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2 years ago

14

Free brainlyest. if get thes quston right Devlin recorded the number of days of rain for his town over three months. The bar gra

ph below shows his data.
Bar Graph 1

Explain why his data is more about climate than weather.

Question 12 options:

2 answers:

m_a_m_a [10]2 years ago

8 0

Answer:

Weather is the conditions (temperature, wind etc.) at a given time, like on that day. Climate, which is what his data would show, is the conditions over an extended period of time like the 3 months he collected data

mel-nik [20]2 years ago

5 0

Answer:

Climographs show monthly average temperatures and precipitation totals on a single graph. Rather than showing measured data for specific time periods—like the graphs you generated in Part A—climographs show long term averages for all 12 months of the year.

Explanation:

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A converging lens of focal length 20 cm is used to form a real image 1.0 m away from the lens. How far from the lens is the obje

Galina-37 [17]

Answer:

0.25 m

Explanation:

We can solve the problem by using the lens equation:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where

f is the focal length

p is the distance of the object from the lens

q is the distance of the image from the lens

In this problem, we have

f = +20 cm=+0.20 m (the focal length is positive for a converging lens)

q = +1.0 m (the image distance is positive for a real image)

Solving the equation for p, we find

$\frac{1}{p}=\frac{1}{f}-\frac{1}{q}=\frac{1}{0.20 m}-\frac{1}{1 m}=4 m^{-1}\\p=\frac{1}{4 m^{-1}}=0.25 m$

6 0

2 years ago

Find the position of the center of mass of the system of the sun and Jupiter? (Since Jupiter is more massive than the rest of th

8090 [49]

Answer:

$r_{cm} = 0.074 m$ from the position of the center of the Sun

Explanation:

As we know that mass of Sun and Jupiter is given as

$M_s = 1.98 \times 10^{30} kg$

$M_j = 1.89 \times 10^{27} kg$

distance between Sun and Jupiter is given as

$r = 7.78 \times 10^{11} m$

now let the position of Sun is origin and position of Jupiter is given at the position same as the distance between them

so we will have

$r_{cm} = \frac{M_s r_1 + M_j r_2}{M_s + M_j}$

$r_{cm} = \frac{1.98 \times 10^{30} (0) + (1.89 \times 10^{27})(7.78 \times 10^{11})}{1.98 \times 10^{30} + 1.89 \times 10^{27}}$

$r_{cm} = 0.074 m$ from the position of the center of the Sun

3 0

3 years ago

An ideal monatomic gas at 275 K expands adiabatically and reversibly to six times its volume. What is its final temperature (in

Gwar [14]

The final temperature is 83 K.

<u>Explanation</u>:

For an adiabatic process,

$T {V}^{\gamma - 1} = \text{constant}$

$\cfrac{{T}_{2}}{{T}_{1}} = {\left( \cfrac{{V}_{1}}{{V}_{2}} \right)}^{\gamma - 1}$

Given:-

${T}_{1} = 275 \; K$

${T}_{2} = T \left( \text{say} \right)$

${V}_{1} = V$

${V}_{2} = 6V$

$\gamma = \cfrac{5}{3} \;$ (the gas is monoatomic)

$\therefore \cfrac{T}{275} = {\left( \cfrac{V}{6V} \right)}^{\frac{5}{3} - 1}$

$\Rightarrow \cfrac{T}{275} = {\left( \cfrac{1}{6} \right)}^{\frac{2}{3}}$

T = 275 $\times$ 0.30

T = 83 K.

3 0

3 years ago

Two identical balls collide one was initially at rest. after the collision one of the balls has a velocity of 3 m/s south. the o

Butoxors [25]

Two identical balls collide<span> head on. The </span>initial velocity<span> of </span>one<span> is 0.75 </span>m/s<span> east, while that of the </span>other one<span> is 0.43 </span>m/s west<span>.</span>

4 0

2 years ago

A 1.90-m-long barbell has a 25.0 kg weight on its left end and a 37.0 kg weight on its right end. if you ignore the weight of th

gregori [183]

Answer: The center of gravity is 1.1338 m away from the left side of the barbell

Explanation:

Length of the barbell = 1.90 m

The distance center of gravity from left = x

Mass on the left side = 25 kg

The distance center of gravity from right = 1.90 - x

Mass on the right side = 37 kg

At the balance point: $m_1x_1=m_2x_2$

$25 kg\times x=37 kg\times (1.90-x)$

$x=1.1338 m$

The center of gravity is 1.1338 m away from the left side of the barbell

7 0

3 years ago

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