10 PTS!

2 answers:

5 0

" This manual applies to Compact Liquid Fuel Pumps & Dispensers The liquid pressure range is from 0.5 - 20m These totals can be displayed by pressing the CLEAR button on the preset keypad five times in When connecting to sites powered by. "

konstantin123 [22]3 years ago

5 0

(Remeber as force increases, pressure increases.)
When you push the button, the pressure of the liquid inside the dispenser increases. Hence, the more force you apply to the button, the more quickly the liquid will eject.

~Hope this helped! Gl~

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A comet is in an elliptical orbit around the Sun. Its closest approach to the Sun is a distance of 4.8 1010 m (inside the orbit

creativ13 [48]

Answer:

Explanation:

From the given information:

Distance $d_i = 4.8 \times 10^{10} \ m$

Speed of the comet $V_i = 9.1 \times 10^{4} \ m/s$

At distance $d_2 = 6 \times 10^{12} \ m$

where;

mass of the sun = $1.98 \times 10^{30}$

$G = 6.67 \times 10^{-11}$

To find the speed $V_f$ :

Using the formula:

$E_f = E_i + W \\ \\ where; \ \ W = 0 \ \ \text{since work done by surrounding is zero (0)}$

$E_f = E_i + 0 \\ \\ K_f + U_f = K_i + U_i \\ \\ = \dfrac{1}{2}mV_f^2 + \dfrac{-GMm}{d^2} = \dfrac{1}{2}mV_i^2+ \dfrac{-GMm}{d_i} \\ \\ V_f = \sqrt{V_i^2 + 2 GM \Big [ \dfrac{1}{d_2}- \dfrac{1}{d_i}\Big ]}$

$V_f = \sqrt{(9.1 \times 10^{4})^2 + 2 (6.67\times 10^{-11}) *(1.98 * 10^{30} ) \Big [ \dfrac{1}{6*10^{12}}- \dfrac{1}{4.8*10^{10}}\Big ]}$

$\mathbf{V_f =53.125 \times 10^4 \ m/s}$

3 0

2 years ago

A spring whose spring constant is 270 lbf/in has an initial force of 100 lbf acting on it. Determine the work, in Btu, required

cricket20 [7]

Answer:

0.02585 BTU

Explanation:

Given: Spring constant, k = 270 lbf/in

Initial force, f =100 lbf

Compression, x = 1 in

Work done can be calculated as follows:

$W = \int {(f+kx)} \, dx \\W = fx + \frac{1}{2}kx^2\\W= (100 lbf)(1 in)+ \frac{1}{2}(270 lbf/in)(1 in)^2\\W= 100+135 = 235 lbf in\\W=235 \times 0.00011 BTU = 0.02585 BTU$