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3 years ago

8

during which of the following months is the sun likely to illuminate Earth equally from pole to pole? December January June Sept

ember

2 answers:

Helga [31]3 years ago

8 0

The sun will most likely illuminate Earth equally on both poles during September and March. On these months, Earth will not be tilted toward or away from the sun. This results to both poles receiving nearly the same amount of light, which means that the length of days and nights are equal all over Earth. This event is referred to as Equinoxes. During March, it is called the Vernal Equinox, while during September, it is called the Autumnal Equinox.

Andrews [41]3 years ago

8 0

Answer:

September

Explanation:

For others who visit later this is the correct answer i took the test :)

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A sphere of radius r = 5cm carries a uniform volume charge density rho = 400 nC/m^3. Q. What is the total charge Q of the sphere

Tanzania [10]

Answer:

The total charge Q of the sphere is $2.094\times10^{-10}\ C$ .

Explanation:

Given that,

Radius = 5 cm

Charge density $J= 400\ nC/m^3$

We need to calculate the total charge Q of the sphere

Using formula of charge

$q=\rho V$

Where, $\rho$ = charge density

V = volume

Put the value into the formula

$q=\rho\times(\dfrac{4}{3}\pi r^3)$

Put the value into the formula

$q=\dfrac{4}{3}\times\pi\times400\times10^{-9}\times(5\times10^{-2})^3$

$q=2.094\times10^{-10}\ C$

Hence, The total charge Q of the sphere is $2.094\times10^{-10}\ C$ .

6 0

3 years ago

the specific heat of water is 4.2 j/c. if it takes 31,500 joules to heat to warm 750 g of water, what was the temperature change

defon

The amount of heat needed to increase the temperature of a substance by $\Delta T$ is given by
$Q=mC_s \Delta T$
where
m is the mass of the substance
$C_s$ the specific heat capacity
$\Delta T$ the increase in temperature

In our problem, the mass of the water is m=750 g, the specific heat is $C_s = 4.2 J/g ^{\circ}C$ and the amount of heat supplied is $Q=31500 J$ , so if we re-arrange the previous formula we find the increase in temperature of the water:
$\Delta T= \frac{Q}{m C_s}= \frac{31500 J}{(750 g)(4.20 J/g^{\circ} C)}=10^{\circ}C$

7 0

3 years ago

*please refer to photo* An electric field of magnitude 5.25 ✕ 10^5N/C points due south at a certain location. Find the magnitude

kvv77 [185]

Answer:

Approximately $3.86\; {\rm N}$ (given that the magnitude of this charge is $-7.35\; {\rm \mu C}$ .)

Explanation:

If a charge of magnitude $q$ is placed in an electric field of magnitude $E$ , the magnitude of the electrostatic force on that charge would be $F = E\, q$ .

The magnitude of this charge is $q = 7.35\; {\rm \mu C}$ . Apply the unit conversion $1\; {\rm \mu C} = 10^{-6}\; {\rm C}$ :

$\begin{aligned} q &= 7.35\; {\mu C} \times \frac{10^{-6}\; {\rm C}}{1\; {\mu C}} = 7.35\times 10^{-6}\; {\rm C}\end{aligned}$ .

An electric field of magnitude $E = 5.25\times 10^{5}\; {\rm N \cdot C^{-1}}$ would exert on this charge a force with a magnitude of:

$\begin{aligned}F &= E\, q \\ &= 5.25 \times 10^{5}\; {\rm N \cdot C^{-1}} \times (-7.35\times 10^{-6}\; {\rm C}) \\ &\approx 3.86\; {\rm N}\end{aligned}$ .

Note that the electric charge in this question is negative. Hence, electrostatic force on this charge would be opposite in direction to the the electric field. Since the electric field points due south, the electrostatic force on this charge would point due north.

4 0

2 years ago

A DNA macromolecule is made up of two strands of DNA, which are connected to form a DNA double helix. The bonds between

Arturiano [62]

Answer:

where is the multiple choice?

6 0

3 years ago

Jin knows that the initial internal energy of a closed system is 78 J and the final internal energy is 180 J. He also knows that

lawyer [7]

As we know by the first law of thermodynamics

$Q = \Delta U + W$

here we know that

Q = heat given to the system

$\Delta U = U_f - U_i$

W = work done by the system

now here we can say

$\Delta U = 180 - 78 = 102 J$

$W = 64 J$

now we can say that heat will be given as

$Q = 64 + 102 = 166 J$

now here we can say that Jin does the error in his first step while calculation of change in internal energy as he had to subtract it while he added the two energy

So best describe Jin's Error is

<em>B )For step 1, he should have subtracted 78 J from 180 J to find the change in internal energy. </em>

8 0

3 years ago