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brilliants [131]
3 years ago
11

A positive point charge Q is fixed on a very large horizontal frictionless tabletop. A second positive point charge q is release

d from rest near the stationary charge and is free to move. Which statement best describes the motion of q after it is released?
(A) As it moves farther and farther from Q, its speed will keep increasing.
(B) As it moves farther and farther from Q, its speed will decrease.
(C) As it moves farther and farther from Q, its acceleration will keep increasing.
(D) Its speed will be greatest just after it is released. Its acceleration is zero just after it is released.
Physics
1 answer:
Scilla [17]3 years ago
7 0

Answer:

(A) As it moves farther and farther from Q, its speed will keep increasing.

Explanation:

When a positive charge Q is fixed on a horizontal frictionless tabletop and a second charge q is released near to it then according to the Coulombs law the force acting on it decreases with the square of the distance between them.

Mathematically:

F=\frac{1}{4\pi.\epsilon_0} \times \frac{Q.q}{r^2}

where:

r = distance between the charges

\epsilon_0= permittivity of free space

By the Newtons' second law of motion if the we know that the acceleration is directly proportional to the force applied. So as  the distance between the charges increases the its acceleration also decreases therefore now the charge feels less acceleration but still continues to accelerate with a fading magnitude.

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A soccer ball of diameter 22.6cm and mass 426g rolls up a hill without slipping, reaching a maximum height of 5m above the base
-Dominant- [34]

Answer with Explanation:

We are given that

Diameter=d=22.6 cm

Mass,m=426 g=426\times 10^{-3} kg

1 kg=1000 g

Radius,r=\frac{d}{2}=\frac{22.6}{2}=11.3 cm=11.3\times 10^{-2} m

1m=100 cm

Height,h=5m

I=\frac{2}{2}mr^2

a.By law of conservation of energy

\frac{1}{2}I\omega^2+\frac{1}{2}mv^2=mgh

\frac{1}{2}\times \frac{2}{3}mr^2\omega^2+\frac{1}{2}mr^2\omega^2=mgh

v=\omega r

gh=\frac{1}{3}r^2+\frac{1}{2}r^2=\frac{5}{6}r^2\omega^2

\omega^2=\frac{6}{5r^2}gh

\omega=\sqrt{\frac{6gh}{5r^2}}=\sqrt{\frac{6\times 9.8\times 5}{5(11.3\times 10^{-2})^2}}=67.86 rad/s

Where g=9.8m/s^2

b.Rotational kinetic energy=\frac{1}{2}I\omega^2=\frac{1}{2}\times \frac{2}{3}mr^2\omega^2=\frac{1}{2}\times \frac{2}{3}(426\times 10^{-3})(11.3\times 10^{-2})^2(67.86)^2=8.35 J

Rotational kinetic energy=8.35 J

6 0
3 years ago
Interactive Solution 9.1 presents a model for solving this problem. The wheel of a car has a radius of 0.380 m. The engine of th
Vilka [71]

Answer:

The magnitude of the static frictional force is 1200 N

Explanation:

given information :

radius, r = 0.380 m

applied-torque, τ1 = 456 N

The car has a constant velocity, thus the acceleration is zero

α = 0

Στ = I α

τ1 - τ2 = I α

τ2 = counter-torque

τ1 - τ2 = 0

τ1 = τ2

r x F_{s} = τ1

F_{s} = the static frictional force (N)

F_{s} = τ1 /r

  = 456 N/0.380 m

  = 1200 N

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Remaining Time: 1 hour, 49 minutes, 34 seconds.
ohaa [14]

Explanation:

i think C . it is twice the size of the object

4 0
3 years ago
The vector values are 4.0 km due East and 3.0 km due north the resultant vector is 5.0 km long and 37 north of East this values
klemol [59]

Answer:

This values shows a right angle triangle

Explanation:

Given;

a vector 4.0 km due East

a 3.0 km due north

the resultant vector is 5.0 km

The resultant vector can be obtained by Pythagoras theorem if the vectors form a right angle triangle.

R² = 4² + 3²

R² = 16 + 9

R² = 25

R = √25

R = 5 km    (right angle triangle proved)

Therefore, this values shows a right angle triangle

4 0
3 years ago
Find the volume of a rectangular prism that is 8m long, 4m wide, and 300cm high
konstantin123 [22]
  • L=8m
  • B=4m
  • H=300cm=3m

\\ \bull\tt\longmapsto Volume=LBH

\\ \bull\tt\longmapsto Volume=8(4)(3)

\\ \bull\tt\longmapsto Volume=96m^3

6 0
2 years ago
Read 2 more answers
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