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3 years ago

In a wheel and axle, the radius of the wheel is 20 cm and the radius of the axle is 4 cm. What is the mechanical advantage?

Physics

2 answers:

aliya0001 [1]3 years ago

4 0

The answer is 5. To find the advantage you just divide 20 by 4.

d1i1m1o1n [39]3 years ago

3 0

Answer: the answer is 5 on e2020 just the test

Explanation:

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One recently discovered extrasolar planet, or exoplanet, orbits a star whose mass is 0.70 times the mass of our sun. This planet

Stels [109]

0.078 times the orbital radius r of the earth around our sun is the exoplanet's orbital radius around its sun.

Answer: Option B

<u>Explanation:</u>

Given that planet is revolving around the earth so from the statement of centrifugal force, we know that any

$\frac{G M m}{r^{2}}=m \omega^{2} r$

The orbit’s period is given by,

$T=\sqrt{\frac{2 \pi}{\omega r^{2}}}=\sqrt{\frac{r^{3}}{G M}}$

Where,

$T_{e}$ = Earth’s period

$T_{p}$ = planet’s period

$M_{s}$ = sun’s mass

$r_{e}$ = earth’s radius

Now,

$T_{e}=\sqrt{\frac{r_{e}^{3}}{G M_{s}}}$

As, planet mass is equal to 0.7 times the sun mass, so

$T_{p}=\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}$

Taking the ratios of both equation, we get,

$\frac{T_{e}}{T_{p}}=\frac{\sqrt{\frac{r_{e}^{3}}{G M_{s}}}}{\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}}$

$\frac{T_{e}}{T_{p}}=\sqrt{\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}}$

$\left(\frac{T_{e}}{T_{p}}\right)^{2}=\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}$

$\left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}=\frac{r_{e}^{3}}{r_{p}^{3}}$

$\frac{r_{e}}{r_{p}}=\left(\left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}$

Given $T_{p}=9.5 \text { days }$ and $T_{e}=365 \text { days }$

$\frac{r_{e}}{r_{p}}=\left(\left(\frac{365}{9.5}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=\left(\frac{133225}{90.25} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=(2108.82)^{\frac{1}{3}}$

$r_{p}=\left(\frac{1}{(2108.82)^{\frac{1}{3}}}\right) r_{e}=\left(\frac{1}{12.82}\right) r_{e}=0.078 r_{e}$

7 0

3 years ago

A helix with 17 turns has height H and radius R. Charge is distributed on the helix so that the charge density increases like (i

marishachu [46]

The total charge on the helix is

Q = $\frac{221}{2}\sqrt{(\frac{H}{17})^2+(2\piR)^2$

Height of one turn = $\frac{H}{n}$

Horizontal distance in one turn = $2\pi R$

<h3>length of spring in one turn is</h3>

$l = \sqrt{h^2+(2\piR)^2}\\\\
l = \sqrt{(\frac{H}{n})^2+(2\piR)^2}
$

So, the total length of spring

$L = n*l\\\\
L = n * \sqrt{(\frac{H}{n})^2+(2\piR)^2}
$

Therefore, charge on spring

$Q = \frac{1}{2}*L*(13-0)\\\\
Q = \frac{1}{2}*n\sqrt{(\frac{H}{n})^2+(2\piR)^2}*13\\\\
Q = \frac{1}{2}*17*13\sqrt{(\frac{H}{17})^2+(2\piR)^2}\\\\
Q = \frac{221}{2}\sqrt{(\frac{H}{17})^2+(2\piR)^2\\\\
$

For more information on linear charge density, visit

brainly.com/question/15359786

8 0

2 years ago

A radio transfers 270 j of energy electrically, 94.5 j is transferred by sound. Work out the efficiency of this radio. Give your

MakcuM [25]

Answer:

35%

Explanation:

Given data

Amount of energy transferred (Input) = 270J

Amount of energy converted to sound (Output)= 94.5J

Efficiency = output/input*100

Efficiency= 94.5/270*100

Efficiency=0.35*100

Efficiency=35%

Hence the efficiency is 35%

6 0

3 years ago

According to the law of conservation of mass does the total number of atoms change during aerobic respiration

vichka [17]

NO. although the arrangement of the atoms will change, the total number stays constant.

3 0

3 years ago

I need help/this is a major grade

Debora [2.8K]

Answer:

1e , 2j , 3c , 4d , 5k , 6a , 7i , 8b , 9m , 10h ,11g , 12f , 13l .

8 0

3 years ago