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disa [49]
2 years ago
5

Two small insulating spheres with radius 3.50×10−2 m are separated by a large center-to-center distance of 0.555 m. One sphere i

s negatively charged, with net charge -2.50 μC, and the other sphere is positively charged, with net charge 4.20 μC. The charge is uniformly distributed within the volume of each sphere.
What is the magnitude E of the electric field midway between the spheres?
Take the permittivity of free space to be epsilon_0 = 8.85×10−12 C^2/(N \cdot m^2).
E =

\rm{N/C}
Physics
1 answer:
xz_007 [3.2K]2 years ago
4 0

Answer:

E = 7.83 \times 10^5 N/C

Explanation:

Since we know that two sphere is oppositely charged so net electric field at the mid point of two balls will be sum of the electric field due to each ball at the mid point

So we know that

E = \frac{kq_1}{r^2} + \frac{kq_2}{r^2}

here we know that

q_1 = 4.20 \mu C

q_2 = 2.50 \mu C

r = \frac{0.555}{2}

so we have

E = \frac{(9\times 10^9)(4.20 + 2.50) \times 10^{-6}}{0.2775^2}

E = 7.83 \times 10^5 N/C

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A dockworker applies a constant horizontal force of 80.0 N to a block of ice on a smooth horizontal floor. The frictional force
Tamiku [17]

Answer:

(a) 91 kg (2 s.f.)    (b) 22 m

Explanation:

Since it is stated that a constant horizontal force is applied to the block of ice, we know that the block of ice travels with a constant acceleration and but not with a constant velocity.

(a)

                                                   s \ = \ ut \ + \ \displaystyle\frac{1}{2} at^{2} \\ \\ a \ = \ \displaystyle\frac{2(s \ - \ ut)}{t^{2}} \\ \\ a \ = \ \displaystyle\frac{2(11 \ - \ 0)}{5^{2}} \\ \\ a \ = \ \displaystyle\frac{22}{25} \\ \\ a \ = \ 0.88 \ \mathrm{m \ s^{-2}}

     Subsequently,

                                                  F \ = \ ma \\ \\ m \ = \ \displaystyle\frac{F}{a} \\ \\ m \ = \ \displaystyle\frac{80 \ \mathrm{kg \ m \ s^{-2}}}{0.88 \ \mathrm{m \ s^{-2}}} \\ \\ m \ = \ 91 \mathrm{kg} \ \ \ \ \ \ (2 \ \mathrm{s.f.})

*Note that the equations used above assume constant acceleration is being applied to the system. However, in the case of non-uniform motion, these equations will no longer be valid and in turn, calculus will be used to analyze such motions.

(b) To find the final velocity of the ice block at the end of the first 5 seconds,

                                                    v \ = \ u \ + \ at \\ \\ v \ = \ 0 \ + \ (0.88 \mathrm{m \ s^{-2}})(5 \ \mathrm{s}) \\ \\ v \ = \ 4.4 \ \mathrm{m \ s^{-1}}

     According to Newton's First Law which states objects will remain at rest

     or in uniform motion (moving at constant velocity) unless acted upon by

     an external force. Hence, the block of ice by the end of the first 5

     seconds, experiences no acceleration (a = 0) but travels with a constant

     velocity of 4.4 m \ s^{-1}.

                                                    s \ = \ ut \ + \ \displaystyle\frac{1}{2}at^{2} \\ \\ s \ = \ (4.4 \ \mathrm{m \ s^{-2}})(5 \ \mathrm{s}) \ + \ \displaystyle\frac{1}{2}(0)(5^{2}) \\ \\ s \ = \ 22 \ \mathrm{m}

      Therefore, the ice block traveled 22 m in the next 5 seconds after the

      worker stops pushing it.

4 0
2 years ago
A wrench 0.500 m long is applied to a nut with a force of 80.0 N. Because of the cramped space, the force must be exerted upward
riadik2000 [5.3K]

Answer:

Torque, \tau=34.6\ N.m

Explanation:

It is given that,

Length of the wrench, l = 0.5 m

Force acting on the wrench, F = 80 N

The force is acting upward at an angle of 60.0° with respect to a line from the bolt through the end of the wrench. We need to find the torque is applied to the nut. We know that torque acting on an object is equal to the cross product of force and distance. It is given by :

\tau=Fr\ sin\theta

\tau=80\times 0.5\ sin(60)

\tau=34.6\ N.m

So, the torque is applied to the nut is 34.6 N.m. Hence, this is the required solution.

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