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Mashcka [7]
3 years ago
6

A.it is moving to the right

Physics
1 answer:
Bas_tet [7]3 years ago
7 0

Answer:

I think c buti don't think I'm correct

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Which of the following is a process that occurs at divergent boundaries?
PIT_PIT [208]
The appropriate answer is 4. The formation of Mid-Ocean ridges. There are two types of divergent margins: continental separation and oceanic floor spreading. Where the sea floor separates magma upwells and new oceanic crust is formed. Mid ocean ridges are home to hydrothermal vents. These support a unique ecosystem of marine organisms. The Mid- Atlantic Ridge where the African and Eurasian plates separate from the North and South American plates.
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A cart was pulled for a distance of 1 kilometer and the amount of work accomplished equaled 40,000 joules. With what force was t
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Answer:

40 N

Explanation:

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3 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
Sugar is made of carbon, hydrogen, and oxygen atoms. Sugar is
Aneli [31]
D. A solution because it dissolves when mixed with water
7 0
3 years ago
Read 2 more answers
A horse of mass 242 kg pulls a cart of mass 224 kg. The acceleration of gravity is 9.8 m/s 2 . What is the largest acceleration
Rufina [12.5K]

To solve this problem it is necessary to apply the concepts related to Newton's second Law and the force of friction. According to Newton, the Force is defined as

F = ma

Where,

m= Mass

a = Acceleration

At the same time the frictional force can be defined as,

F_f = \mu N

Where,

\mu = Frictional coefficient

N = Normal force (mass*gravity)

Our values are given as,

m_h = 242 kg\\m_c = 224 kg\\\mu = 0.894\\

By condition of Balance the friction force must be equal to the total net force, that is to say

F_{net} = F_f

m_{total}a = \mu m_hg

(m_h+m_c)a = \mu*m_h*g

Re-arrange to find acceleration,

a= \frac{\mu*m_h*g}{(m_h+m_c)}

a = \frac{0.894*242*9.8}{(242+224)}

a = 4.54 m/s^2

Therefore the acceleration the horse can give is 4.54m/s^2

3 0
3 years ago
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