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2 years ago

I NEED HELP ASAP

Physics

2 answers:

pogonyaev2 years ago

8 0

Answer:

Meteor, Streak of Light and it has not yet reached earths atmosphere.

Explanation:

A meteoroid is a celestial object which is very smaller than an asteroid. These objects are produced as a collision impact from mars or moon and float freely in space without any specific orbit. When they come inside the Earth's gravitational field, they are attracted by the Earth's gravity to Earth's crust. These objects in Earth's atmosphere are called meteors. As they travel through Earth's atmosphere, they do face a huge friction from Earth's atmosphere which let them burn and that is visible as the tail of the meteor.

Most of them are so small that they are burnt away in the atmosphere. But some are bigger and they reach the Earth's surface and are called as meteorites.

(not my words)

dimaraw [331]2 years ago

5 0

Answer:

Explanation:

Meteor because it reached earth's atmosphere

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A bullet fired into a fixed target loses half of its velocity after penetrating 3 cm. How much further it will penetrate before

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${\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{First \: penetrating \: length\:(s_{1}) = 3 \: cm}$

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${\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{Left \: Penetration \: length \: before \: it \: comes \: to \: rest \:( s_{2} )}$

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${\mathfrak{\underline{\purple{\:\:\: Calculation:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{Let \: Initial \: velocity = v\:m/s} \\\\$

$\:\:\:\:\bullet\:\:\:\sf{Left \: velocity \: after \: s_{1} \: penetration = \dfrac{v}{2} \:m/s} \\\\$

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☯ As we know that,

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$\dashrightarrow\:\: \sf{ {v}^{2} = {u}^{2} + 2as }$

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$\dashrightarrow\:\: \sf{ \bigg(\dfrac{v}{2} \bigg)^{2} = {v}^{2} + 2a s_{1}}$

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$\dashrightarrow\:\: \sf{ \dfrac{ {v}^{2} }{4} = {v}^{2} + 2 \times a \times 0.03 }$

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$\dashrightarrow\:\: \sf{ \dfrac{ {v}^{2} }{4} - {v}^{2} = 0.06 \times a }$

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$\dashrightarrow\:\: \sf{\dfrac{ - 3{v}^{2} }{4} = 0.06 \times a }$

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$\dashrightarrow\:\: \sf{a = \dfrac{ - 3 {v}^{2} }{4 \times 0.06} }$

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$\dashrightarrow\:\: \sf{ a = \dfrac{ - 25 {v}^{2} }{2}\:m/s^{2} ......(1) }$

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$\:\:\:\:\bullet\:\:\:\sf{ Initial\:velocity=v\:m/s} \\\\$

$\:\:\:\:\bullet\:\:\:\sf{ Final \: velocity = 0 \: m/s }$

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$\dashrightarrow\:\: \sf{ {v}^{2} = {u}^{2} + 2as}$

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$\dashrightarrow\:\: \sf{{0}^{2} = {v}^{2} + 2 \times \dfrac{ - 25 {v}^{2} }{2} \times s }$

$\\$

$\dashrightarrow\:\: \sf{ - {v}^{2} = - 25 {v}^{2} \times s }$

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$\dashrightarrow\:\: \sf{ s = \dfrac{ - {v}^{2} }{ - 25 {v}^{2} }}$

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$\dashrightarrow\:\: \sf{ s = \dfrac{1}{25} }$

$\\$

$\dashrightarrow\:\: \sf{ s = 0.04 \: m }$

$\\$

☯ For left penetration (s₂)

$\\$

$\dashrightarrow\:\: \sf{s = s_{1} + s_{2} }$

$\\$

$\dashrightarrow\:\: \sf{ 0.04 = 0.03 + s_{2}}$

$\\$

$\dashrightarrow\:\: \sf{ s_{2} = 0.04 - 0.03 }$

$\\$

$\dashrightarrow\:\: \sf{s_{2} = 0.01 \: m = {\boxed{\sf{\purple{1 \: cm }}} }}$

$\\$

$\star\:\sf{Left \: penetration \: before \: it \: come \: to \: rest \: is \:{\bf{ 1 \: cm}}} \\$

4 0

2 years ago

You can use any coordinate system you like in order to solve a projectile motion problem. To demonstrate the truth of this state

posledela

Answer:

a) y₂ = 49.1 m , t = 1.02 s , b) y = 49.1 m , t= 1.02 s

Explanation:

a) We will solve this problem with the missile launch kinematic equations, to find the maximum height, at this point the vertical speed is zero

$v_{y}$ ² = $v_{oy}$ ² - 2 g (y –yo)

The origin of the coordinate system is on the floor and the ball is thrown from a height

y-yo = $v_{oy}² /2 g
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 y - 0 = 5.10 m
 
The height from the ground is the height that rises from the reference system plus the depth of the ground from the reference system
 y₂ = 5.1 + 44
 y₂ = 49.1 m
Let's use the other equation to find the time
 [tex]v_{y}$ = $v_{oy}$ - g t

t = $v_{oy}$ / g

t = 10 / 9.8

t = 1.02 s

b) the maximum height

y- 44.0 = $v_{y}$ ² / 2 g

y - 44.0 = 5.1

y = 5.1 +44.0

y = 49.1 m

The time is the same because it does not depend on the initial height

t = 1.02 s

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2 years ago

A Ping-Pong ball has a mass of 2.3 g and a terminal speed of 9.4 m/s. The drag force is of the form bv^2 What is the value of b?

julia-pushkina [17]

Drag Force = bv^2 = ma; a = g = 9.81 m/s^2

b = mg/v^2 = (0.0023×9.81)/(9.4^2)

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The answer is A

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