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4 years ago

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This problem uses the same concepts as Multiple-Concept Example 17, except that kinetic, rather than static, friction is involve

d. A crate is sliding down a ramp that is inclined at an angle 30.6 ° above the horizontal. The coefficient of kinetic friction between the crate and the ramp is 0.360. Find the acceleration of the moving crate.

1 answer:

coldgirl [10]4 years ago

8 0

Answer: $1.94 m/s^2$

Explanation:

Given

inclination $\theta =30.6^{\circ}$

Coefficient of kinetic Friction $\mu =0.36$

Forces on crate is

$mg\sin \theta -f_r=ma$

$f_r=\mu mg\cos \theta$

$mg\sin \theta -\mu mg \cos \theta =ma$

$a=g\sin \theta -\mu g\cos \theta$

$a=1.94 m/s^2$

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An object of mass 0.50 kg is released from the top of a building of height 8 m. The object experiences a horizontal constant for

liberstina [14]

To solve this problem with the given elements we will apply the linear motion kinematic equations. We will start by calculating the time taken, with the vertical displacement data. Subsequently, with the components of the acceleration, we will obtain the magnitude of the total acceleration, to finally obtain the horizontal displacement with the data already found.

PART A) From vertical movement we know that the acceleration is equivalent to gravity and the displacement is 8m so the time taken to carry out the route would be

$h = \frac{1}{2} gt^2$

Here,

$h = 8 m$

$g = 9.8 m/s^2$

Replacing,

$8 = 0.5 * 9.8 * t^2$

$t = 1.277 sec$

PART B) Now, Magnitude of acceleration

$a = \sqrt{a_x^2 + a_y^2}$

$a_x = \frac{1.9}{0.5} = 3.8 m/s^2$

$a_y = g = 9.8 m/s^2$

Thus, magnitude of net acceleration

$a = \sqrt{3.82^2 + 9.8^2}= 10.51 m/s^2$

PART C) Finally the displacement along horizontal direction is:

$s =v_0 t + \frac{1}{2} a t^2$

$s = 0 + \frac{1}{2} (3.8)(1.277)^2$

$s = 3.098 m$

Therefore the distance traveled along the horizontal direction before it hits the ground is 3.098m

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3 years ago

Explain the purpose of a geologic time scale

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Geologic time scales are essential tools for geologists and biologists alike because they help show the history of the entire world in terms of relationships between different plate movements.

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Read 2 more answers

The model below shows a carbide ion. 8 light grey and 6 dark grey balls sit at the center with 2 concentric black rings around t

azamat

Answer:

This carbide ion change into a neutral carbon atom.

Explanation:

Given that,

Light grey balls = 8

dark grey balls = 6

medium grey balls = 2

Charge on dark grey ball = +1

Charge on light grey ball = 0

Charge on medium grey ball = -1

We need to change this carbide ion to a neutral carbon atom

Using given data

A dark grey ball has one positive charge. It means, it is a proton.

A light grey ball has has zero charge. It means, it is a neutron.

A medium grey ball has one negative charge.It means, it is a electron.

We know that,

The number of proton and number of electron is equal in a neutral atom.

So, No charge is present on the given model of atom.

Therefore, we can say that the atom is neutral.

We know that,

The sum of total number of protons presents in an atom it is called atomic number.

So, The total number of protons is 8, the total number of electrons is 8 and the total number of neutrons is 9 in given atom.

Hence, This carbide ion change into a neutral carbon atom.

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4 years ago

A 1.0 kg object is attached to a string 0.50 m. It is twirled in a horizontal circle above the ground at a speed of 5.0 m/s. A b

aivan3 [116]

<span>50 N The centripetal force upon an object is expressed as F = mv^2/r So let's substitute the known values and calculate F = mv^2/r F = 1.0 kg * (5.0 m/s)^2 / 0.5 m F = 1.0 kg * 25 m^2/s^2 / 0.5 m F = 25 kg*m^2/s^2 / 0.5 m F = 50 kg*m/s^2 F = 50 N So the answer is 50 N which matches one of the available choices.</span>

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3 years ago

PLEASEEEE HEEEEELP!!!!!

8090 [49]

Answer:

Before:

$p_{truck}=16400\ kg.m/s$

$p_{car}=10000\ kg.m/s$

After:

$p_{truck}=8000\ kg.m/s$

$p_{car}=8400\ kg.m/s$

$v_{fcar}=8.4\ m/s$

$F=9333.33 \ Nw$

Explanation:

<u>Conservation of Momentum</u>

Two objects of masses m1 and m2 moving at speeds v1o and v2o respectively have a total momentum of

$p_1=m_1v_{1o}+m_2v_{2o}$

After the collision, they have speeds of v1f and v2f and the total momentum is

$p_2=m_1v_{1f}+m_2v_{2f}$

Impulse J is defined as

$J=F.t$

Where F is the average impact force and t is the time it lasted

Also, the impulse is equal to the change of momentum

$J=\Delta p$

As the total momentum is conserved:

$p_1=p_2$

$m_1v_{1o}+m_2v_{2o}=m_1v_{1f}+m_2v_{2f}$

We can compute the speed of the second object by solving the above equation for v2f

$\displaystyle v_{2f}=\frac{ m_1v_{1o}+m_2v_{2o} -m_1v_{1f} }{ m_2 }$

The given data is

$m_1=2000\ kg$

$m_2=1000\ kg\\v_{1o}=8.2\ m/s\\v_{2o}=0\ m/s\\v_{1f}=4\ m/s$

a) The impulse will be computed at the very end of the answer

b) Before the collision

$p_{truck}=2000\cdot 8.2=16400\ kg.m/s$

$p_{car}=1000\cdot 0=0\ kg.m/s$

c) After collision

$p_{truck}=2000\cdot 4=8000\ kg.m/s$

Compute the car's speed:

$\displaystyle v_{2f}=\frac{ 16400+0 -8000 }{ 1000 }$

$v_{2f}=8.4\ m/s$

And the car's momentum is

$p_{car}=1000\cdot 8.4=8400\ kg.m/s$

The Impulse J of the system is zero because the total momentum is conserved, i.e. \Delta p=0.

We can compute the impulse for each object

$J_1=\Delta p_1=2000(4-8.2)=-8400 \N.s$

The force can be computed as

$\displaystyle F=\frac{J}{t}=-\frac{8400}{0.9}=-9333.33 \ Nw$

The force on the car has the same magnitude and opposite sign

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4 years ago