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tiny-mole [99]
3 years ago
7

Nisha was making a list of things that are obtained from plants and Amol was making a list of things that can dissolve in water.

Nisha's List (obtained from plants)- Jute, wood, rubber.
Amol's list (can be dissolved in water)- Vinegar, lemon juice, cooking soda Which of the following could be added to both lists?Immersive Reader
(1 Point)
Oil
Sugar
Cotton
Table Salt
Physics
1 answer:
Allisa [31]3 years ago
6 0

Answer:

Sugar

Explanation:

Nisha's list:                               Amol's list

  Jute                                          Vinegar

  Wood                                       Lemon juice

  Rubber                                   Cooking soda

Nisha's list is made up of things that can be derived from plants. From the list given, oil, sugar and cotton can also be obtained from plants.

Most plant materials are organic matter.

Amol's list is made up of things that can dissolve in water. Sugar and table salt can also be dissolved in water.

Water is able to dissolve these materials because they are polar compounds. One rule of solubility is that like dissolves like.

New list:

Nisha's list:                               Amol's list

  Jute                                          Vinegar

  Wood                                       Lemon juice

  Rubber                                   Cooking soda

  Sugar                                       Sugar

                                   

Only sugar from the list can be added to both lists. It can be obtained from plant and can also dissolve in water.

                                             

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Angular momentum = 0.7 kg.m²/s

Angular velocity = 583.3 rad/s

Explanation:

1. The torque τ  is related to the angular momentum L by the relation

τ = ΔL/Δt

ΔL = τΔt

τ = 10 N. m

Δt = 70 ms = 70 × 10⁻³s

ΔL = (10 N. m) × (70 × 10⁻³s) = 700 × 10⁻³ kg.m²/s = 0.7 kg.m²/s

2. The rotational inertia I relates the angular momentum L to the angular velocity w

L = Iw

w = L/I

L =  0.7 kg.m²/s

I = 1.2 × 10⁻³ kg.m²

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3 0
2 years ago
A ball is thrown directly downward with an initial speed of 7.70 m/s, from a height of 30.2 m. After what time interval does it
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Answer:

t = 1.82

Explanation:

Given

u = 7.70m/s -- initial velocity

s = 30.2m --- height

Required

Determine the time to hit the ground

This will be solved using the following motion equation.

s = ut + \frac{1}{2}gt^2

Where

g = 9,8m/s^2

So, we have:

30.2 = 7.70t + \frac{1}{2} * 9.8 * t^2

30.2 = 7.70t + 4.9 * t^2

Subtract 30.2 from both sides

30.2 -30.2  = 7.70t + 4.9 * t^2 - 30.2

0  = 7.70t + 4.9 * t^2 - 30.2

0  = 7.70t + 4.9t^2 - 30.2

7.70t + 4.9t^2 - 30.2  = 0

4.9t^2 + 7.70t - 30.2  = 0

Solve using quadratic formula:

t = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}

Where

a = 4.9;\ b = 7.70;\ c = -30.2

t = \frac{-7.70\±\sqrt{7.70^2 - 4*4.9*-30.2}}{2*4.9}

t = \frac{-7.70\±\sqrt{651.21}}{9.8}

t = \frac{-7.70\±25.52}{9.8}

Split the expression

t = \frac{-7.70+25.52}{9.8} or t = \frac{-7.70-25.52}{9.8}

t = \frac{17.82}{9.8} or t = -\frac{33.22}{9.8}

Time can't be negative;  So, we have:

t = \frac{17.82}{9.8}

t = 1.82

Hence, the time to hit the ground is 1.82 seconds

7 0
2 years ago
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