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3 years ago

Which figure represents the electric field lines of two negative point charges? Help

Physics

1 answer:

Yuliya22 [10]3 years ago

3 0

Answer:

The question is incomplete because the options are not given. We'll solve this question with the options.

There are two properties of negative charges which we have to consider to find the figure of its electrical field.

Firstly, for a negative charge, the electrical field lines are always directed radially outwards and they don't intersect.

Secondly, we know that similar charges repel each other, so there will be no electrical field present directly between these two negative charges. Electrical field line will be present between two charges only when there is a force of attraction.

Taking both of these facts into consideration, the electrical field line between two negatives is represented by the figure below.

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Assume that a satellite orbits mars 150km above its surface. Given that the mass of mars is 6.485 X 10^23kg, and the radius of m

Kisachek [45]

<span>3598 seconds The orbital period of a satellite is u=GM p = sqrt((4*pi/u)*a^3) Where p = period u = standard gravitational parameter which is GM (gravitational constant multiplied by planet mass). This is a much better figure to use than GM because we know u to a higher level of precision than we know either G or M. After all, we can calculate it from observations of satellites. To illustrate the difference, we know GM for Mars to within 7 significant figures. However, we only know G to within 4 digits. a = semi-major axis of orbit. Since we haven't been given u, but instead have been given the much more inferior value of M, let's calculate u from the gravitational constant and M. So u = 6.674x10^-11 m^3/(kg s^2) * 6.485x10^23 kg = 4.3281x10^13 m^3/s^2 The semi-major axis of the orbit is the altitude of the satellite plus the radius of the planet. So 150000 m + 3.396x10^6 m = 3.546x10^6 m Substitute the known values into the equation for the period. So p = sqrt((4 * pi / u) * a^3) p = sqrt((4 * 3.14159 / 4.3281x10^13 m^3/s^2) * (3.546x10^6 m)^3) p = sqrt((12.56636 / 4.3281x10^13 m^3/s^2) * 4.458782x10^19 m^3) p = sqrt(2.9034357x10^-13 s^2/m^3 * 4.458782x10^19 m^3) p = sqrt(1.2945785x10^7 s^2) p = 3598.025212 s Rounding to 4 significant figures, gives us 3598 seconds.</span>

8 0

3 years ago

7. A block of copper of unknown mass has an initial temperature of 65.4oC. The copper is immersed in a beaker containing 95.7g o

dolphi86 [110]

Answer:

37.34372 kg

Explanation:

m = Mass

$\Delta T$ = Change in temperature

1 denotes water

2 denotes copper

c = Heat capacity

Heat is given by

$Q=mc\Delta T$

In this case the heat transfer will be equal

$m_1c_1\Delta T_1=m_2c_2\Delta T_2\\\Rightarrow m_2=\frac{m_1c_1\Delta T_1}{c_2\Delta T_2}\\\Rightarrow m_2=\frac{95.7\times 4.18(24.2-22.7)}{0.39(65.4-24.2)}\\\Rightarrow m_2=37.34372\ kg$

Mass of copper block is 37.34372 kg

5 0

2 years ago

Calculate delta g for the reaction if the partial pressures of the initial mixture are pcl5 = .0029 atm, pcl3 = .27 atm, and pcl

AnnyKZ [126]

Answer: equation for the reaction is given below

PCL2+PCL3=PCL5

Where pcl2=0.40atm,pcl3=0.27atm

Pcl5=0.0029atm

Using ∆G=-RTin(PCL5/PCl2*PCL3)

Where R=8.314J/K/mol and T=298K

∆G=-8.314*298in(0.0029/0.40*.27)

∆G=8962.6J/mol

Explanation:

7 0

3 years ago

You are asked to design a cylindrical steel rod 50.0 cm long, with a circular cross section, that will conduct 170.0 J/s from a

zepelin [54]

Answer:

You are asked to design a cylindrical steel rod 50.0 cm long, with a circular cross section, that will conduct 170.0 J/s from a furnace at 350.0 ∘C to a container of boiling water under 1 atmosphere.

Explanation:

Given Values:

L = 50 cm = 0.5 m

H = 170 j/s

To find the diameter of the rod, we have to find the area of the rod using the following formula.

Here Tc = 100.0° C

k = 50.2

H = k × A × $\frac{[T_{H -}T_{C} ] }{L}$