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Taya2010 [7]
3 years ago
14

A runner makes one lap around a 200m track of 25s. What were the runners average speed and average velocity?

Physics
1 answer:
kobusy [5.1K]3 years ago
5 0
Here is the answer. In order for us to get the speed, we are just going to use the formula speed is equal to distance over time. Given that the distance s 200m and the time is 25s, this would give us the answer of 8m/sec. Hope this answers your question.
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1. What is the momentum of a 1550 kg car that is traveling leftward at a velocity of 15 m/s?
Alik [6]

Answer:

Momentum, p = 23250 kg m/s

Explanation:

Given that

Mass of a car, m = 1550 kg

Speed pf car, v = 15 m/s

We need to find the momentum of the car. The formula for the momentum of an object is given by :

p = mv

Substituting all the values in the above formula

p = 1550 kg × 15 m/s

p = 23250 kg m/s

So, the momentum of the car is 23250 kg m/s.

3 0
3 years ago
Holding onto a tow rope moving parallel to a frictionless ski slope, a 70.1 kg skier is pulled up the slope, which is at an angl
lara [203]

Answer:

given,

mass of the skier = 70.1 Kg

angle with horizontal, θ = 8.6°

magnitude of the force,F = ?

a) Applying newton's second law

   m g sin\theta - F_{rope} = ma

 velocity is constant, a = 0

   m g sin\theta - F_{rope} =0

   F_{rope} = m g sin\theta

   F_{rope} = 70.1\times 9.8\times sin 8.6^0

  F_{rope}= 102.73\ N

b) now, when acceleration, a = 0.135 m/s²

   F_{rope}-m g sin\theta = ma

 velocity is constant, a = 0.135 m/s₂

   F_{rope} = m g sin\theta+ma

   F_{rope} = 70.1\times 9.8\times sin 8.6^0+70.1\times 0.135

  F_{rope}= 112.19\ N

7 0
3 years ago
When a river flows into an ocean, it slows down and deposits materials in its alluvial fan/delta
nikklg [1K]
When a river flows into an ocean, it slows down and deposits materials in its delta
8 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
Why is the scientific method described as cyclic?
Troyanec [42]
Because the scientific method can go around in a circle as many times as neccisary to get the results you need
8 0
3 years ago
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