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3 years ago

What happened to a apple's weigh as the plane rose up toward the sky?

1 answer:

5 0

Okay, I don't know if this question is supposed to be a trick question or not. The weight of the apple does not change as the plane travels up the atmosphere, but the MASS changes. Weight doesn't change no matter what environment you're in, but the mass changes in different environments. In this case, the weight is constant but the mass is decreasing as you go higher up.

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What are the two most important traits between a whale and a shark?

nordsb [41]

Answer:

Both are aquatic animals and are hunters

Explanation:

8 0

3 years ago

What name is used for scientists who study the weather in an area over a long time, such as 200 years? Hydrologists weather repo

andrew11 [14]

Answer:

I think it is meteorologist because they also study of the atmosphere, atmospheric phenomena, and atmospheric effects on our weather.

4 0

3 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 horses are pulling a box in two different directions as shown in the below image. The image shows one 30.0 N force due north

Novay_Z [31]

Explanation:

'What is the magnitude of the force needed to stop the horses and bring the box into equilibrium?' ≈42N; according to the vectors rules.

'Where would you locate the rope to apply the force?' - in point D.

PS. zoom out the attached picture.

4 0

3 years ago

A 4 kg rock is dropped from 5 m. There is no friction. What kind of energy does is have before? What kind of energy does it have

denpristay [2]

1) The total mechanical energy of the rock is:
$E=U+K$
where U is the gravitational potential energy and K the kinetic energy.

Initially, the kinetic energy is zero (because the rock starts from rest, so its speed is zero), and the total mechanical energy of the rock is just gravitational potential energy. This is equal to
$E_i=U=mgh$
where $m=4 kg$ is the mass, $g=9.81 m/s^2$ is the gravitational acceleration and $h=5 m$ is the height.
Putting the numbers in, we find the potential energy
$U=mgh=(4 kg)(9.81 m/s^2)(5 m)=196.2 J$

2) Just before hitting the ground, the potential energy U is zero (because now h=0), and all the potential energy of the rock converted into kinetic energy, which is equal to:
$E_f=K= \frac{1}{2}mv^2$
where v is the speed of the rock just before hitting the ground. Since the mechanical energy of the rock must be conserved, then the kinetic energy K before hitting the ground must be equal to the initial potential energy U of the rock:
$K=U=196.2 J$

3) For the work-energy theorem, the work W done by the gravitational force on the rock is equal to the variation of kinetic energy of the rock, which is:
$W=196.2 J-0 J=196.2 J$

6 0

3 years ago