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aleksandrvk [35]

3 years ago

5

Meteorite On October 9, 1992, a 27-pound meteorite struck a car in Peekskill, NY, creating a dent about 22 cm deep. If the initi

al speed of the meteorite was 550 m/s, what was the average force exerted on the meteorite by the car?

2 answers:

Rom4ik [11]3 years ago

8 0

Answer:

8427375 N

Explanation:

From newton's equation of motion,

F = ma................. Equation 1

Where F = average force exerted on the meteorite by the car, m = mass of the meteorite, a = acceleration.

We can find the value of acceleration from newton's equation of motion

using,

v² = u²+2as.................. Equation 2

Where v = final velocity, u = initial velocity, a = acceleration, s = distance.

Make a the subject of the equation

a = (v²-u²)/2s............... Equation 3

Note: When the meteorite struck the car, it was brought to rest

Given: u = 550 m/s, v = 0 m/s ( brought to rest), s = 22 cm = 0.22 m

Substitute into equation 3

a = (0²-550²)/(2×0.22)

a = -302500/0.44

a = -687500 m/s²

Also given: m = 27 pound = (27×0.454) = 12.258 kg

Substitute into equation 1

F = 12.258(-687500)

F = -8427375 N

Note: The negative sign tells that the force act in opposite direction to the motion of the meteorite

Viktor [21]3 years ago

7 0

Answer:

The average force exerted on the meteorite by the car is 8.435x10⁶N

Explanation:

Given:

m = 27 lb = 12.27 kg

D = 22 cm = 0.22 m

v1 = 550 m/s

The acceleration of the meteorite is:

$v_{2} ^{2} =v_{1} ^{2}+2aD\\0=550^{2} +2a(0.22)\\a=-\frac{550^{2} }{2*0.22} =-687500m/s^{2}$

Negative sign indicates opposite to the initial velocity. The average force is equal to:

Magnitude of a = 687500 m/s²

F = m*a = 12.27 * 687500 = 8.435x10⁶N

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How many non-square numbers lie between the squares of 12 and 13?

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Hint: Here, we can see that 12 and 13 are consecutive numbers. So, all numbers between squares of 12 and 13 are non-square numbers. Therefore, first find squares of 12 and 13 and then subtract square of 12 from square of 13, we get numbers of non-square numbers. At the last subtract 1 from the result obtained as both extremes numbers are not included.

Complete step-by-step answer:

In these types of questions, a simple concept of numbers should be known that is between squares of two consecutive numbers all numbers are non-square numbers. Also one tricky point should remember that whenever we find the difference between two numbers we get a number of numbers between them including anyone of the extreme numbers. So we subtract 1 to exclude both extreme numbers.

Square of 12 = 122=144 and square of 12 = 132=169

As 12 and 13 are consecutive numbers so all numbers between their squares will be non-square numbers.

Therefore, 169 – 144 = 25

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In the Olympic shot-put event, an athlete throws the shot with an initial speed of 12.0m/s at a 40.0? angle from the horizontal.

HACTEHA [7]

A) Horizontal range: 16.34 m

B) Horizontal range: 16.38 m

C) Horizontal range: 16.34 m

D) Horizontal range: 16.07 m

E) The angle that gives the maximum range is $41.9^{\circ}$

Explanation:

A)

The motion of the shot is a projectile motion, so we can analyze separately its vertical motion and its horizontal motion.

The vertical motion is a uniformly accelerated motion, so we can use the following suvat equation to find the time of flight:

$s=u_y t + \frac{1}{2}at^2$ (1)

where

s = -1.80 m is the vertical displacement of the shot to reach the ground (negative = downward)

$u_y = u sin \theta$ is the initial vertical velocity, where

u = 12.0 m/s is the initial speed

$\theta=40.0^{\circ}$ is the angle of projection

So

$u_y=(12.0)(sin 40.0^{\circ})=7.7 m/s$

$a=g=-9.8 m/s^2$ is the acceleration due to gravity (downward)

Substituting the numbers, we get

$-1.80 = 7.7t -4.9t^2\\4.9t^2-7.7t-1.80=0$

which has two solutions:

t = -0.21 s (negative, we ignore it)

t = 1.778 s (this is the time of flight)

The horizontal motion is instead uniform, so the horizontal range is given by

$d=u_x t$

where

$u_x = u cos \theta=(12.0)(cos 40^{\circ})=9.19 m/s$ is the horizontal velocity

t = 1.778 s is the time of flight

Solving, we find

$d=(9.19)(1.778)=16.34 m$

B)

In this second case,

$\theta=42.5^{\circ}$

So the vertical velocity is

$u_y = u sin \theta = (12.0)(sin 42.5^{\circ})=8.1 m/s$

So the equation for the vertical motion becomes

$4.9t^2-8.1t-1.80=0$

Solving for t, we find that the time of flight is

t = 1.851 s

The horizontal velocity is

$u_x = u cos \theta = (12.0)(cos 42.5^{\circ})=8.85 m/s$

So, the range of the shot is

$d=u_x t = (8.85)(1.851)=16.38 m$

C)

In this third case,

$\theta=45^{\circ}$

So the vertical velocity is

$u_y = u sin \theta = (12.0)(sin 45^{\circ})=8.5 m/s$

So the equation for the vertical motion becomes

$4.9t^2-8.5t-1.80=0$

Solving for t, we find that the time of flight is

t = 1.925 s

The horizontal velocity is

$u_x = u cos \theta = (12.0)(cos 45^{\circ})=8.49 m/s$

So, the range of the shot is

$d=u_x t = (8.49)(1.925)=16.34$ m

D)

In this 4th case,

$\theta=47.5^{\circ}$

So the vertical velocity is

$u_y = u sin \theta = (12.0)(sin 47.5^{\circ})=8.8 m/s$

So the equation for the vertical motion becomes

$4.9t^2-8.8t-1.80=0$

Solving for t, we find that the time of flight is

t = 1.981 s

The horizontal velocity is

$u_x = u cos \theta = (12.0)(cos 47.5^{\circ})=8.11 m/s$

So, the range of the shot is

$d=u_x t = (8.11)(1.981)=16.07 m$

E)

From the previous parts, we see that the maximum range is obtained when the angle of releases is $\theta=42.5^{\circ}$ .

The actual angle of release which corresponds to the maximum range can be obtained as follows:

The equation for the vertical motion can be rewritten as

$s-u sin \theta t + \frac{1}{2}gt^2=0$

The solutions of this quadratic equation are

$t=\frac{u sin \theta \pm \sqrt{u^2 sin^2 \theta+2gs}}{-g}$

This is the time of flight: so, the horizontal range is

$d=u_x t = u cos \theta (\frac{u sin \theta \pm \sqrt{u^2 sin^2 \theta+2gs}}{-g})=\\=\frac{u^2}{-2g}(1+\sqrt{1+\frac{2gs}{u^2 sin^2 \theta}})sin 2\theta$

It can be found that the maximum of this function is obtained when the angle is

$\theta=cos^{-1}(\sqrt{\frac{2gs+u^2}{2gs+2u^2}})$

Therefore in this problem, the angle which leads to the maximum range is

$\theta=cos^{-1}(\sqrt{\frac{2(-9.8)(-1.80)+(12.0)^2}{2(-9.8)(-1.80)+2(12.0)^2}})=41.9^{\circ}$

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