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2 years ago

The mechanical advantage of a wheel and axle is the radius of the wheel divided by the radius of the axle. TRUE or FALSE.

Physics

1 answer:

viva [34]2 years ago

4 0

True! The mechanical advantage of the wheel and axle is equal to the ratio of the radius of the wheel over the radius of the axle.

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How much work is done by the force lifting a

Ann [662]

The work done in lifting the hamburger is equal to the increase in gravitational potential energy of the hamburger, given by

$W=\Delta U=mg \Delta h$

where

m=0.1 kg is the mass of the hamburger

$g=9.81 m/s^2$ is the gravitational acceleration

$\Delta h=0.3 m$ is the increase in height of the hamburger

Substituting numbers into the equation, we find

$W=(0.1 kg)(9.81 m/s^2)(0.3 m)=0.3 J$

So, the correct answer is

(3) 0.3 J

3 0

3 years ago

A car drives 20 miles north and then 3 miles south. what is the displacement or the car

kobusy [5.1K]

An:
Displacement is 16.

5 0

2 years ago

Read 2 more answers

Determine the focal length of a plano-concave lens (refractive index n =1.5) with 24 cm radius of curvature on its curve surface

tatyana61 [14]

Answer:

Option 3: -48 cm

Explanation:

We are given:

refractive index; n = 1.5

radius of curvature; r2 = 24 cm

Formula for the focal length is given as;

1/f = (n - 1) × [(1/r1) - (1/r2)]

As r1 tends to infinity, 1/r1 = 0

Thus,we now have;

1/f = (n - 1) × (-1/r2)

Plugging in the relevant values;

1/f = (1.5 - 1) × (-1/24)

1/f = -0.02083333333

f = -1/0.02083333333

f = -48 cm

3 0

2 years ago

An undiscovered planet, many light-years from Earth, has one moon, which has a nearly circular periodic orbit. If the distance f

EleoNora [17]

Answer:

364days

Explanation:

Pls see attached file

Explanation:

8 0

3 years ago

What is the strength of the electric field 0.1 mmmm below the center of the bottom surface of the plate

Aleksandr [31]

Complete Question

A thin, horizontal, 12-cm-diameter copper plate is charged to 4.4 nC . Assume that the electrons are uniformly distributed on the surface. What is the strength of the electric field 0.1 mm above the center of the top surface of the plate?

Answer:

The values is $E =248.2 \ N/C$

Explanation:

From the question we are told that

The diameter is $d = 12 \ cm = 0.12 \ m$

The charge is $Q = 4.4 nC = 4.4 *10^{-9} \ C$

The distance from the center is $k = 0.1 \ mm = 1*10^{-4} \ m$

Generally the radius is mathematically represented as

$r = \frac{d}{2}$

=> $r = \frac{0.12}{2}$

=> $r = 0.06 \ m$

Generally electric field is mathematically represented as

$E = \frac{Q}{ 2\epsilon_o } [1 - \frac{k}{\sqrt{r^2 + k^2 } } ]$

substituting values

$E = \frac{4.4 *10^{-9}}{ 2* (8.85*10^{-12}) } [1 - \frac{(1.00 *10^{-4})}{\sqrt{(0.06)^2 + (1.0*10^{-4})^2 } } ]$

$E =248.2 \ N/C$

4 0

3 years ago