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3 years ago

A block of mass 11.0 kg slides from rest down a frictionless 38.0° incline and is stopped by a strong spring with

Physics

1 answer:

ivanzaharov [21]3 years ago

8 0

Answer:

11.72 mm

Explanation:

The gravitational potential energy equals the potential energy of the spring hence

$PE_{gravitational}=PE_{spring}$

$mgh=0.5kx^{2}$ where m is the mass of object, g is the acceleration due to gravity, h is the height, k is the spring constant and x is the extension of the spring

$mgdsin\theta=0.5kx^{2}$ where \theta is the angle of inclination and d is the sliding distance

Making x the subject then

$x=\sqrt {\frac {2mgdsin\theta}{k}}$

Substituting the given values then

$x=\sqrt{\frac {2\times 11\times 9.81\times 3\times sin 38}{2.9\times 10^{4}}}= 0.117240716\approx 11.72 mm$

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3 years ago

A cannon shoots an artillery shell with a initial velocity of 400 meters/second at an indirect fire angle of 60 on level ground

notsponge [240]

Answer: 14139.19 m

Explanation:

This situation is related to parabolic motion and can be solved using the following equations:

$x=V_{o}cos \theta t$ (1)

$y=y_{o}+V_{o} sin \theta t+\frac{g}{2}t^{2}$ (2)

Where:

$x$ is the horizontal distance (where the artillery shell lands)

$V_{o}=400 m/s$ is the initial velocity

$\theta=60\°$ is the angle

$t$ is the time

$y=0 m$ is the final height

$y_{o}=0 m$ is the initial height

$g=-9.8 m/s^{2}$ is the acceleration due gravity, always directed downwards

So, let's begin by isolating $t$ from (2):

$0=V_{o} sin \theta t+\frac{g}{2}t^{2}$ (3)

$t=-\frac{2 V_{o}sin \theta}{g}$ (4)

Substituting (4) in (1):

$x=V_{o}cos \theta (-\frac{2 V_{o}sin \theta}{g})$ (5)

Rewriting (5) and taking into account $sin(2\theta)=2 sin \theta cos \theta$ :

$x=-\frac{V_{o}^{2}sin(2\theta)}{g}$ (6)

$x=-\frac{(400 m/s)^{2}sin(2(60\°))}{-9.8 m/s^{2}}$ (7)

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3 years ago

Which diagram best illustrates what happens when electromagnetic waves strike a reflective material?

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Answer:

The second diagram does that

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3 years ago

Read 2 more answers

An object of mass 4 kg is initially at rest on a frictionless ice rink. It suddenly explodes into three pieces (I have no idea w

Rina8888 [55]

We have two events that occurred on different axes, the most convenient is to perform the operations on the two axes, and then look for the resulting force.

Our values are,

$m_1=1kg\\m_2 = 2kg\\v_1' = 1.2m/s\\v_1''= 0.8m/s$

PAR A) For the X axis, we apply momentum conservation, which is given by,

Total momentum before = Total momentum After

We start from rest, so in X the initial speeds are 0,

$m_1v_1+m_2v_2=m_1v_1'+m_2v_2'$

$0+0 = (1*1.2)+(1)v_2'$

$V_2' = -1.2m/s$

Now we apply for the conservation of the moment, it is part of the rest, so,

$m_1v_1+m_2v_2=m_1v_1''+m_2v_2''$

$0+0=2*0.8+1*v_2''$

$v_2'' = -1.6$

To find the total speed, we simply apply pitagoras,

$V = \sqrt{v_2'^2+v_2^2''}$

$V = \sqrt{1.2^2+1.6^2}$

$V = 2m/s$

PART B) The address is given by,

$tan\theta = \frac{V_2''}{V_2'}$

$\theta = tan^{-1} \frac{V_2''}{V_2'}$

$\theta = tan^{-1} \frac{1.6}{1.2}$

$\theta = 53.31\°$ (Below -x axis)

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4 years ago

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forsale [732]

Explanation:

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On the other hand, when a charged object touches a neutral object and then the charged object is removed will lead to the development of a charge on the neutral object. This charge is known as residual charge.

Thus, we can conclude that when a charged body is brought close to an uncharged body without touching it, an induced charge may result on the uncharged body. When a charged body is brought into contact with an uncharged body and then is removed, a residual charge may result on the uncharged body.

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4 years ago