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PilotLPTM [1.2K]
2 years ago
10

5. Is the following measurement, 9.7 x 104 km East, a vector quantity or a scalar quantity?

Physics
1 answer:
Neporo4naja [7]2 years ago
3 0

Answer:

Vector quantity

Explanation:

What is a vector quantity?

-> A vector quantity has both direction and magnitude

What is scalar quantity?

-> S scalar quantity has only magnitude

[] Based on these definitions, 9.7 x 104 km East is a vector quantity because it includes a direction, East.

Have a nice day!

     I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly. (ノ^∇^)

- Heather

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The duration (i.e time) of the impact, given that the head impact was 1770 N is 0.005 s

<h3>How do I determine the duration (i.e time)?</h3>

Impulse is defined as the change in momentum of an object. It is expressed as:

Impulse = change in momentum

Impulse = final momentum – Initial momentum

Impule = m(v - u)

Impulse = force × time

Impulse = Ft

Thus,

Ft = m(v - u)

Where

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  • t is the time
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Now, we can obtain the duration (i.e time) of the impact. Details below:

  • Force (F) = 1770 N
  • Initial velocity = 2.09 m/s
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  • Final velocity = 0 m/s
  • Duration (t) =?

Ft = m(v + u) since the collision came to a stop

1770 × t = 4.12 × (0 + 2.09)

1770 × t = 4.12 × 2.09

Divide both sides by 1770

t = (4.12 × 2.09) / 1770

t = 0.005 s

Thus, the duration is 0.005 s

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A tank has the shape of an inverted circular cone with height 16m and base radius 3m. The tank is filled with water to a height
rewona [7]

Answer:

W=17085KJ

Explanation:

From the question we are told that:

Height H=16m

Radius R=3

Height of water H_w=9m

Gravity g=9.8m/s

Density of water \rho=1000kg/m^3

Generally the equation for Volume of water is mathematically given by

 dv=\pi*r^2dy

 dv=\frac{\piR^2}{H^2}(H-y)^2dy

Where

   y is a random height taken to define dv

Generally the equation for Work done to pump water is mathematically given by

 dw=(pdv)g (H-y)

Substituting dv

 dw=(p(=\frac{\piR^2}{H^2}(H-y)^2dy))g (H-y)

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Therefore

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 W=\int(\frac{\rho*g*R^2}{H^2}(H-y)^3)dy

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 W=\frac{1000*9.8*3.142*3^2}{9^2}[((9-y)^3)}^9_0

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 W=17085KJ

 

'

'

4 0
3 years ago
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