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Arte-miy333 [17]
3 years ago
7

The small piston of a hydraulic lift has an area of 0.20 m2. A car weighing 1200 N sits on a rack mounted on the large piston. T

he large piston has an area of 0.90 m2. What is the mechanical advantage of the hydraulic lift?
Physics
1 answer:
lora16 [44]3 years ago
7 0

Answer:

4.5

Explanation:

Given the following

Weight of the car = 1200N (Load)

Let us get the effort by using the formula;

F1/A1 = F2/A2

Substitute the given values

1200/0.9 = F2/0.2

Cross multiply

0.9F2 = 1200 * 0.2

0.9F2 = 240

F2 = 240/0.9

F2 = 266.67N

Hence the effort is 266.67N

Mechanical Advntage = Load/Effort

MA = 1200/266.67

MA = 4.5

Hence he mechanical advantage of the hydraulic lift is 4.5

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Use Coulomb law: F = k * q1*q2 / (r^2), where k = 9.00 * 10^9 N.m^2/C^2

F = 9.00 * 10^9 N.m^2/C^2 * 2.4*10^-8 C * 1.8*10^-6 C / [0.008m]^2 = 38.88 * 10^ -5 N

F = 39 * 10 -5N


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3 years ago
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Which of the following would most accurately describe lightning?
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4 years ago
. If an 8 N force pulled a 3.0 kg object to the left and a 5 N force pushed it to the right,
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Answer:

8-5=3

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3N to the left

Explanation:

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1 year ago
1.
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Answer:

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7 0
3 years ago
If the frequency of the 13C signal of TMS is 201.16 MHz, the two 13C signals of acetic acid at 179.0 and 20.0 ppm are separated
Lelu [443]

The difference in frequency of the two signals is 1.33 \times 10^{10} \ kHz.

The given parameters;

  • <em>frequency of the 13 C signal = 201.16 MHz</em>

The energy of the 13 C signal located at 20 ppm is calculated as follows;

E = hf\\\\E_1 = h \frac{c}{\lambda} \\\\E_1 =  \frac{(6.626 \times 10^{-34})\times 3\times 10^8}{20 \times 10^{-6}} \\\\E_1 = 9.94 \times 10^{-21} \ J

The energy of the 13 C signal located at 179 ppm is calculated as follows;

E_2 = \frac{hc}{\lambda} \\\\E_2 = \frac{(6.626\times 10^{-34})\times (3\times 10^{8})}{179 \times 10^{-6} } \\\\E_2 = 1.11 \times 10^{-21} \ J

The difference in frequency of the two signals is calculated as follows;

E_1- E_2 = hf_1 - hf_2\\\\E_1 - E_2 = h(f_1 - f_2)\\\\f_1 - f_2 = \frac{E_1 - E_2 }{h} \\\\f_1 - f_2 = \frac{(9.94\times 10^{-21}) - (1.11 \times 10^{-21})}{6.626\times 10^{-34}} \\\\f_1 - f_2 = 1.33 \times 10^{13} \ Hz\\\\f_1 - f_2 = 1.33\times 10^{10} \ kHz

Thus, the difference in frequency of the two signals is 1.33 \times 10^{10} \ kHz.

Learn more here:brainly.com/question/14016376

4 0
3 years ago
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