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AfilCa [17]
3 years ago
13

Runner A is initially 6.0 km west of a flagpole and is running with a constant velocity of 9.0 km/h due east. Runner B is initia

lly 5.0 km east of the flagpole and is running with a constant velocity of 8.0 km/h due west. What will be the distance of the two runners from the flagpole when their paths cross
Physics
1 answer:
Eduardwww [97]3 years ago
7 0

Answer:

From the data we know that runner A and runner B are 11 km apart from the start because (6+5) km

So the runner from the east direction has distance as unknown km, rate= 9 k/h ; time= d/r=x/9 hr

So runner towards the west will be

distance = 11-x, rate= 8 k/h, time = d/r = (11-x)/8

So equating east and west time we have

x/9= (11-x)/8

8x=99-9x

17x=99

x=5.92 km

That is the distance covered by runner towards the east and he will meet the runner toward the west at

6-5.92=0.08 km west of the flagpole.

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What kind of light results when the total spectrum of refracted light is recombined? violet green white red?
ElenaW [278]
Hello
The final light will be white. In fact, each color of the visible spectrum is an electromagnetic wave with its own specific frequency and wavelength. White, instead, does not have a specific frequency: it is the sum of all the different wavelengths of the visible spectrum. Therefore, when recombining the spectrum of the refracted light all the different frequencies recombine together, and their sum gives white light.

(edited)
4 0
3 years ago
I need the solution to this
posledela

Answer:

He could jump 2.6 meters high.

Explanation:

Jumping a height of 1.3m requires a certain initial velocity v_0. It turns out that this scenario can be turned into an equivalent: if a person is dropped from a height of 1.3m in free fall, his velocity right before landing on the ground will be v_0. To answer this equivalent question, we use the kinematic equation:

v_0 = \sqrt{2gh}=\sqrt{2\cdot 9.8\frac{m}{s^2}\cdot 1.3m}=5.0\frac{m}{s}

With this result, we turn back to the original question on Earth: the person needs an initial velocity of 5 m/s to jump 1.3m high, on the Earth.

Now let's go to the other planet. It's smaller, half the radius, and its meadows are distinctly greener. Since its density is the same as one of the Earth, only its radius is half, we can argue that the gravitational acceleration g will be <em>half</em> of that of the Earth (you can verify this is true by writing down the Newton's formula for gravity, use volume of the sphere times density instead of the mass of the Earth, then see what happens to g when halving the radius). So, the question now becomes: from which height should the person be dropped in free fall so that his landing speed is 5 m/s ? Again, the kinematic equation comes in handy:

v_0^2 = 2g_{1/2}h\implies \\h = \frac{v_0^2}{2g_{1/2}}=\frac{25\frac{m^2}{s^2}}{2\cdot 4.9\frac{m}{s^2}}=2.6m

This results tells you, that on the planet X, which just half the radius of the Earth, a person will jump up to the height of 2.6 meters with same effort as on the Earth. This is exactly twice the height he jumps on Earth. It now all makes sense.

6 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
Please help with this question.
EastWind [94]

Answer:

41.16 Joules

Explanation:

Potential energy at a given instant is a function of mass and height of an object. The formula is

E_p = mgh = 2.80kg\cdot 9.8\frac{m}{s^2}\cdot 1.50m = 41.16 J

4 0
3 years ago
A positively charged metal sphere, A, is held close to but not touching and identical uncharged sphere, sphere B. Sphere A is no
Yuri [45]

Answer:

The sphere C carries no net charge.

Explanation:

  • When brougth close to the charged sphere A, as charges can move freely in  a conductor, a charge equal and opposite to the one on the sphere A, appears on the sphere B surface facing to the sphere A.
  • As sphere B must remain neutral (due to the principle of conservation of charge) an equal charge, but of opposite sign, goes to the surface also, on the opposite part of the sphere.
  • If sphere A is removed, a charge movement happens in the sphere B, in such a way, that no net charge remains on the surface.
  • If in such state, if  the sphere B (assumed again uncharged completely, without any local charges on the surface), is touched by an initially uncharged sphere C, due to the conservation of  charge principle, no net  charge can be built on sphere C.
3 0
3 years ago
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