Ask question

Ask question

All categories

3 years ago

12

A Meteorite Strikes On October 9, 1992, a 27-pound meteorite struck a car in Peekskill, NY, leaving a dent 22 cm deep in the tru

nk.If the meteorite struck the car with a speed of 130 m/s, what was the magnitude of its deceleration, assuming it to be constant?

1 answer:

vagabundo [1.1K]3 years ago

7 0

Answer:

Acceleration of the meteorite, $a=-38409.09\ m/s^2$

Explanation:

It is given that,

A Meteorite after striking struck a car, v = 0

Initial speed of the Meteorite, u = 130 m/s

Distance covered by Meteorite, s = 22 cm = 0.22 m

We need to find the magnitude of its deceleration. It can be calculated using the third equation of motion as :

$a=\dfrac{v^2-u^2}{2s}$

$a=\dfrac{-(130)^2}{2\times 0.22}$

$a=-38409.09\ m/s^2$

So, the deceleration of the Meteorite is $-38409.09\ m/s^2$ . Hence, this is the required solution.

You might be interested in

A particle is moving along a straight line such that its' acceleration is defined as

ANTONII [103]

Answer:

1. $s=-vt^2$

2. $v=\frac{s}{t}$

3. $a=-2\times \frac{s}{t}$

Explanation:

Given:

Acceleration as a Function of velocity, $a=-2v\ m.s^{-2}$

Velocity, $v=20\ m.s^{-1}$ at displacement s=0 and t=0

1.

<u>Particles position as a function of time:</u>

<u>Using equation of motion:</u>

$s=u.t+\frac{1}{2} a.t^2$

where:

$u =$ initial velocity

$v=$ final velocity

so,

$s=0+\frac{1}{2}\times (-2v) t^2$

$s=-vt^2$

2.

<u>Particle velocity as a function of time:</u>

$v=\frac{s}{t}$

3.

<u>Particles acceleration as a function of time:</u>

$a=-2\times \frac{s}{t}$

7 0

3 years ago

Two students are messing around with a nerf gun during a fire drill. They manage to fire a foam dart directly upwards from an in

Westkost [7]

Answer:

v₀= 19.23 m/s

Explanation:

Look at the attached graphic

Foam dart Kinematic 1-2 (upward movement), vf₂=0 ,

We calculate t₁ and y₁ to reach the highest point (2)

vf₂=v₀-gt₁

t₁=v₀/g

v f₂²=v₀²-2g*y₁

2g*y₁ =v₀² , y₁=v₀²/2g

Foam dart Kinematic 2-3 (downward movement), v₀₂=0 ,

We calculate t₂ and y₂ from the highest point (2) to

touch the ground (3)

vf₃=v₀₂+gt₂ , v₀₂=0

gt₂=vf₃ , t₂=vf₃/g

vf₃²=v₀₂²+2g*y₂

$v_{f3} =\sqrt{2*g*y_{2} }$

y₂= 1.5+y₁

y₂= 1.5+v₀²/2g

$v_{f3} =\sqrt{2g(1.5+v_{o}^{2}/2g) }$

$vf_{3} =\sqrt{v_{o}^{2}+29.4 }$

We propose the equation for the total time (Four seconds) :

t₁ = time it takes from position 1 to position 2 (going up)

t₂: and time that takes from position 2 to position 3 ( going down)

t₁+t₂=4

v₀/g+ vf₃/g =4

v₀ + vf₃ =4g

$v_{o} +\sqrt{v_{o} ^{2}+29.4 } =4*g$ : we move v to the other side and square both sides of the equation

v₀²+29.4=(4g-v₀²)²

v₀²+29.4=(4g)²-8gv₀+v₀² we eliminate v₀²

29.4=(4g)²-8gv₀

8*g*v₀=(4*g)²-29.4

v₀=1507.24/78,4

v₀= 19.23 m/s

6 0

3 years ago

A hawk flying at 11 m/s at an altitude of 132 m accidentally drops its prey. The parabolic trajectory of the falling prey is des

WITCHER [35]

Answer:

s = 153.34 m

Explanation:

given,

speed of hawk flying = 11 m/s

altitude = 132 m

equation of parabolic trajectory

$y = 132 -\dfrac{x^2}{33}$ ............at y = 0 x =66

$ds=\sqrt{dx^2+dy^2}$

$ds=dx\sqrt{1+(\dfrac{dy}{dx})^2}$

$\dfrac{dy}{dx}=\dfrac{-2x}{33}$

$ds=dx\sqrt{1+(\dfrac{-2x}{33} )^2}$

$ds=dx\sqrt{1+(\dfrac{4x^2}{1089} )}$

integrating

$s = \int _0^66\sqrt{1+(\dfrac{4x^2}{1089} )}$

using formula

$\sqrt{x^2+a^2}=\dfrac{x}{2}\sqrt{x^2+a^2}+\dfrac{a^2}{2}log|x+\sqrt{x^2+a^2}|$

s = 153.34 m

3 0

3 years ago

Kepler's laws and explanations

IgorC [24]

Answer:

<em>The first law states that</em> every planet describes an elliptical path about the sun as a single focus.

<em>The</em><em> </em><em>second</em><em> </em><em>law</em><em> </em><em>states</em><em> </em><em>that</em><em> </em>The line joining the planet to the sun sweeps out equal areas in equal time intervals.

<em>The</em><em> </em><em>third</em><em> </em><em>law</em><em> </em><em>states</em><em> </em><em>that</em><em> </em>The squares of the period of revolution is proportional to the cubes of the mean distance between the planet and the sun

5 0

3 years ago

Read 2 more answers

A local AM radio station broadcasts at a frequency of 627 kHz. Calculate the wavelength at which it is broadcasting. Wavelength

rewona [7]

Answer:

Wavelength = 478.46 m

Explanation:

It is given that,

A local AM radio station broadcasts at a frequency of 627 kHz, f = 627000 Hz

We need to find the wavelength at which it is broadcasting. The wavelength is given by :

$\lambda=\dfrac{c}{f}\\\\=\dfrac{3\times 10^8}{627\times 10^3}\\\\=478.46\ m$

So, the wavelength is 478.46 m.

6 0

4 years ago