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2 years ago

15

A small rubber ball is launched by a compressed-air cannon from ground level with an initial speed of 17.3 m/s directly upward.

Choose upward as the positive direction in your analysis. (a) What is the maximum height above the ground that the ball reaches?

1 answer:

elena-s [515]2 years ago

7 0

Answer:

h=15.27m

Explanation:

Since at maximum height the vertical velocity must be null it's better to use the formula:

$v_f^2=v_i^2+2ad$

We will use this formula for the vertical direction, choosing the upward direction as the positive one, so we have:

$0=v_i^2+2ah$

or

$h=-\frac{v_i^2}{2a}$

which for our values is:

$h=-\frac{(17.3m/s)^2}{2(-9.8m/s^2)}=15.27m$

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An object of mass m = 5.0 kg hangs from a cord around a light pulley: The length of the cord between the oscillator and the pull

puteri [66]

Answer:

$\mu=0.0049Kg/m$

Explanation:

When a standing wave is formed with six loops means the normal mode of the wave is n=6, the frequency of the normal mode is given by the expression:

$f_n=\frac{nv}{2L}$

Where $L$ is the length of the string and $v$ the velocity of propagation. Use this expression to find the value of $v$ .

$f_6=\frac{6v}{2L}\\(150)=\frac{6v}{2(2)} \\150=\frac{3v}{2} \\3v=150(2)\\ v=\frac{300}{3} \\v=100m/s$

The velocity of propagation is given by the expression:

$v=\sqrt{\frac{T}{\mu }$

Where $\mu$ is the desirable variable of the problem, the linear mass density, and $T$ is the tension of the cord. The tension is equal to the weight of the mass hanging from the cord:

$T=W=mg=(5)(9.81)=49.05N$

With the value of the tension and the velocity you can find the mass density:

$v=\sqrt{\frac{T}{\mu}$

$v^2=\frac{T}{\mu}\\ \mu=\frac{T}{v^2} =\frac{49.05}{(100)^2} =\frac{49.05}{10000} =0.0049Kg/m$

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2 years ago

As charges move in a closed loop, they

mamaluj [8]

As charges move in a closed loop, they gain as much energy as they lose.

<h3>What is principle of conservation of energy?</h3>

According to the principle of conservation of energy, in a closed or isolated system, the total energy of the system is always conserved.

The energy gained by the particles or charges in a closed system is equal to the energy lost by the charges.

Thus, we can conclude the following based on principles of conservation of energy;

As charges move in a closed loop, they gain as much energy as they lose.

Learn more about conservation of energy here: brainly.com/question/166559

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2 years ago

A particle's trajectory is described by x = (0.5t^3-2t^2) meters and y = (0.5t^2-2t), where time is in seconds. What is the part

mestny [16]

Differentiate the components of position to get the corresponding components of velocity :

$v_x = \dfrac{\mathrm dx}{\mathrm dt} = \left(1.5\dfrac{\rm m}{\mathrm s^3}\right) t^2 - \left(4\dfrac{\rm m}{\mathrm s^2}\right)t$

$v_y = \dfrac{\mathrm dy}{\mathrm dt} = \left(1\dfrac{\rm m}{\mathrm s^2}\right)t-2\dfrac{\rm m}{\rm s}$

At <em>t</em> = 5.0 s, the particle has velocity

$v_x = \left(1.5\dfrac{\rm m}{\mathrm s^3}\right) (5.0\,\mathrm s)^2 - \left(4\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s) = 17.5\dfrac{\rm m}{\rm s}$

$v_y = \left(1\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s)-2\dfrac{\rm m}{\rm s} = 3.0\dfrac{\rm m}{\rm s}$

The speed at this time is the magnitude of the velocity :

$\sqrt{{v_x}^2 + {v_y}^2} \approx \boxed{17.8\dfrac{\rm m}{\rm s}}$

The direction of motion at this time is the angle $\theta$ that the velocity vector makes with the positive <em>x</em>-axis, such that

$\tan(\theta) = \dfrac{3.0\frac{\rm m}{\rm s}}{17.5\frac{\rm m}{\rm s}} \implies \theta = \tan^{-1}\left(\dfrac{3.0}{17.5}\right) \approx \boxed{9.73^\circ}$

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2 years ago

A pitcher throws a baseball with a mass of 143 g horizontally at a speed of 38.8 m/s (87 mi/h). The hitter's bat is in contact w

SIZIF [17.4K]

Answer:

F = −10093.41 N

Explanation:

Given that,

Mass of a baseball, m = 143 g = 0.143 kg

Initial speed of the baseball, u = +38.8 m/s

The hitter's bat is in contact with the ball for 1.20 ms and then travels straight back to the pitcher's mound at a speed of 45.9 m/s, v = -45.9 m/s

We need to find the average force exerted on the ball by the bat. So, Force is given by :

$F=ma$

a is acceleration

$F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{0.143\times (-45.9-(38.8))}{1.2\times 10^{-3}}\\\\F=-10093.41\ N$

So, the average force exerted on the ball by the bat has a magnitude of 10093.41 N.

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3 years ago

When you lift a box off the ground gravity opposes your lifting force.

ivolga24 [154]

Answer:

TRUE?

Explanation:

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2 years ago