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4 years ago

What information can scientists obtian from tree rings

Physics

1 answer:

Alexandra [31]4 years ago

5 0

What information can scientists obtain from tree rings?

Answer's <u>I chose</u>:

<h3>how narrow the rings are</h3><h3>how the climate changed in the tree’s life</h3><h3>how wide the rings are</h3>

Please <u>correct</u> me if there are <em>more </em>or <em>less</em>

Please give a brainliest and a thanks.

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Which of the following is a compound?<br><br> (A) Brass<br> (B) Rust<br> (C) Iron<br> (D) Steel

Naya [18.7K]

Rust is a compound of iron and oxygen.

7 0

4 years ago

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As you finish listening to your favorite compact disc (CD), the CD in the player slows down to a stop. Assume that the CD spins

devlian [24]

Answer:

10.8rev

Explanation:

Using

Wf²-wf = 2 alpha x theta

0²- 56.36x56.36/ 2(-20.13) x theta

Theta = 68.09 rad

But 68.09/2π

>= 10.8 revolutions

Explanation:

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4 years ago

A race car moving along a circular track has a centripetal acceleration of 15.4 m/s? If the car has

Helen [10]

Answer:

r = 58.44 [m]

Explanation:

To solve this problem we must use the following equation that relates the centripetal acceleration with the tangential velocity and the radius of rotation.

a = v²/r

where:

a = centripetal acceleration = 15.4 [m/s²]

v = tangential speed = 30 [m/s]

r = radius or distance [m]

r = v²/a

r = 30²/15.4

r = 58.44 [m]

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3 years ago

Jared made a study chart about heat, temperature, kinetic energy, and thermal energy. A chart with 2 columns and 5 rows. The fir

Maurinko [17]

Answer:

Switch the term "heat" with the term "thermal energy."

Explanation:

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3 years ago

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Suppose you apply a ﬂame and heat one liter of water, raising its temperature 10°C. If you transfer the same heat energy to two

polet [3.4K]

(a) the water density is $d=1000 kg/m^3$ , and 1 liter corresponds to a volume of $V=1 L=0.001 m^3$ . Therefore we can find the mass of the water in the first case:
$m=dV=(1000 kg/m^3)(0.001 m^3)=1 kg$

The amount of heat supplied to the water to raise its temperature by $\Delta T=10 ^{\circ} C$ is
$Q=m C_s \Delta T$ (1)
where
$C_s = 4.18 kJ/(kg ^{\circ} C}$ is the specific heat capacity of the water.
Using the data, we find
$Q=(1 kg)(4.18 kJ/(kg ^{\circ} C)(10^{\circ} C)=41.8 kJ$

We want to find the increase in temperature if we transfer the same amount of heat Q to 2 liters of water. The mass of 2 liters of water is
$m=dV=(1000 kg/m^3)(0.002 m^3)=2 kg$
And so by re-arranging equation (1) we can calculate the new increase of temperature:
$\Delta T_2 = \frac{Q}{m C_s} = \frac{41.8 kJ}{(2 kg)(4.18 kJ/(kg ^{\circ} C)}=5 ^{\circ} C$

(b) Now we have 3 liters of water. SImilarly to point (a), the mass is now
$m=dV=(1000 kg/m^3)(0.003 m^3)=3 kg$
And so, the increase in temperature if we use the same amount of heat as before is
$\Delta T_3= \frac{Q}{m C_s} = \frac{41.8 kJ}{(3 kg)(4.18 kJ/(kg ^{\circ} C)}=3.3 ^{\circ} C$

5 0

3 years ago