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3 years ago

Which of the following is true? *

Physics

1 answer:

LUCKY_DIMON [66]3 years ago

4 0

Answer: D. Protons and neutrons

Explanation: PLEASE GIVE BRAINLIEST IT HELPS ALOT

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What happens to the work done when a force is doubled and the distance moved remain the same?

san4es73 [151]

Answer:

Work done gets doubled.

Explanation:

The work done by a force is given by :

W = Fd

Where

F is force and d is distance move

If the force is doubled and the distance moved remain the same, it would mean that the work done becomes double of the initial work done.

4 0

2 years ago

I NEED HELP PLEASE, THANKS! :)

mrs_skeptik [129]

Answer:

1. Largest force: C; smallest force: B; 2. ratio = 9:1

Explanation:

The formula for the force exerted between two charges is

$F=K\dfrac{ q_{1}q_{2}}{r^{2}}$

where K is the Coulomb constant.

q₁ and q₂ are also identical and constant, so Kq₁q₂ is also constant.

For simplicity, let's combine Kq₁q₂ into a single constant, k.

Then, we can write

$F=\dfrac{k}{r^{2}}$

1. Net force on each particle

Let's

Call the distance between adjacent charges d.
Remember that like charges repel and unlike charges attract.

Define forces exerted to the right as positive and those to the left as negative.

(a) Force on A

$\begin{array}{rcl}F_{A} & = & F_{B} + F_{C} + F_{D}\\& = & -\dfrac{k}{d^{2}} - \dfrac{k}{(2d)^{2}} +\dfrac{k}{(3d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(-1 - \dfrac{1}{4} + \dfrac{1}{9} \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-36 - 9 + 4}{36} \right)\\\\& = & \mathbf{-\dfrac{41}{36} \dfrac{k}{d^{2}}}\\\\\end{array}$

(b) Force on B

$\begin{array}{rcl}F_{B} & = & F_{A} + F_{C} + F_{D}\\& = & \dfrac{k}{d^{2}} - \dfrac{k}{d^{2}} + \dfrac{k}{(2d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1}{4} \right)\\\\& = &\mathbf{\dfrac{1}{4} \dfrac{k}{d^{2}}}\\\\\end{array}$

(C) Force on C

$\begin{array}{rcl}F_{C} & = & F_{A} + F_{B} + F_{D}\\& = & \dfrac{k}{(2d)^{2}} + \dfrac{k}{d^{2}} + \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( \dfrac{1}{4} +1 + 1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1 + 4 + 4}{4} \right)\\\\& = & \mathbf{\dfrac{9}{4} \dfrac{k}{d^{2}}}\\\\\end{array}$

(d) Force on D

$\begin{array}{rcl}F_{D} & = & F_{A} + F_{B} + F_{C}\\& = & -\dfrac{k}{(3d)^{2}} - \dfrac{k}{(2d)^{2}} - \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( -\dfrac{1}{9} - \dfrac{1}{4} -1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-4 - 9 -36}{36} \right)\\\\& = & \mathbf{-\dfrac{49}{36} \dfrac{k}{d^{2}}}\\\\\end{array}$

(e) Relative net forces

In comparing net forces, we are interested in their magnitude, not their direction (sign), so we use their absolute values.

$F_{A} : F_{B} : F_{C} : F_{D} = \dfrac{41}{36} : \dfrac{1}{4} : \dfrac{9}{4} : \dfrac{49}{36}\ = 41 : 9 : 81 : 49\\\\\text{C experiences the largest net force.}\\\text{B experiences the smallest net force.}\\$

2. Ratio of largest force to smallest

$\dfrac{ F_{C}}{ F_{B}} = \dfrac{81}{9} = \mathbf{9:1}\\\\\text{The ratio of the largest force to the smallest is $\large \boxed{\mathbf{9:1}}$}$

7 0

3 years ago

Two balls are thrown against a wall. Ball 1 has a much higher speed than ball 2.

Sunny_sXe [5.5K]

Let both the balls have the same mass equals to m.

Let $v_1$ and $v_2$ be the speed of the ball1 and the ball2 respectively, such that

$v_1>v_2\;\cdots(i)$

Assuming that both the balls are at the same level with respect to the ground, so let h be the height from the ground.

The total energy of ball1= Kinetic energy of ball1 + Potential energy of ball1. The Kinetic energy of any object moving with speed, $v$ , is $\frac 12 m v^2$

and the potential energy is due to the change in height is $mgh$ [where $g$ is the acceleration due to gravity]

So, the total energy of ball1,

$=\frac 12 m v_1^2 + mgh\;\cdots(ii)$

and the total energy of ball1,

$=\frac 12 m v_2^2 + mgh\;\cdots(iii)$ .

Here, the potential energy for both the balls are the same, but the kinetic energy of the ball1 is higher the ball2 as the ball1 have the higher speed, refer equation (i)

So, $\frac 12 m v_1^2 >\frac 12 m v_2^2$

Now, from equations (ii) and (iii)

The total energy of ball1 hi higher than the total energy of ball2.

6 0

2 years ago

In which direction does centripetal force act on an object

Gwar [14]

In a circle or circular motion

5 0

3 years ago

Read 2 more answers

A decrease in the magnitude of velocity is called

vagabundo [1.1K]

Any change in the speed or direction of motion is called "acceleration". You'll hear "deceleration" used for slowing down but that's not technically correct.

6 0

3 years ago