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3 years ago

What increases your ability to see at night

Physics

1 answer:

PtichkaEL [24]3 years ago

8 0

The answer would be increased number <span />of rod cells.Hope this helps :)

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Manufactured bolts can’t be too long or too short or they might not fit where they need to. Help determine which bolts will work

Sever21 [200]

What is the longest the bolt can be and still be acceptable

5 0

3 years ago

A 645 g block is released from rest at height h0 above a vertical spring with spring constant k = 530 N/m and negligible mass. T

Nina [5.8K]

Answer:

a)5.88J

b)-5.88J

c)0.78m

d)0.24m

Explanation:

a) W by the block on spring is given by

W= $\frac{1}{2}$ kx² = $\frac{1}{2}$ (530)(0.149)² = 5.88 J

b) Workdone by the spring = - Workdone by the block = -5.88J

c) Taking x = 0 at the contact point we have U top = U bottom

So, mg $h_o$ = $\frac{1}{2}$ kx² - mgx

And, $h_o$ = ( $\frac{1}{2}$ kx² - mgx )/(mg) = $[\frac{1}{2} (530)(0.149^2)-(0.645)(9.8)(0.149)$ ]/(0.645x9.8)

$h_o$ = 0.78m

d) Now, if the initial initial height of block is 3 $h_o$

$h_o$ = 3 x 0.78 = 2.34m

then, $\frac{1}{2}$ kx² - mgx - mg $h_o$ =0

$\frac{1}{2}$ (530)x² - [(0.645)(9.8)x] - [(0.645)(9.8)(2.34) = 0

265x² - 6.321x - 14.8 = 0

a=265

b=-6.321

c=-14.8

By using quadratic eq. formula, we'll have the roots

x= 0.24 or x=-0.225

Considering only positive root:

x= 0.24m (maximum compression of the spring)

4 0

3 years ago

Learning Check: Ranking Elements

loris [4]

Yes
62748476182973838292

8 0

3 years ago

Please show steps as to how to solve this problem <br> Thank you!

bezimeni [28]

Explanation:

Let x = distance of $F_1$ from the fulcrum and let's assume that the counterclockwise direction is positive. In order to attain equilibrium, the net torque $\tau_{net}$ about the fulcrum is zero:

$\tau_{net} = -F_1x + F_2d_2 = 0$

$-m_1gx + m_2gd_2 = 0$

$m_1x = m_2d_2$

Solving for <em>x</em>,

$x = \dfrac{m_2}{m_1}d_2$

$\:\:\:\:=\left(\dfrac{105.7\:\text{g}}{65.7\:\text{g}} \right)(13.8\:\text{cm}) = 22.2\:\text{cm}$

4 0

3 years ago

A suspension bridge with weight uniformly distributed along its length has twin towers that extend 65 meters above the road surf

Fynjy0 [20]

Answer:

16.25 m

Explanation:

we know that the equation pf parabola

$y=kx^2$

from bellow figure the coordinate of parabola is (600,65) that is y=600 and x=65

putting the the value of y and x in the equation of parabola

$65=k600^2$

k=0.0001805

now the equation is

$y=0.0001805x^2$

we have to find the value of y at x=300m

so $y=0.0001805\times 300^2$

y=16.25 m

4 0

3 years ago