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Fed [463]
4 years ago
10

Two tiny beads are 25 cm apart with no other charges or fields present. Bead A carries 10 µC of charge and bead B carries 1 µC.

Which one of the following statements is true about the magnitudes of the electric forces on these beads?
The force on A is 10 times the force on
b. The force on A is 100 times the force on
b. The force on A is exactly equal to the force on
b. The force on B is 100 times the force on
a. The force on B is 10 times the force on
a.
Physics
1 answer:
melomori [17]4 years ago
8 0
The correct answer is (B) The force on A is exactly equal to the force on B. Two tiny beads are 25 cm apart with no other charges or fields present. Bead A carries 10 µC of charge and bead B carries 1 µC. The statement that<span> is true about the magnitudes of the electric forces on these beads is the force on A is exactly equal to B.</span>
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An object 1 of mass m1 is separated by some distance d from an object 2 of mass 2m1 . An object 3 of mass m3 is to be placed bet
Harrizon [31]

If the potential energy of the three-object system is to be a maximum (closest to zero), should object 3 be placed closer to object 1, closer to object 2, or halfway between them?

If the potential energy of the three-object system is to be a maximum (closest to zero), should object 3 be placed closer to object 1, closer to object 2, or halfway between them?

 

Object 3 should be placed closer to object 1.

 

Object 3 should be placed on a halfway between object 2 and object 1.

 

Object 3 should be placed closer to object 2.

 

Solution

I think that Object 3 should be placed closer to object 2.

6 0
3 years ago
A car with 2 × 10^3 kg moving at a speed of 10 m/s collides and sticks with car B of mass of 3 × 10^3 kg initially at rest. How
stepan [7]

Answer:

6 \times 10^4 \; \rm J.

Explanation:

KE lost = Total KE before Collision - Total KE after Collision.

For each car, the KE before collision can simply be found with the equation:

\displaystyle \mathrm{KE} = \frac{1}{2}\, m \cdot v^2, where

  • m is the mass of the car, and
  • v is the speed of the car.

The 2 \times 10^3\; \rm kg car would have an initial KE of:

\displaystyle \frac{1}{2} \times 2 \times 10^3 \times 10^2 = 10^5\; \rm J.

The 3 \times 10^3\; \rm kg car was initially not moving. Hence, its speed and kinetic energy would zero before the collision.

To find the velocity of the two cars after the collision, apply the conservation of momentum.

The momentum p of an object is equal to its mass m times its velocity v. In other words, p = m\cdot v.

Let the mass of the two cars be denoted as m_1 and m_2, and their initial speeds v_1 and v_2. Since the two cars are stuck to each other after the collision, their final speeds would be the same. Let that speed be denotes as v_3.

Initial momentum of the two-car system:

\begin{aligned}& m_1 \cdot v_1 + m_2 \cdot v_2 \\ &= 2 \times 10^3 \times 10 + 3 \times 10^3 \times 0 \\ &= 2 \times 10^4\; \rm kg \cdot m \cdot s^{-1}\end{aligned}.

After the collision, both car would have a velocity of v_3 (since they were stuck to each other.) As a result, the final momentum of the two-car system would be:

m_1\cdot v_3 + m_2 \cdot v_3 = (m_1 + m_2)\, v_3.

Since momentum is conserved during the collision, the momentum of the system after the collision would also be 2 \times 10^4 \; \rm kg \cdot m \cdot s^{-1}. That is: (m_1 + m_2) \, v_3 = 2 \times 10^4 \; \rm kg \cdot m \cdot s^{-1}.

Solve for v_3:

\begin{aligned} v_3 &= \frac{(m_1 + m_2)\, v_3}{m_1 + m_2} \\ &= \frac{2 \times 10^4}{2 \times 10^3 + 3 \times 10^3} \\ &= \frac{2 \times 10^4}{5 \times 10^3} \\ &= 4 \; \rm m \cdot s^{-1}\end{aligned}.

Hence, the total kinetic energy after the collision would be:

\begin{aligned} &\frac{1}{2}\, m_1 \, v^2 + \frac{1}{2}\, m_2\, v^2 \\ &= \frac{1}{2}\, (m_1 + m_2)\, v^2 \\ &= \frac{1}{2} \times \left(2 \times 10^3 + 3 \times 10^3\right) \times 4^2 \\ &= 4 \times 10^4\; \rm J\end{aligned}.

The amount of kinetic energy lost during the collision would be:

\begin{aligned}&\text{KE After Collision} - \text{KE Before Collision} \\ &= 10^5 - 4 \times 10^4 \\&= 6\times 10^4\; \rm J \end{aligned}.

5 0
3 years ago
The magnetic field lines of a bar magnet spread out from the north end to the south end. South end to the north end. Edges to th
Gwar [14]
<h3><u>Answer;</u></h3>

the north end to the south end.

<h3><u>Explanation;</u></h3>
  • Magnetic field lines from a bar magnet form lines that are closed. The direction of magnetic field is taken to be outward from the North pole of the magnet and in to the South pole of the magnet.
  • A magnetic field refers to  the area surrounding a magnet where a force is exerted on certain objects. These lines are spread out of the north end of the magnet.
  • The magnetic field lines resemble a bubble.
3 0
3 years ago
Read 2 more answers
a man exerts 3000.00N of force to push a car 35.00 meters in 90.00 seconds.... 1. what is the work done 2.what is the power gene
vaieri [72.5K]
1).  Work = (force) x (distance)

Work = (3,000 newtons) x (35 meters) = 105,000 newton-meters = <em>105,000 joules</em>

2).  Power = (work) / (time)

Power = (105,000 joules) / (90 seconds)
                                         = 1,166-2/3 joules per second =<em> 1,166 and 2/3 watts</em> .

<em>Note:</em>
That's no ordinary man.
1,166 watts is the same as roughly 1.6 horsepower.
Not too many people can sustain 1 horsepower or more for 90 seconds.




4 0
4 years ago
What should you do if someone on foot approaches your car while you are in it?
tatyana61 [14]
It depends. If you are driving and the person doesn't look like a serial killer, you should stop.
6 0
3 years ago
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