Why is it important to observe the necessary precautions in dancing??​

Answer:

2.30625m

Explanation:

The acceleration of the ball is the acceleration of gravity g.

By definition

g = constant = + 9.8 m?s^2  and is directed towards the centre of the earth

The ball does not slow down while falling

The position function is:

P(t) = 1/2 a t^2 + Vo t

In this specific case: P(t) = 4.9 t^2 + 8t

P(0.25s) = 4.9 (0.25)^2 + 8*0.25 = 2.30625m

The total moment in an islated system remains constant through time. a) V₂ = 109.43 m/s and V₃ = 33.33 m/s. b) h = 56.67 m. c) d = 480 m.

First let us remember that momentum or lineal momentum is a motion quantity. In physics it is the fundamental quantity that characterizes the motion of any object.

The momentum is a vectorial quantity that can be calculated as the product of the object mass by its lineal velocity ⇒ mv

The total lineal moment of a system constituted by a group of objects is the sum of the vectorial moments of each of the objects.

In an isolated system -the one that does not interact with the exterior environment-, the moment remains constant through time. This is the law of conservation of momentum. The initial moment is equal to the final moment ⇒ Pinitial = Pfinal

Objects involved in crushes, explosions, collitions, and others, are considered to be isolated systems.  

.......................

So, in the exposed example we know that,

a) We need to calculate the speed of the 0.7 and 0.3 kg objects.

So, 0.7 kg object moves in the original forward direction, meaning that we can use the following formula to calculate its velocity,

m₀v₀ = m₂v₂

Where

So first, we need to get the horizontal component of projectile velocity, V₀.

V₀ = V cosθ

V₀ = 50 cos40º

V₀ = 38.3 m/s

Now we can calculate the velocity of the 0.7 kg object.

m₀v₀ = m₂v₂

2 kg x 38.3 m/s = 0.7 kg x v₂

v₂ = (2 x 38.3) / 0.7

v₂ = 109.43 m/s

Now we need to calculate the velocity of the 0.3 kg object. Since this part goes exactly in the opposite direction of the 10 kg object, their addition will equal zero. So we can use the following equation to calculate the velocity of the 0.3 kg object.

m₁v₁ + m₃v₃ = 0

Where

m₁v₁ + m₃v₃ = 0

(1 kg x 10m/s) + (0.3 kg x v₃ ) = 0

(0.3 kg x v₃ ) = - (1 kg x 10m/s)

v₃ = - (1 kg x 10m/s) / 0.3

v₃ = - 33.33 m/s

The - sign is indicating the opposite direction concerning the 1 kg object.

ANSWER:

.........................

b) We need to get the high from the break-up point where the 0.3-kg piece go before rest

At rest,

1/2 m₃v₃² = m₃gh

Where,

1/2 m₃v₃² = m₃gh

h = 1/2 (v₃²) /g

h = 1/2 (33.33²) / 9.8

h = 56.67 m

ANSWER: The high from the break-up point where the 0.3-kg piece go before coming to rest is 56.67 m.

........................

c) Finally, we need to get where does the 0.7-kg piece land

ANSWER: The distance at which the 0.7 piece lands, concerning the point from which it was fired is 480 m.

You can learn more about law of conservation of momentum at

https://brainly.com/question/17140635

https://brainly.com/question/1113396

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Answer:

It could be a stringed instrument or an open pipe.

Explanation:

Let v be the speed of sound, y be wavelength and f be frequency.

v = yf

f= v/y; v is constant.

In a stringed instrument, the fundamental frequency note is heard when the length of the string, l = y/2; y= 2l

f′= v/2l

The second harmonic is heard when l= y

f"= v/y

...

f'''= 3v/2l

We can infer that f"= 2f'

f'''= 3f'

This is similar to the values in the question as;

523.26 =2(261.63) and so on.

Same thing happens with open pipes.

Answer:

b) cooking food

c) heating water

Explanation:

cooking and heating use thermal energy to transfer heat from the source to another object, in the examples provided; the food and the water. on the other hand, moving objects uses kinetic energy and lightbulbs utilize electrical energy.

Answer:

cooking food

heating water

i think making a light bulb work

Explanation:

Answer:

20 000 - 40 000 K    from the graph

Explanation:

Look straight down from blue giants and read the temps

Answer:

True

False

False

Explanation:

The first answer is inferred from Newton's third law of motion. As the rocket burns fuel and ejects the products downwards, a thrust is created, which acts downwards; pushing the rocket upwards.

The second answer is gotten from the same law. Pushing against the atmosphere (action) will create a reaction (this reaction is from the atmosphere; action and reaction are equal and opposite) which acts downwards. Hence the rocket doesn't move up.

Third. When the rocket is in space (vacuum), it moves at a uniform velocity since no force acts on it. This is gotten from Newton first law of motion. Here, the rocket moves into the orbit, fuelled by the force of it's thrust (when it was on the ground).

Answer:

the reflected ray will be 110 degree