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3 years ago

11

What does the word resolution mean in the sentence "By 1915, the British Roentgen Society had adopted a resolution to protect pe

ople from overexposure to X-rays"?

1 answer:

iris [78.8K]3 years ago

5 0

Resolution in this sentence refers to a solution to the problem

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A softball is thrown from the origin of an x-y coordinate system with an initial speed of 18 m/s at an angle of 35∘ above the ho

podryga [215]

<span>What we need to first do is split the ball's velocity into vertical and horizontal components. To do that multiply by the sin or cos depending upon if you're looking for the horizontal or vertical component. If you're uncertain as to which is which, look at the angle in relationship to 45 degrees. If the angle is less than 45 degrees, the larger value will be the horizontal speed, if the angle is greater than 45 degrees, the larger value will be the vertical speed. So let's calculate the velocities sin(35)*18 m/s = 0.573576436 * 18 m/s = 10.32437585 m/s cos(35)*18 m/s = 0.819152044 * 18 m/s = 14.7447368 m/s Since our angle is less than 45 degrees, the higher velocity is our horizontal velocity which is 14.7447368 m/s. To get the x positions for each moment in time, simply multiply the time by the horizontal speed. So 0.50 s * 14.7447368 m/s = 7.372368399 m 1.00 s * 14.7447368 m/s = 14.7447368 m 1.50 s * 14.7447368 m/s = 22.1171052 m 2.00 s * 14.7447368 m/s = 29.48947359 m Rounding the results to 1 decimal place gives 0.50 s = 7.4 m 1.00 s = 14.7 m 1.50 s = 22.1 m 2.00 s = 29.5 m</span>

6 0

3 years ago

A 4.87-kg ball of clay is thrown downward from a height of 3.21 m with a speed of 5.21 m/s onto a spring with k = 1570 N/m. The

Yuki888 [10]

Answer:

Approximately $0.560\; {\rm m}$ , assuming that:

the height of $3.21\; {\rm m}$ refers to the distance between the clay and the top of the uncompressed spring.
air resistance on the clay sphere is negligible,
the gravitational field strength is $g = 9.81\; {\rm m\cdot s^{-2}}$ , and
the clay sphere did not deform.

Explanation:

Notations:

Let $k$ denote the spring constant of the spring.
Let $m$ denote the mass of the clay sphere.
Let $v$ denote the initial speed of the spring.
Let $g$ denote the gravitational field strength.
Let $h$ denote the initial vertical distance between the clay and the top of the uncompressed spring.

Let $x$ denote the maximum compression of the spring- the only unknown quantity in this question.

After being compressed by a displacement of $x$ , the elastic potential energy $\text{PE}_{\text{spring}}$ in this spring would be:

$\displaystyle \text{PE}_{\text{spring}} = \frac{1}{2}\, k\, x^{2}$ .

The initial kinetic energy $\text{KE}$ of the clay sphere was:

$\displaystyle \text{KE} = \frac{1}{2}\, m \, v^{2}$ .

When the spring is at the maximum compression:

The clay sphere would be right on top of the spring.
The top of the spring would be below the original position (when the spring was uncompressed) by $x$ .
The initial position of the clay sphere, however, is above the original position of the top of the spring by $h = 3.21\; {\rm m}$ .

Thus, the initial position of the clay sphere ( $h = 3.21\; {\rm m}$ above the top of the uncompressed spring) would be above the max-compression position of the clay sphere by $(h + x)$ .

The gravitational potential energy involved would be:

$\text{GPE} = m\, g\, (h + x)$ .

No mechanical energy would be lost under the assumptions listed above. Thus:

$\text{PE}_\text{spring} = \text{KE} + \text{GPE}$ .

$\displaystyle \frac{1}{2}\, k\, x^{2} = \frac{1}{2}\, m\, v^{2} + m\, g\, (h + x)$ .

Rearrange this equation to obtain a quadratic equation about the only unknown, $x$ :

$\displaystyle \frac{1}{2}\, k\, x^{2} - m\, g\, x - \left[\left(\frac{1}{2}\, m\, v^{2}\right)+ (m\, g\, h)\right] = 0$ .

Substitute in $k = 1570\; {\rm N \cdot m^{-1}}$ , $m = 4.87\; {\rm kg}$ , $v = 5.21\; {\rm m\cdot s^{-1}}$ , $g = 9.81\; {\rm m \cdot s^{-2}}$ , and $h = 3.21\; {\rm m}$ . Let the unit of $x$ be meters.

$785\, x^{2} - 47.775\, x - 219.453 \approx 0$ (Rounded. The unit of both sides of this equation is joules.)

Solve using the quadratic formula given that $x \ge 0$ :

$\begin{aligned}x &\approx \frac{-(-47.775) + \sqrt{(-47.775)^{2} - 4 \times 785 \times (-219.453)}}{2 \times 785} \\ &\approx 0.560\; {\rm m}\end{aligned}$ .

(The other root is negative and is thus invalid.)

Hence, the maximum compression of this spring would be approximately $0.560\; {\rm m}$ .

5 0

2 years ago

Help, please!!!!!<br> I just need help putting the vocab word in the box where it belongs.

Ber [7]

here's the first part but for the 2nd one all I know is that the word "compression" goes on the spirals that are closer together.

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8 0

2 years ago

What is a suspension bridge- in your own words please and ty

PIT_PIT [208]

A bridge supported by vertical cables which then leads to more support from larger cables.

4 0

3 years ago

If the elephant were then allowed to fall straight down, how fast would it be moving when it landed back on the ground?

Oxana [17]

Answer:

0 mph because it is probably dead

6 0

3 years ago