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3 years ago

In which situation is maximum work considered to be done by a force?

Physics

2 answers:

AfilCa [17]3 years ago

8 0

Answer:

When the direction of the force is parallel to the displacement of the object

Explanation:

The work done by a force when moving an object is given by:

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement of the object

$\theta$ is the angle between the force's direction and the displacement

From the formula, we see that the work done is maximum when

$cos \theta=1$

which occurs when

$\theta=0^{\circ}$

so, when the force is parallel to the displacement.

ch4aika [34]3 years ago

7 0

The force and the motion at witch the force is driving it have a directly proportional relationship, meaning the have to both go the same way

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If the amount of work done on a book was 10 J and the force required to move the book was 2.5 N, what was the distance the book

olga55 [171]

Answer:

Explanation:4meters

Work=10J

Force=2.5N

Distance=work ➗ force

Distance=10 ➗ 2.5

Distance=4meter

5 0

3 years ago

Compare the gravitational acceleration on the following objects compared to the Sun using:

arsen [322]

The gravitational acceleration of White dwarf compared to Sun is 13,675.86.

The gravitational acceleration of Neutron star compared to Sun is 6.79 x 10⁻²⁴.

The gravitational acceleration of Star Betelgeuse compared to Sun is 8.5 x 10¹⁰.

<h3>Mass of the planets</h3>

Mass of sun = 2 x 10³⁰ kg

Mass of white dwarf = 2.765 x 10³⁰ kg

Mass of Neutron star = 5.5 x 10¹² kg

Mass of star Betelgeuse = 2.188 x 10³¹ kg

<h3>Radius of the planets</h3>

Radius of sun = 696,340 km

Radius of white dwarf = 7000 km

Radius of Neutron star = 11 km

Radius of star Betelgeuse = 617.1 x 10⁶ km

<h3>Gravitational acceleration of White dwarf compared to Sun</h3>

$\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{2.765 \times 10^{30}}{2\times 10^{30}} \times [\frac{696,340,000}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 13,675.86$

<h3>Gravitational acceleration of Neutron star compared to Sun</h3>

$\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{5.5 \times 10^{12}}{2\times 10^{30}} \times [\frac{11,000}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 6.79\times 10^{-24}$

<h3>Gravitational acceleration of Star Betelgeuse compared to Sun</h3>

$\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{2.188 \times 10^{31}}{2\times 10^{30}} \times [\frac{617.1 \times 10^9}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 8.5\times 10 ^{10}$

Learn more about acceleration due to gravity here: brainly.com/question/88039

3 0

3 years ago

The dielectric strength of rutile is 6.0 × 106 V/m, which corresponds to the maximum electric field that the dielectric can sust

VLD [36.1K]

Answer:

6.48*10⁻⁷ C

Explanation:

By definition, the capacitance of a capacitor is expressed as follows:

$C =\frac{Q}{V}$

where Q is the charge on one of the plates of the capacitor, and V the potential difference between the plates.
The maximum electric field, the potential difference, and the distance between plates are related by the following expression:

$V = E_{max} * d$

Replacing by the givens, we can find V as follows:

$V = 6.0e6 V/m * 0.00108 m= 6480 V$

Now, we can find the maximum charge Qmax, as follows:

$Q_{max} = C*V = 1e-10F* 6480 V = 6.48e-7 C$

The maximum charge that the capacitor can hold is 6.48*10⁻⁷ C.

5 0

3 years ago

Read 2 more answers

Identical isolated conducting spheres 1 and 2 have equal charges and are separated by a distance that is large compared with the

scoundrel [369]

Answer:

F = 0.1575 N

Explanation:

When the third sphere touches the first sphere, the charge is distributed between both spheres, then now the first sphere has only half of his original charge.

In this moment then

Sphere one has a charge = Q/2

Sphere three has a charge = Q/2

Now when the third sphere touches the second sphere again the charge is distributed in a manner that both sphere has the same charge.

How the total charge is Q = Q/2 + Q = 3/2Q, when the spheres are separated each one has 3/4Q

Sphere two has a charge = 3/4Q

Sphere three has a charge = 3/4Q

The electrostatic force that acts on sphere 2 due to sphere 1 is:

F = $\frac{kQ_{1}Q_{2} }{r^{2} }$