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3 years ago

A winding mountain road is an example of a(n) ____.

Physics

1 answer:

Paul [167]3 years ago

6 0

<span>❅☃ Lever
A winding mountain road is a example of a lever.</span>

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A student initially 10.0 m East of his school walks 17.5 m West. The magnitude of the student's displacement, relative to the sc

Fantom [35]

Answer:

1. 7.5 m

2. towards west side

explanation:

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2 years ago

The hottest climates on Earth are located near the Equator because this region

Ber [7]

You should select Choice-4 .

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3 years ago

Formula for deriving initial temperature in linear expansivity

jekas [21]

Answer:

∆L=aL∆T

Explanation:

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2 years ago

Three identical resistors are connected in parallel. The equivalent resistance increases by 630 when one resistor is removed and

strojnjashka [21]

Answer:

each resistor is 540 Ω

Explanation:

Let's assign the letter R to the resistance of the three resistors involved in this problem. So, to start with, the three resistors are placed in parallel, which results in an equivalent resistance $R_e$ defined by the formula:

$\frac{1}{R_e}=\frac{1}{R} } +\frac{1}{R} } +\frac{1}{R} \\\frac{1}{R_e}=\frac{3}{R} \\R_e=\frac{R}{3}$

Therefore, R/3 is the equivalent resistance of the initial circuit.

In the second circuit, two of the resistors are in parallel, so they are equivalent to:

$\frac{1}{R'_e}=\frac{1}{R} +\frac{1}{R}\\\frac{1}{R'_e}=\frac{2}{R} \\R'_e=\frac{R}{2} \\$

and when this is combined with the third resistor in series, the equivalent resistance ( $R''_e$ ) of this new circuit becomes the addition of the above calculated resistance plus the resistor R (because these are connected in series):

$R''_e=R'_e+R\\R''_e=\frac{R}{2} +R\\R''_e=\frac{3R}{2}$

The problem states that the difference between the equivalent resistances in both circuits is given by:

$R''_e=R_e+630 \,\Omega$

so, we can replace our found values for the equivalent resistors (which are both in terms of R) and solve for R in this last equation:

$\frac{3R}{2} =\frac{R}{3} +630\,\Omega\\\frac{3R}{2} -\frac{R}{3} = 630\,\Omega\\\frac{7R}{6} = 630\,\Omega\\\\R=\frac{6}{7} *630\,\Omega\\R=540\,\Omega$

8 0

3 years ago

A girl weighing 50 kgf wears sandals of pencil heel of area of cross section 1 cm^2, stands on the floor.An elephant weighing 20

Klio2033 [76]

Answer:

$\boxed{{\boxed{\blue{ 12.5}}}}$

Explanation:

Given, for girl : Weight or force;

$\rm \: F_1 = 50 \: kgf$

Area of both heels;

$\rm \: A_1 = \; 2 ×1 \; cm^2 = 2 \: cm^2$

$\rm \: Pressure \: P_1 = \cfrac{F_1}{ A_1 } = \dfrac{50 \: kgf}{2 \: cm {}^{2} } = 25 \: kgf \: cm {}^{ - 1}$

For elephant, Weight = Force $\rm F_2$ = 2000 kg•f

Area of 4 feet;

$\rm \: A_2 = \; 4 \times 250 \; cm^2 = 1000 \: cm^2$

$\rm \: Pressure \: P_2 = {F_2}/{A_2} \; = \cfrac{2 \cancel{0 00 }\: kgf}{1 \cancel{000} \: cm^2} = 2 \: kgf \: { \:cm}^{- 1}$

Now;

$\rm = \dfrac{Pressure \: Exerted \: by \: the \: Girl}{Pressure \: exerted \: by \: the \: elephant}$

$= \rm \: P_1/P_2$

$\implies \rm\cfrac{25 \: kgf \: \: cm {}^{ - 2} }{2 \: kgf \: cm {}^{ - 2} } = \rm\cfrac{25 \: \cancel{kgf \: \: cm {}^{ - 2}} }{2 \: \cancel{ kgf \: cm {}^{ - 2}} } = \boxed{12.5}$

Thus, the girl's pointed heel sandals exert 12.5 times more pressure P than the pressure P exerted by the elephant.

I aspire this helps!

3 0

2 years ago