State the relation between 1kgf and si unit of force

I have a doubt on heat and temperature. I need help with this science question and here the question: The water in a lake just r

Volgvan

Answer:

you can't go ice skating on it because if it just reached the temp then you need to wait for about 2 hours

Explanation:

5 0

2 years ago

What type of motion is shown with this graph? (5 points)

Vladimir [108]

It should be Constant speed. The line goes straight & doesn’t change within the graph.

7 0

2 years ago

What is a fact about incline planes?

SVETLANKA909090 [29]

They are incline hope this helps!

8 0

2 years ago

Galactic Alliance Junior Mission Officer (GAJMO) Bundit Nermalloy is predicting the kinetic energy of a supply spacecraft, which

antiseptic1488 [7]

Answer:

the ship's energy is greater than this and the crew member does not meet the requirement

Explanation:

In this exercise to calculate kinetic energy or final ship speed in the supply hangar let's use the relationship

W =∫ F dx = ΔK

Let's replace

∫ (α x³ + β) dx = ΔK

α x⁴ / 4 + β x = ΔK

Let's look for the maximum distance for which the variation of the energy percent is 10¹⁰ J

x (α x³ + β) = $K_{f}$ - K₀

$K_{f}$ = K₀ + x (α x³ + β)

Assuming that the low limit is x = 0, measured from the cargo hangar

Let's calculate

$K_{f}$ = 2.7 10¹¹ + 7.5 10⁴ (6.1 10⁻⁹ (7.5 10⁴) 3 -4.1 10⁶)

Kf = 2.7 10¹¹ + 7.5 10⁴ (2.57 10⁶ - 4.1 10⁶)

Kf = 2.7 10¹¹ - 1.1475 10¹¹

Kf = 1.55 10¹¹ J

In the problem it indicates that the maximum energy must be 10¹⁰ J, so the ship's energy is greater than this and the crew member does not meet the requirement

We evaluate the kinetic energy if the System is well calibrated

W = x F₀ = $K_{f}$ –K₀

$K_{f}$ = K₀ + x F₀

We calculate

$K_{f}$ = 2.7 10¹¹ -7.5 10⁴ 3.5 10⁶

$K_{f}$ = (2.7 -2.625) 10¹¹

$K_{f}$ = 7.5 10⁹ J

5 0

2 years ago

A spherical capacitor contains a charge of 3.00 nC when connected to a potential difference of 230 V. If its plates are separate

Assoli18 [71]

Answer:

Part(a): the capacitance is 0.013 nF.

Part(b): the radius of the inner sphere is 3.1 cm.

Part(c): the electric field just outside the surface of inner sphere is $\bf{2.81 \times 10^{4}~n~C^{-1}}$ .

Explanation:

We know that if 'a' and 'b' are the inner and outer radii of the shell respectively, 'Q' is the total charge contains by the capacitor subjected to a potential difference of 'V' and ' $\epsilon_{0}$ ' be the permittivity of free space, then the capacitance (C) of the spherical shell can be written as

$C = \dfrac{4 \pi \epsilon_{0}}{(\dfrac{1}{a} - \dfrac{1}{b})}~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)$

Part(a):

Given, charge contained by the capacitor Q = 3.00 nC and potential to which it is subjected to is V = 230V.

So the capacitance (C) of the shell is

$C &=& \dfrac{Q}{V} = \dfrac{3 \times 10^{-90}~C}{230~V} = 1.3 \times 10^{-11}~F = 0.013~nF$

Part(b):

Given the inner radius of the outer shell b = 4.3 cm = 0.043 m. Therefore, from equation (1), rearranging the terms,

$&& \dfrac{1}{a} = \dfrac{1}{b} + \dfrac{1}{C/4 \pi \epsilon_{0}} = \dfrac{1}{0.043} + \dfrac{1}{1.3 \times 10^{-11} \times 9 \times 10^{9}} = 31.79\\&or,& a = \dfrac{1}{31.79}~m = 0.031~m = 3.1~cm$

Part(c):

If we apply Gauss' law of electrostatics, then

$&& E~4 \pi a^{2} = \dfrac{Q}{\epsilon_{0}}\\&or,& E = \dfrac{Q}{4 \pi \epsilon_{0}a^{2}}\\&or,& E = \dfrac{3 \times 10^{-9} \times 9 \times 10^{9}}{0.031^{2}}~N~C^{-1}\\&or,& E = 2.81 \times 10^{4}~N~C^{-1}$

3 0

3 years ago

State the relation between 1kgf and si unit of force​

State the relation between 1kgf and si unit of force