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4 years ago

7

Toddlers are extremely active, naturally curious, and very top heavy making them prone to falls. The disproportionately large he

ad and small legs common to 2-3 year-olds is due to which pattern of growth?

1 answer:

Serhud [2]4 years ago

6 0

Answer:

No

Explanation:

no

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Why does it take mars over 500 days to orbit the sun?

kakasveta [241]

Answer:

Mar's orbital path is more than that of Earth, thus it takes more number of days to orbit around the sun.

Explanation:

Mars takes over 500 days to orbit all the way around the sun than Earth because its distance from the sun (228 million kilometers) is greater than that of Earth (150 million kilometers) which takes it 365 days.

Planets that orbit closer to the sun take shorter time to orbit around the sun because the cover a shorter orbital distance and orbit faster than those planets further from the sun.

<u>For example</u>

Using Earth's distance from the sun, 150 million kilometers and the number of days taken to orbits the sun ,365 days and the distance Mars is from the Earth, 228 million kilometers, you can approximate the time Mar takes to orbit the sun as:

Earth 150 million kilometers = 365 days

Mars 228 million kilometers= ?

Cross product ; (228 *365) /150 =555 -----(a value closer to that in the question)

6 0

3 years ago

PLEASE HELP! I'LL GIVE BRAINLEST

erma4kov [3.2K]

Answer:

8.162

Explanation:

W=mg

= 2.2 x 3.71

= 8.162 N

5 0

3 years ago

Which of the following statements about price elasticity of demand is true?

STatiana [176]

Answer:

B. The elastic portion of a straight-line, downward-sloping demand curve corresponds to the segment above the midpoint.

Explanation:

Elasticity measures the sensitivity of one variable to another. Specifically it is a figure that indicates the percentage variation that a variable will experience in response to a variation of another one percent.

The elasticity of demand measures the reaction of demand when one of the factors that affects it varies.

<u>Elasticity - Price of demand.</u>

easure the sensitivity of the quantity demanded to price variations. It indicates the percentage variation that the quantity demanded of a good will experience if its price rises by 1 percent.

<u> Elastic Demand </u>

The demand quantity is relatively sensitive to price variations, so the total expenditure on the product decreases when the price rises, the price elasticity takes value greater than -∞ but less than -1

7 0

3 years ago

Julio blows air across his hot bowl of soup. The tiny ripples he creates are similar to _____.

Allushta [10]

The tiny ripples on the soup are not only similar to wind-generated
waves ... they ARE wind-generated waves. This is a big part of the
reason why they bear such an uncanny resemblance.

8 0

4 years ago

Read 2 more answers

. Consider the equation =0+0+02/2+03/6+04/24+5/120, where s is a length and t is a time. What are the dimensions and SI units of

Olegator [25]

Answer:

See Explanation

Explanation:

Given

$s=s_0+v_0t+\frac{a_0t^2}{2}+ \frac{j_0t^3}{6}+\frac{S_0t^4}{24}+\frac{ct^5}{120}$

Solving (a): Units and dimension of $s_0$

From the question, we understand that:

$s \to L$ --- length

$t \to T$ --- time

Remove the other terms of the equation, we have:

$s=s_0$

Rewrite as:

$s_0=s$

This implies that $s_0$ has the same unit and dimension as $s$

Hence:

$s_0 \to L$ --- dimension

$s_o \to$ Length (meters, kilometers, etc.)

Solving (b): Units and dimension of $v_0$

Remove the other terms of the equation, we have:

$s=v_0t$

Rewrite as:

$v_0t = s$

Make $v_0$ the subject

$v_0 = \frac{s}{t}$

Replace s and t with their units

$v_0 = \frac{L}{T}$

$v_0 = LT^{-1}$

Hence:

$v_0 \to LT^{-1}$ --- dimension

$v_0 \to$ $m/s$ --- unit

Solving (c): Units and dimension of $a_0$

Remove the other terms of the equation, we have:

$s=\frac{a_0t^2}{2}$

Rewrite as:

$\frac{a_0t^2}{2} = s_0$

Make $a_0$ the subject

$a_0 = \frac{2s_0}{t^2}$

Replace s and t with their units [ignore all constants]

$a_0 = \frac{L}{T^2}\\$

$a_0 = LT^{-2$

Hence:

$a_0 = LT^{-2$ --- dimension

$a_0 \to$ $m/s^2$ --- acceleration

Solving (d): Units and dimension of $j_0$

Remove the other terms of the equation, we have:

$s=\frac{j_0t^3}{6}$

Rewrite as:

$\frac{j_0t^3}{6} = s$

Make $j_0$ the subject

$j_0 = \frac{6s}{t^3}$

Replace s and t with their units [Ignore all constants]

$j_0 = \frac{L}{T^3}$

$j_0 = LT^{-3}$

Hence:

$j_0 = LT^{-3}$ --- dimension

$j_0 \to$ $m/s^3$ --- unit

Solving (e): Units and dimension of $s_0$

Remove the other terms of the equation, we have:

$s=\frac{S_0t^4}{24}$

Rewrite as:

$\frac{S_0t^4}{24} = s$

Make $S_0$ the subject

$S_0 = \frac{24s}{t^4}$

Replace s and t with their units [ignore all constants]

$S_0 = \frac{L}{T^4}$

$S_0 = LT^{-4$

Hence:

$S_0 = LT^{-4$ --- dimension

$S_0 \to$ $m/s^4$ --- unit

Solving (e): Units and dimension of $c$

Ignore other terms of the equation, we have:

$s=\frac{ct^5}{120}$

Rewrite as:

$\frac{ct^5}{120} = s$

Make $c$ the subject

$c = \frac{120s}{t^5}$

Replace s and t with their units [Ignore all constants]

$c = \frac{L}{T^5}$

$c = LT^{-5}$

Hence:

$c \to LT^{-5}$ --- dimension

$c \to m/s^5$ --- units

4 0

3 years ago