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3 years ago

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A shot-putter released the shot at an angle of 41.5 degrees and a height of 1.9 m with an initial velocity of 13.3 m/s. How far

(in meters) will the shot go if we assume they do not reach over the stop board?

1 answer:

liq [111]3 years ago

3 0

Answer:

x = 17.88[m]

Explanation:

We can find the components of the initial velocity:

$(v_{x})_{o} = 13.3*cos(41.5)=9.96[m/s]\\(v_{y})_{o} = 13.3*sin(41.5)=8.81[m/s]$

We have to remember that the acceleration of gravity will be worked with negative sign, since it acts in the opposite direction to the movement in direction and the projectile upwards.

g = - 9.81[m/s^2]

Now we must find the time it takes for the projectile to hit the ground, as the problem mentions that it does not impact on the board.

$y=y_{o} +(v_{y} )_{o} *t-0.5*g*(t)^{2} \\0=1.9+(8.81*t)-(4.905*t^{2})\\-1.9=8.81*t*(1-0.5567*t)\\t=0\\t=1.796[s]$

With this time we can calculate the horizontal distance:

$x=(v_{x})_{o} *t\\x=9.96*1.796\\x=17.88[m]$

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When a carpenter shuts off his circular saw, the 10.0 inch diameter blade slows from 4250 rpm to 0.00 in 4.00 s. (a) What is the

MaRussiya [10]

Answer:

(a) $\alpha=-111.26rad/s$

(b) $s=4450.6in$

(c) $8.66in$

Explanation:

First change the units of the velocity, using these equivalents $1rev=2\pi rad$ and $1 min =60s$

$4250rpm(\frac{2\pi rad}{1rev})(\frac{1 min}{60 s} )=445.06rad/s$

The angular acceleration $\alpha$ the time rate of change of the angular speed $\omega$ according to:

$\alpha=\frac{\Delta \omega}{\Delta t}$

$\Delta \omega=\omega_i-\omega_f$

Where $\omega_i$ is the original velocity, in the case the velocity before starting the deceleration, and $\omega_f$ is the final velocity, equal to zero because it has stopped.

$\alpha=\frac{\Delta \omega}{\Delta t} =\frac{\omega_i-\omega_f}{4}\frac{0-445.06}{4} =\frac{-445.06}{4} =-111.26rad/s$

b) To find the distance traveled in radians use the formula:

$\theta = \omega_i t + \frac{1}{2} \alpha t^2$

$\theta = 445.06 (4) + \frac{1}{2}(-111.26) (4)^2=1780.24-890.12=890.12rad$

To change this result to inches, solve the angular displacement $\theta$ for the distance traveled $s$ ( $r$ is the radius).

$\theta=\frac{s}{r} \\s=\theta r$

$s=890.12(5)=4450.6in$

c) The displacement is the difference between the original position and the final. But in every complete rotation of the rim, the point returns to its original position. so is needed to know how many rotations did the point in the 890.16 rad of distant traveled:

$\frac{890.12}{2\pi}=141.6667$

The real difference is in the 0.6667 (or 2/3) of the rotation. To find the distance between these positions imagine a triangle formed with the center of the blade (point C), the initial position (point A) and the final position (point B). The angle $\gamma=\frac{2\pi}{3}=\frac{360^o}{3}=120$ is between the two sides known. Using the theorem of the cosine we can find the missing side of the the triangle(which is also the net displacement):

$c^2=a^2+b^2-2abcos(\gamma)$

$c^2=5^2+5^2-2(5)(5)cos(\frac{2\pi}{3} )\\c^2=25+25+25\\c^2=75\\c=5\sqrt{3}=8.66in$

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3 years ago

A spherically-spreading EM wave comes from a 104.0 W source. At a distance of 9.6 m, what is the intensity of the wave?

Nookie1986 [14]

Answer:

Approximately 0.0898 W/m².

Explanation:

The intensity of light measures the power that the light delivers per unit area.

The source in this question delivers a constant power of $\rm 104.0\; W$ . If the source here is a point source, that $\rm 104.0\; W$ of power will be spread out evenly over a spherical surface that is centered at the point source. In this case, the radius of the surface will be 9.6 meters.

The surface area of a sphere of radius $r$ is equal to $4\pi r^{2}$ . For the imaginary 9.6-meter sphere here, the surface area will be:

$\rm 4\pi \times (9.6\; m)^{2} \approx 1158.12\; m^{2}$ .

That $\rm 104.0\; W$ power is spread out evenly over this 9.6-meter sphere. The power delivered per unit area will be:

$\displaystyle\rm \frac{104.0\; W}{1158.12\; m^{2}}\approx 0.0898\; W\cdot m^{-2}$ .

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Planets A and B have the same mass, but planet A is half the size of planet B.

pychu [463]

Answer:

c. You would weigh less on planet A because the distance between

you and the planet's center of gravity would be smaller.

Explanation:

The statement that best describes your weight on each planet is that you would weigh less on planet A because the distance between you and the planet's center of gravity would be smaller.

This is based on Newton's law of universal gravitation which states that "the force of gravity between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distances between them".

Since weight is dependent on the force of gravity and mass, the planet with more gravitational pull will have masses on them weighing more.

Since the distance between the person and the center of the planet is smaller, therefore, the weight will be lesser.

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The anwser is A or D

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(I) In a ballistic pendulum experiment, projectile 1 results in a maximum height h of the pendulum equal to 2.6 cm. A second pro

Kipish [7]

Answer:

The second projectile was 1.41 times faster than the first.

Explanation:

In the ballistic pendulum experiment, the speed (v) of the projectile is given by:

$v = \frac{m + M}{m} \cdot \sqrt{2gh}$

<em>where m: is the mass of the projectile, M: is the mass of the pendulum, g: is the gravitational constant and h: is the maximum height of the pendulum. </em>

To know how many times faster was the second projectile than the first, we need to take the ratio for the velocities for the projectiles 2 and 1:

$\frac{v_{2}}{v_{1}} = \frac{\frac{m_{2} + M}{m_{2}} \cdot \sqrt{2gh_{2}}}{\frac{m_{1} + M}{m_{1}} \cdot \sqrt{2gh_{1}}}$ (1)

<em>where m₁ and m₂ are the masses of the projectiles 1 and 2, respectively, and h₁ and h₂ are the maximum height reached by the pendulum by the projectiles 1 and 2, respectively. </em>

Since the projectile 1 has the same mass that the projectile 2, we can simplify equation (1):

$\frac{v_{2}}{v_{1}} = \frac{\sqrt{h_{2}}}{\sqrt{h_{1}}}$

$\frac{v_{2}}{v_{1}} = \frac{\sqrt{5.2 cm}}{\sqrt{2.6 cm}}$

$\frac{v_{2}}{v_{1}} = 1.41$

Therefore, the second projectile was 1.41 times faster than the first.

I hope it helps you!

8 0

3 years ago