condition, E=0
c) r = 2.8 R
As this distance falls between the sphere and the inner radius of the shell, we can calculate the electric field, applying Gauss' law to a gaussian surface of radius equal to r= 2.80 R.
First we need to find the total charge of the sphere, as follows:
Q = ρ*V =
In the same way that for a) we can write the following expression:
E*A = Q/ε₀, where Q= -16.2 nC, A= 4*π*(2.8R)² and ε₀ =8.85*10⁻¹² C²/N*m²
We can solve for E, as follows:
⇒ E = -0.21*10³ N/C
d) r= 7.30 R
In order to find the electric field at this distance, which falls beyond the outer radius of the shell, we need to find the total charge on the outer surface.
As the sphere has a charge of -16.2 nC, and the total charge of the conducting shell is 66.7nC, in order to make E=0 inside the shell, the total charge enclosed by a gaussian surface with a radius larger than the inner radius of the shell and shorter than the outer one, must be zero, which means that a charge of +16.2 nC must be distributed on the inner surface of the shell.
This leaves an excess charge on the outer surface of the shell as follows:
Qsh = 66.7 nC - 16.2 nC = 50.5 nC
Now, we can repeat the same process than for a) and c) as follows:
E*A = Q/ε₀, where Q= 50.5 nC, A= 4*π*(7.3R)² and ε₀ =8.85*10⁻¹² C²/N*m²
We can solve for E, as follows:
⇒ E = 0.1*10⁻³ N/C
5. Left
Explanation: centripetal acceleration is the acceleration that acts on an object which is moving by circular motion. The direction of the centripetal acceleration is always towards the centre of the circular trajectory: that means that for a car following a bend, the direction of the centripetal acceleration is towards the centre of the bend, and so it is towards the left if the bend is towards the left, and towards the right if the bend is towards the right.
In this example, the road in this bend is going to the left, so centripetal acceleration is toward the left as well.
6. Greater
Explanation: Inertia represents the tendency of an object at staying in motion if it is moving, and at staying at rest if the object is at rest. We can also see inertia as a measure of "how difficult it is" to put an object in motion. Due to this definition, we can also notice that inertia is directly proportional to the mass of the object: the greater the mass, the greater the inertia, because it is more difficult to put the object in motion.
So, in this example, the greater the boulder's mass, the greater inertia it has.
7. Direction
Explanation: displacement is a vector quantity that represents the difference in position between an object and a starting point. Because it is a vector, displacement consists of two elements:
- A magnitude, which is equal to the object's distance from the starting point
- A direction, again compared to the starting point
So, displacement depends on the object's distance and direction from the starting point.
8. Negative
Explanation: acceleration is defined as the rate of change of velocity; in formula:
where v is the final velocity, u the initial velocity, and t the time taken to accelerate from u to v. From the formula, we see that:
- When the final velocity (v) is greater than the initial velocity (u), acceleration is positive
- When the final velocity (v) is smaller than the initial velocity (u), acceleration is negative
In this problem, we have a car slowing down: it means that the final velocity is smaller than the initial velocity, so in this case acceleration is negative.