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3 years ago

The application of force over distance is known as

Physics

1 answer:

rodikova [14]3 years ago

6 0

The application of force over distance is known as [WORK]

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Name human body organs where magnetism is significant

ser-zykov [4K]

It has been hypothesized, and some studies have supported the conjecture,
that certain species of insects and birds are able to sense the direction of external
magnetic fields.

I don't think there is any such notion where human beings are concerned.

5 0

3 years ago

Two coffee cups one at 80 degrees c another at 60 degrees c are put into a large cup what is the final temperature of the coffee

8_murik_8 [283]

If you're trying to sum it up, it would be 140 degrees c*, otherwise it would probably be 80 degrees c since its more hotter than 60 degrees

8 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

2 years ago

A force of 20. Newtons to the left exerted on a cart for 10. Seconds. For what period of time must a 50.-newton force to the rig

FinnZ [79.3K]

Impulse = (force) x (time)

The first impulse was (20 N) x (10 sec) = 200 meters/sec

The second one is (50 N) x (time) and we want it equal to the first one, so

(50 N) x (time) = 200 meters/sec

Divide each side by 50N : Time = 200/50 = <em>4 seconds</em>

By the way, the quantity we're playing with here is the cart's <em>momentum</em>.

6 0

3 years ago

yaroslaw [1]

Answer:

12 J

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 7.6 kg

Distance (d) = 6 m

Velocity (v) = 5 m/s

Force (F) = 2 N

Workdone (Wd) =.?

Workdone can be defined as the product of force and distance moved in the direction of the force. Mathematically, it is expressed as:

Workdone = Force × distance

Wd = F × d

With the above formula, we can obtain the workdone as follow:

Distance (d) = 6 m

Force (F) = 2 N

Workdone (Wd) =.?

Wd = F × d

Wd = 2 × 6

Wd = 12 J

Thus, the workdone is 12 J

6 0

2 years ago