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2 years ago

14

A mother and her young child want to play on a seesaw at a playground. The child sits on the end of one side of the seesaw. Wher

e should the mother sit to balance the seesaw?(1 point) at the opposite side of the seesaw on the end at the opposite side of the seesaw towards the middle on the same side of the seesaw towards the middle on the same end as her child

1 answer:

EleoNora [17]2 years ago

4 0

Answer:middle

Explanation:

Because it will make the seasaw balanced

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Which of the following prefixes represents the largest value

Zigmanuir [339]

C should be right ok

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3 years ago

Suppose astronomers built a 20-meter telescope. How much greater would its light-collecting area be than that of the 10-meter Ke

nirvana33 [79]

Answer:

4 times greater

Explanation:

<u>Step 1:</u> Calculate light-collecting area of a 20-meter telescope (A₁) by using area of a circle.

Area of circle = π*r² = $\frac{\pi d^{2}}{4}$

Where d is the diameter of the circle = 20-m

$A_{1} = \frac{\pi d^{2}}{4}$

$A_{1} = \frac{\pi (20^{2})}{4}$

A₁ = 314.2 m²

<u>Step 2:</u> Calculate light-collecting area of a 10-meter Keck telescope (A₂)

$A_{2} = \frac{\pi d^{2}}{4}$

Where d is the diameter of the circle = 10-m

$A_{2} = \frac{\pi (10^{2})}{4}$

A₂ = 78.55 m²

<u>Step 3</u>: divide A₁ by A₂

$= \frac{314.2 m^2}{78.55 m^2}$

= 4

Therefor, the 20-meter telescope light-collecting area would be 4 times greater than that of the 10-meter Keck telescope.

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3 years ago

Someone please help me for a brainliest but it has to be correct

Katarina [22]

The answer is 34 you have to add the numbers and divide them by how many numbers there are

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3 years ago

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The eyes of amphibians such as frogs have a much flatter cornea but a more strongly curved (almost spherical) lens than do the e

Lapatulllka [165]

Answer:

0.2cm towards the retina.

Explanation:

the focal length of the frog eye is

(1/f) = (1/10) + (1/0.8)

f = 0.74cm

Since the distance of the object is 15cm Hence

(1/0.74) = (1/15) + (1/V)

V = 0.78cm

Therefore the distance the retina is to move is

0.78cm - 0.8cm = 0.02cm towards the retina.

3 0

3 years ago

If 500g of water at 20 degrees C is mixed with 750g of water at 30 degrees C, what will the temperature of the mixture be?

hjlf

Answer:

$26 ^\circ C$

Explanation:

Given that the temperature of $500g$ of water and $750 g$ of water are

at $20^{\circ}C$ and $30^\circ C$ respectively.

Let $m_1=500g$ , $T_1= 20^\circ C$

and $m_2=750g$ , $T_2= 30^\circC$ .

The specific heat capacity of water is,

$C= 4.186 J/g ^\circ C$ .

Let the final temperature of the mixture be $T^\circ C$ .

As there is no energy loss, so, the energy loss by the water at higher temperature, i.e. $30^\circ C$ , will be equal to the energy gain by the water at lower temperature, i.e. $20^{\circ}C$ .

$m_2C (T_2-T)=m_1C(T-T_1)$

$\Rightarrow m_2 (T_2-T)=m_1(T-T_1)$ [ both sides divided by $C$ ]

$\Rightarrow m_2T_2-m_2T=m_1T-m_1T_1$

$\Rightarrow m_1T+m_2T=m_1T_1+m_2T_2$

$\Rightarrow T=\frac{m_1T_1+m_2T_2}{m_1+m_2}$

Now, putting the given value in the above equation, we have

$\Rightarrow T=\frac {500\times 20+750\times 30}{500+750}$

$\Rightarrow T=26^\circ C.$

Hence, the temperature of the mixture will be $26 ^\circ C$ .

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3 years ago