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alexdok [17]
3 years ago
8

Difference between kilogram and kilometre in points​

Physics
1 answer:
skelet666 [1.2K]3 years ago
7 0

Answer:

Kilogram(kg) is the SI unit for mass while kilometre(km) is a unit for length. They are both similar in that they are 10^3 of a unit, thus kilo. As kilogram represents mass, it is a measure of how much matter is present in an object. While kilometre is a measure of distance/how long or short an object is.

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White dwarfs because you look at where the highest temperature is and where is the highest magnitude. Fun fact, the sun will become a white dwarf.

Explanation:

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Why can gases and liquids both transmit heat by convection?
defon

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A. Their particles are free to move away from each other.

Explanation:

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Read 2 more answers
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
Given s=1/2(v+u)t find t ​
Andrej [43]

Explanation:

s=(v+u)t/2

2s/(v+u)=t

.................

6 0
3 years ago
Suppose we take a 1 m long uniform bar and support it at the 21 cm mark. Hanging a 0.40 kg mass on the short end of the beam res
Ipatiy [6.2K]

Answer:

The mass of the beam is = 29 kg.

Explanation:

A beam with mass 40 kg is shown in figure. Point S is the support point. Point B is the middle point on the beam where mass of the beam acts.

Taking moment about Point S

40 × 21 = M_{beam} × 29

M_{beam} = 29 kg

Therefore the mass of the beam is = 29 kg.

4 0
3 years ago
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