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Evgen [1.6K]
2 years ago
13

1. Describe what conversation of momentum means.

Physics
2 answers:
MakcuM [25]2 years ago
7 0

Explanation:

<em>1</em><em>.</em><em> </em><em>Conservation</em><em> </em><em>of</em><em> </em><em>momentum</em><em>,</em><em> </em><em>general awareness</em><em> </em><em>of</em><em> </em><em>physics</em><em> </em><em>according</em><em> </em><em>to</em><em> </em><em>which</em><em> </em><em>the</em><em> </em><em>quantity</em><em> </em><em>called</em><em> </em><em>momentum</em><em> </em><em>that</em><em> </em><em>characteri</em><em>z</em><em>es</em><em> </em><em>motion</em><em> </em><em>never</em><em> </em><em>changes</em><em> </em><em>in</em><em> </em><em>an</em><em> </em><em>isolated</em><em> </em><em>collection</em><em> </em><em>of</em><em> </em><em>objects</em><em>.</em><em> </em>

Snowcat [4.5K]2 years ago
3 0

Answer:

1. The conversation of momentum is when a person or object has momentum and transfers it to another person or object.

Explanation:

2. Like if someone is on a swing and kicks someone in front of him. He stops and the kid in front of him falls over because the momentum was transferred.

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A hot cube of iron was heated up using 1500 J of thermal energy and was placed in a beaker of water. Before it was heated, the i
Mariana [72]

Answer:

451.13 J/kg.°C

Explanation:

Applying,

Q = cm(t₂-t₁)............... Equation 1

Where Q = Heat, c = specific heat capacity of iron, m = mass of iron, t₂= Final temperature, t₁ = initial temperature.

Make c the subject of the equation

c = Q/m(t₂-t₁).............. Equation 2

From the question,

Given: Q = 1500 J, m = 133 g = 0.113 kg, t₁ = 20 °C, t₂ = 45 °C

Substitute these values into equation 2

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4 0
3 years ago
A bird is about 6.26.2 in.​ long, with a​ thin, dark bill and a​ wide, white wing stripe. If the bird can fly 9292 mi with the w
Trava [24]

Answer:

209 mph

Explanation:

V = Speed of bird in still air

v = Speed of wind = 44 mph

Consider the motion of the bird with the wind

D_{1} = distance traveled with the wind = 9292 mi

t_{1} = time taken to travel the distance with wind

Time taken to travel the distance with wind is given as

t_{1} = \frac{D_{1}}{V + v}

t_{1} = \frac{9292}{V + 44}                              eq-1

Consider the motion of the bird with the wind

D_{2} = distance traveled against the wind = 6060 mi

t_{2} = time taken to travel the distance against wind

Time taken to travel the distance against wind is given as

t_{2} = \frac{D_{2}}{V + v}

t_{2} = \frac{6060}{V - 44}                              eq-2

As per the question,

Time taken with the wind = Time taken against the wind

t_{1} = t_{2}

\frac{9292}{V + 44} = \frac{6060}{V - 44}

(9292) (V - 44) = (6060) (V + 44)

9292V - 408848 = 6060V + 266640

3232V = 675488

V = 209 mph

5 0
3 years ago
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