**Answer:**

76 days is the longest number of days in which the temperature in phoenix Arizona were at 100 f or hotter. That was in the year 1993

**Explanation:**

Phoenix is the capital of the southwestern USA state of Arizona

The longest stretch of consecutive 100 f or more is 76 days in the year 1993 (10 June 1993 - 24 August 1993)

There were 129 days with 100 f plus in 2003 but not consecutively.

There were 143 days of 100 f plus in 1989, but not consecutively.

29 days was recorded of 110 plus temperatures in 2019 but not consecutively.

Longest stretch of consecutive 110 f was 18 days (12 June 1974 to 29 June 1974)

The top hottest days in Phoenix

(a) 122 f (26 June, 1990)

(b) 121 f (28 July 1995)

(c) 120 f (25 June 1990)

(d) 119 f (29 June 2013)

Average method hope this helps

**Answer:**

a) + Q charge is inducce that compensates for the internal charge

b) There is no excess charge on the external face q_net = 0

c) E=0

**Explanation:**

Let's analyze the situation when a negative charge is placed inside the cavity, it repels the other negative charges, leaving the necessary positive charges to compensate for the -Q charge. The electrons that migrated to the outer part of the sphere, as it is connected to the ground, can pass to the earth and remain on the planet; therefore on the outside of the sphere the net charge remains zero.

With this analysis we can answer the specific questions

a) + Q charge is inducce that compensates for the internal charge

b) There is no excess charge on the external face q_net = 0

c) If we create a Gaussian surface on the outside of the sphere the net charge on the inside of this sphere is zero, therefore there is no electric field, on the outside

d) If it is very reasonable and this system configuration is called a Faraday Cage

e) We cannot apply this principle to gravity since there are no particles that repel, in all cases the attractive forces.

**Answer:**

A. 2.58m

**Explanation:**

Using the formula;

F = GMm/r²

M and m are the masses

G is the gravitational constant

r is the distance between the masses

Substitute

0.00006 = 6.67*10^-11(2000)(3000)/r²

0.00006r² = 6.67*10^-11*6000000

0.00006r² = 40.02*10^-5

0.00006r² = 0.0004

r² = 0.0004/0.00006

r² = 6.66

r = 2.58m

**Hence the distance between them is 2.58m**

Answer:

14.49 g/cm²

Explanation:

I = Io e^-(ux)

Where:

I = 573

Io = 1045

x = 0.3 inches and

rho = 11.4g/cm^3

Using the conversion constant

1 inch = 2.54 cm;

0.3 inches = 0.3 * 2.54 cm

0.3 inches = 0.762 cm

I/Io = e^-(ux), or say

Io/I = e^(ux), taking the In of both sides

ln(Io/I) = ux, making u subject of formula

u = 1/x * ln(Io/I)

u = 1/0.762 * ln(1045/573)

u = 1.312 * 0.6

u = 0.787

Next, we say that

u/rho = 0.7872/11.4 = 0.069

And finally, we make

1/(u/rho) to be our final answer

Inverse of the answer is = 14.49 g/cm²

Therefore, the um^-1 in g/cm^2? is 14.49