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zaharov [31]
3 years ago
13

25. A car is accelerating when it is (a) traveling on a straight, flat road at 50 miles per hour. (b) traveling on a straight up

hill road at 30 miles per hour. (c) going around a circular track at a steady 100 miles per hour.
Physics
1 answer:
katovenus [111]3 years ago
3 0

Answer:

(c) going around a circular track at a steady 100 miles per hour.

Explanation:

The acceleration of an object is equal to the rate of change of velocity of the object:

a=\frac{\Delta v}{\Delta t}

where

\Delta v is the change in velocity of the object

\Delta t is the time interval

We notice that velocity is a vector, so it has both a magnitude and a direction. This means that \Delta v, the change in velocity, is either caused by a change in magnitude of velocity, or by a change in the direction.

Therefore, the acceleration is non-zero if at least one of the two is verified:

- There is a change in magnitude of the velocity

- There is a change in direction

Let's now analyze the three statements:

(a) traveling on a straight, flat road at 50 miles per hour. --> here there is no change in magnitude or direction of the velocity, so there is no acceleration.

(b) traveling on a straight uphill road at 30 miles per hour. --> here there is no change in magnitude or direction of the velocity, so there is no acceleration.

(c) going around a circular track at a steady 100 miles per hour. --> here there is no change in magnitude of the velocity, but the direction is changing (circular track), therefore there is acceleration.

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Three beads are placed on the vertices of an equilateral triangle of side d = 3.4cm. The first bead of mass m1=140gis placed on
Vlad [161]

Answer:

Xcm = 1.95 cm  and Ycm = 1.76 cm

Explanation:

The very useful concept of mass center is

     R cm = 1/M  ∑ m_{i}  r_{i}

Where ri, mi are the mass positions of the bodies from some reference point by selecting and M is the total mass of the body.

Let's look for the total mass

     M = m₁ + m₂ + m₃

     M = 140 + 45 + 85

     M = 270 g

Let's look for the position of each point

Point 1. top vertex, if the triangle has as side d

      R₁ = d / 2 i ^ + d j ^

      R₁ = (1.7 cm i ^ + 3.4 j ^) cm

Point 2. left vertex. What is the origin of the system?

      R₂ = 0

Point 3. Right vertex

      R₃ = d i ^

      R₃ = 3.4 i ^ cm

a) The x component of the massage center

      Xcm = 1 / M (m₁ x₁ + m₂ x₂ + m₃ x₃)

      Xcm = 1 / M (m₁ d / 2 + 0 + m₃ d)

      Xcm = d / M (m₁ / 2 + m₃)

b)   Let's write the mass center component x

      Xcm = 1/270 (1.7 140 + 0 + 3.4 85)

      Xcm = 238/270

      Xcm = 1.95 cm

c) let's find the component and center of mass

     Ycm = 1 / M (m₁ y₁ + m₂ y₂ + m₃ y₃)

    Ycm = 1 / M (m₁ d + 0 + 0)

    Ycm = m₁ / M d

d) let's calculate

    Y cm = 1/270 (140 3.4 + 0 + 0)

    Ycm = 1.76 cm

8 0
3 years ago
A center-seeking force related to acceleration is _______ force
34kurt
A because centrifugal is to velocity to how slow or fast something is  and centrifugal has expresssed as ac=v2 / r (1)<span />
5 0
3 years ago
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When a satellite is in orbit around the earth the force of gravity on the satellite-
mojhsa [17]

Answer:

Is always towards the center of the Earth

Explanation:

As a satellite moves around the Earth in a circular orbit, the direction of the force of gravity is always towards the center of the Earth. At an altitude of 100 km, you would be so high that you would see black sky and stars if you looked upwards.

5 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
The owner of a company that manufactures drinking cups decides it would be impressive to build an inground swimming pool that is
garri49 [273]

Answer:

The depth is 5.15 m.

Explanation:

Lets take the depth of the pool = h m

The atmospheric pressure ,P = 101235 N/m²

The area of the top = A m²

The area of the bottom =  a m²

Given that A= 1.5 a

The force on the top of the pool = P A

The total pressure on the bottom = P + ρ g h

ρ =Density of the  water = 1000 kg/m³

The total pressure at the bottom of the pool =  (P + ρ g h) a

The bottom and the top force is same

(P + ρ g h) a = P A

P a +ρ g h a =  P A

ρ g h a =  P A - P a

h=\dfrac{P ( A-a)}{\rho g a}

h=\dfrac{P ( 1.5 a-a)}{\rho g a}

h=\dfrac{P ( 1.5- 1)}{\rho g}

h=\dfrac{101235 ( 1.5- 1)}{1000\times 9.81}\ m

h=5.15 m

The depth is 5.15 m.

7 0
3 years ago
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