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3 years ago

11

A cannon at rest fires a 32.5 kg cannonball forward at 388 m/s. After firing, the cannon recoils at 7.42 m/s. What is the mass o

f the cannon?
(Unit = kg)
Remember: right is +, left is -

1 answer:

Bond [772]3 years ago

3 0

Answer:

1700 kg

Explanation:

Let’s use conservation of momentum

32.5 * 388 = 7.42 * mc

mc = 1699.46

mc = 1700 kg

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an oculist charges 30 for an eye exam, frames and glasses lenses, but 42 for an eye examination, frames, and plastic lenses. If

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The cost of the glass lenses is calculated as follows:

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Cost of plastic lenses = 4y

Cost of eye exam and frames be z.

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2 years ago

A 17.5 kg block is dragged over a rough, horizontal surface by a constant force of 167 N acting at an angle of 30◦ above the hor

torisob [31]

Answer:

The final velocity of the block is 68.85m/s.

Explanation:

The final velocity can be determined by means of the equations for a Uniformly Accelerated Rectilinear Motion:

$v_{f}^{2} = v_{i}^{2} + 2ad$

$v_{f} = \sqrt{v_{i}^{2} + 2ad}$ (1)

But it is necessary to know the acceleration. For a better procedure it will be listed the knowns and unknowns of the problem:

Knowns:

F = 167 N

$\theta$ = 30°

m = 1.75 Kg

d = 23.9 m

$\mu$ = 0.136

Unknowns:

$F_{r}$ = ?

a = ?

The acceleration can be found by means of Newton's second law:

$\sum F_{net} = ma$

Where $\sum F_{net}$ is the net force, m is the mass and a is the acceleration.

$Fx + Fy = ma$ (2)

All the forces can be easily represented in a free body diagram, as it is shown below.

<em>Force in the x axis:</em>

$F_{x} = F + W_{x} - F_{r}$ (3)

<em>Forces in the y axis: </em>

$F_{y} = N - W_{y}$ (4)

<em>Solving for the forces in the x axis:</em>

$F_{x} = F + W_{x} - F_{r}$

Notice that is necessary to found $F_{r}$ :

$F_{r} = \mu N$ (5)

The normal force can be obtained from equation (4)

$N - W_{y} = 0$

$N = W_{y}$

The component of the weight in the y axis can be gotten by means of trigonometry:

$\frac{Adjacent}{Hypotenuse} = cos \theta$

$\frac{W_{y}}{W} = cos \theta$

$W_{y}= W cos \theta$

Remember that the weight is defined as:

$W = mg$

$W_{y}= mgcos \theta$

$N = mg cos \theta$

$F_{r} = \mu mgcos \theta$

$F_{r} = (0.136)(1.75Kg)(9.8m/s^{2})(cos30)$

$F_{r} = 2.01N$

The component of the weight in the x axis can be gotten by means of trigonometry:

$\frac{Opposite}{Hypotenuse} = sen \theta$

$\frac{W_{x}}{W} = sen \theta$

$W_{x} = W sen \theta$

$W_{x} = mgsen \theta$

$W_{x} = (1.75Kg)(9.8m/s^2)(sen30)$

$W_{x} = 8.57N$

Then, replacing $W_{x}$ and $F_{r}$ in equation (3) it is gotten:

$F_{x} = 167N + 8.57N - 2.01N$

$F_{x} = 173.56N$

Solving for the forces in the y axis:

$F_{y} = N - W_{y}$

$F_{y} = mgcos \theta - mgcos \theta$

$F_{y} = 0$

Replacing the values of $F_{x}$ and $F_{x}$ in equation (2) it is gotten:

$F_{x} + 0 = ma$

$F_{x} = ma$

$a = \frac{F_{x}}{m}$

$a = \frac{173.56N}{1.75Kg}$

$a = \frac{173.56Kg.m/s^{2}}{1.75Kg}$

$a = 99.17m/s^{2}$

Now that the acceleration is known, equation (1) can be used:

$v_{f} = \sqrt{v_{i}^{2} + 2ad}$

However, since the block was originally at rest its initial velocity will be zero ( $v_{i} = 0$ ).

$v_{f} = \sqrt{2ad}$

$v_{f} = \sqrt{2(99.17m/s^{2})(23.9m)}$

$v_{f} = \sqrt{2(99.17m/s^{2})(23.9m)}$

$v_{f} = 68.85m/s$

<u>Hence, the final velocity of the block is 68.85m/s.</u>

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4 years ago

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An athlete is running at a constant velocity with a javelin held in his right hand. The force he is applying on the javelin as h

NeTakaya

-- Since our athlete is running at constant velocity, AND the javelin is
staying with him, the speed of the javelin itself must also be constant.

-- Therefore, the force of 5N that he's applying to the javelin can't possibly
be in the direction that he and the javelin are moving, otherwise the javelin
would be accelerating.

-- So we know that the force is perpendicular to the javelin's motion ... like
probably straight up to prevent it from falling to the ground.

-- Since none of the force is in the direction the javelin is moving, the force
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4 years ago

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Un tren se mueve en línea recta con velocidad media de 90 km/h. ¿que distancia recorre en el termino de de 5 horas?

jeka94

Answer:

Distance = 450 kilometres

Explanation:

Given the following data;

Speed = 90 km/h

Time = 5 hours

To find the distance covered by the train;

Mathematically, the speed of an object or body is given by the formula;

$Speed = \frac {distance}{time}$

Making distance the subject of formula, we have;

Distance = speed * time

Substituting into the formula, we have;

Distance = 90 * 5

Distance = 450 kilometres

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3 years ago