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Mama L [17]
1 year ago
13

Think of something from everyday life that follows a two-dimensional path. It could be a kicked football, a bus that's turning a

corner, or a person jogging around a track, etc. Describe your scenario in detail, and then identify the acceleration at each point. When is the acceleration vector not aligned with the direction of travel
Physics
1 answer:
OLEGan [10]1 year ago
4 0

Answer:

Let us consider the case of a bus turning around a corner with a constant velocity, as the bus approaches the corner, the velocity at say point A is Va, and is tangential to the curve with direction pointing away from the curve. Also, the velocity at another point say point B is Vb and is also tangential to the curve with direction pointing away from the curve.<em> </em><em>Although the velocity at point A and the velocity at point B have the same magnitude, their directions are different (velocity is a vector quantity), and hence we have a change in velocity. By definition, an acceleration occurs when we have a change in velocity, so the bus experiences an acceleration at the corner whose direction is away from the center of the corner</em>.

The acceleration is not aligned with the direction of travel because<em> the change in velocity is at a tangent (directed away) to the direction of travel of the bus.</em>

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A charming friend of yours who has been reading a little bit about astronomy accompanies you to the campus observatory and asks
lidiya [134]

Answer:

b

Explanation:

3 0
2 years ago
To eight significant figures, Avogadro's constant is 6.0221367×10^(23)mol−1. Which of the following choices demonstrates correct
Zepler [3.9K]

Answer with Explanation:

We are given Avogadro's constant =6.0221367\times 10^{23}mol^{-1}

There are eight significant figures.

We have to round off.

1.If we round off to four significant figures

The ten thousandth place of Avogadro's constant is less than five therefore, digits on left side of ten thousandth  place remains same and digits on right side of ten thousandth place and ten thousandth place  replace by zero.

 Then ,Avogadro's constant can be written as

6.022\times 10^{23}mol^{-1}

If we round off to 2 significant figures

Hundredth place of given number is less than 5 therefore, digits on left side of hundredth  place remains same and digits on right side of hundredth place and hundredth place replace by zero.

Then,Avogadro's constant can be written as

6.0\times 10^{23}mol^{-1}

If we round off six significant figures

6 is greater than 5 therefore, 1 will be added to 3 and digits on right side of 6 and 6 replace by zero and digits on left side of 6 remains same except 3.

Then, the Avogadro's constant can be written as

6.02214\times 10^{23}mol^{-1}

3 0
2 years ago
A 450g mass on a spring is oscillating at 1.2Hz. The totalenergy of the oscillation is 0.51J. What is the amplitude.
Volgvan

Answer:

A=0.199

Explanation:

We are given that  

Mass of spring=m=450 g==\frac{450}{1000}=0.45 kg

Where 1 kg=1000 g

Frequency of oscillation=\nu=1.2Hz

Total energy of the oscillation=0.51 J

We have to find the amplitude of oscillations.

Energy of oscillator=E=\frac{1}{2}m\omega^2A^2

Where \omega=2\pi\nu=Angular frequency

A=Amplitude

\pi=\frac{22}{7}

Using the formula

0.51=\frac{1}{2}\times 0.45(2\times \frac{22}{7}\times 1.2)^2A^2

A^2=\frac{2\times 0.51}{0.45\times (2\times \frac{22}{7}\times 1.2)^2}=0.0398

A=\sqrt{0.0398}=0.199

Hence, the amplitude of oscillation=A=0.199

4 0
1 year ago
A roller coaster car drops a maximum vertical distance of 35.4 m. Determine the maximum speed of the car at the bottom of that d
marissa [1.9K]

Answer:

The maximum speed of the car at the bottom of that drop is 26.34 m/s.

Explanation:

Given that,

The maximum vertical distance covered by the roller coaster, h = 35.4 m

We need to find the maximum speed of the car at the bottom of that drop. It is a case of conservation of energy. The energy at bottom is equal to the energy at top such that :

mgh=\dfrac{1}{2}mv^2

v=\sqrt{2gh}

v=\sqrt{2\times 9.8\times 35.4}

v = 26.34 m/s

So, the maximum speed of the car at the bottom of that drop is 26.34 m/s. Hence, this is the required solution.

8 0
2 years ago
A 5.00-kg ball is hanging from a long but very light flexible wire when it is struck by a 1.50-kg stone traveling horizontally t
devlian [24]

consider the right direction as positive and left direction as negative.

M = mass of the ball = 5 kg

m = mass of stone = 1.50 kg

V_{bi} = initial velocity of the ball before collision = 0 m/s

V_{si} = initial velocity of the stone before collision = 12 m/s

V_{bf} = final velocity of the ball after collision = ?

V_{sf} = final velocity of the stone after collision = - 8.50 m/s

using conservation of momentum

MV_{bi} + mV_{si} = MV_{bf} + mV_{sf}

(5) (0) + (1.5) (12) = 5 V_{bf} + (1.50) (- 8.50)

V_{bf} = 6.15 m/s

h = height gained by the ball

using conservation of energy

Potential energy gained by ball at Top = kinetic energy at the bottom

Mgh = (0.5) MV_{bf}^{2}

(9.8) h = (0.5) (6.15)²

h = 1.93 m

8 0
2 years ago
Read 2 more answers
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