Newton’s ______ law states that an object in motion stays in motion unless an unbalanced force acts on that object.

Does the rankine degree represent a larger or smaller temperature unit than the kelvin degree

zvonat [6]

The best and most correct answer among the choices provided by the question is the first choice, larger.

Rankine is Fahrenheit + 460 , while Kelvin is Celsius + 273. We all know that Fahrenheit has larger number compared to kelvin , thus rankine is much larger.
Hope my answer would be a great help for you. If you have more questions feel free to ask here at Brainly.

8 0

3 years ago

An object with a mass of 78kg is lifted through a height of 6 meters. How much work is done?

g100num [7]

Work = force x distance.

force = mass x acceleration

work = mass x acceleration x diastance

use acceleration of gravity in this problem

W (J) = m (kg) x a (m/s/s) x d (m)
W = 78 x 9.8 x 6
W = 4586.4

3 0

3 years ago

The electric field on the surface of an irregularly shaped conductor varies from 74.0 kN/C to 14.0 kN/C.

mrs_skeptik [129]

Answer:

(a). The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 123.9 nC/m².

(b). The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 654.9 nC/m².

Explanation:

Given that,

Electric field $E_{1}=74.0\ kN/C$

Electric field $E_{2}=14.0\ kN/C$

When the radius of curvature is greatest, the electric field at the surface will be smaller.

Where the radius of curvature is greatest

(a). We need to calculate the local surface charge density at the point on the surface

Using formula of charge density

$\sigma=\epsilon_{0}E_{2}$

Put the value into the formula

$\sigma=8.85\times10^{-12}\times14\times10^{3}$

$\sigma=1.239\times10^{-7}\ C/m^2$

$\sigma=123.9\times10^{-9}\ C/m^2$

$\sigma=123.9\ nC/m^2$

The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 123.9 nC/m².

(b). We need to calculate the local surface charge density at the point on the surface where the radius of curvature of the surface is smallest

Using formula of charge density

$\sigma=\epsilon_{0}E_{1}$

Put the value into the formula

$\sigma=8.85\times10^{-12}\times74\times10^{3}$

$\sigma=6.549\times10^{-7}\ C/m^2$

$\sigma=654.9\times10^{-9}\ C/m^2$

$\sigma=654.9\ nC/m^2$

The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 654.9 nC/m².

Hence, (a). The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 123.9 nC/m².

(b). The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 654.9 nC/m².

7 0

3 years ago

Katy works at a pet store, and is in charge of tracking the cat food supply for the morning, afternoon, and evening shifts. Ther

Basile [38]

The cats have full bowls in the morning and afternoon, Katy can assume that the cats do not eat in the morning or afternoon. The bowls are replenished in the evening, which suggests that they become empty in the evening, which suggest that the pattern is that the cats eat in the evening so your answer would be C. Hope this helps. ;)

7 0

4 years ago

Read 2 more answers

Solve for (b) how many revolutions it takes for the cd to reach its maximum angular velocity in 1.36s?

Digiron [165]

Angular displacement of the cd is 3.25 rev

Explanation:

The question is incomplete. It is not given what is the maximum angular velocity of the cd.

Here we are going to assume that the maximum angular velocity is:

$\omega = 30 rad/s$