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8090 [49]
3 years ago
7

It is thought that caused past ice age climates.

Physics
1 answer:
lyudmila [28]3 years ago
4 0

Answer:

Explanation:

The change in past ice age climates can be explained by using Milankovitch cycle. According to the cycle past ice can be caused by any of the these reasons:

  1. Earth's orbit around the sun.
  2. Obliquity of the axis.
  3. Eccentricity of it's orbit.
  4. Precedence of earth's orbit.

However, today materials from past are being used such as rocks, fossil remain,etc. to study the earth's climate.

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If an area receives a large amount of insulation is it likely to become warm or cold?
Levart [38]

Answer: heat

Insulation Traps Heat. Keeping the cold air out

Explanation:

5 0
3 years ago
You hear a sound with a frequency of 256 Hz. The amplitude of the sound increases and decreases periodically: it takes 2 seconds
german

To solve this problem it is necessary to take into account the concepts related to frequency and period, and how they are related to each other.

The relationship that defines both agreements is given by the equation,

f_{beat}=\frac{1}{T}

Then the frequency for the previous period given (2sec) is

f_{beat}=\frac{1}{2}

f_{beat} = 0.5Hz

The beat frequency of two frequencies is equal to the difference between the two frequencies, then

f_{beat} = |f_1-f_2|\\f_{beat} = |256Hz-2Hz|\\f_{beat} = 254Hz

<em>Hence option A is incorrect.</em>

We can do this process for 254Hz as f_1 and 258 Hz for f_2 , then

f_{beat} =|254Hz-258Hz|

f_{beat} = 4Hz

<em>Hence option B is incorrect. </em>

We can also do this process for 255Hz as  f_1 and 257 Hz for f_2 , then

f_{beat} =|255Hz-257Hz|

f_{beat} = 2Hz

<em>Hence option C is incorrect. </em>

We can also do this process for 255.5Hz as f_1 and 256.5 Hz for f_2, then

f_{beat} =|255.5Hz-256.5Hz|\\f_{beat} = 1Hz

<em>Hence option D is incorrect. </em>

We can also do this process for 255.75Hz as f_1 and 256.25 Hz for f_2, then

f_{beat} =|255.75Hz-256.25Hz|\\f_{beat} = 0.5Hz

<em>Hence option E is incorrect. </em>

Therefore the sum of the frequencies in the sound wave would be 256.25Hz and 255.75Hz

3 0
3 years ago
A student hangs a weight on a newtonmeter. The energy currently stored in the spring in the newton meter is 0.045N. The student
castortr0y [4]

Answer:

5x10^-3

Explanation:

Hooke's Law states that the force needed to compress or extend a spring is directly proportional to the distance you stretch it.

Hooke's Law can be represented as

<h3> F = kx, </h3>

<em>where F is the force </em>

<em>            k is the spring constant</em>

<em>            x is the extension of the material </em>

<em />

Plug values in the equation

Step 1 find the original extension

0.045 = (400)x

x = 1.125x 10^-4 m d

Step 2 find the new extension

0.045+2 = 400(x)

2.045 = 400x

x = 5.1125x10^-3

Step 3 subtract the new extension with original

Total extension of the spring =  5.1125x10^-3 - 1.125x 10^-4 m = 5x10^-3

8 0
3 years ago
JUST PLZ HELP!!! Why does the lightbulb in the right electrical circuit turn on but not the one on the left?
makkiz [27]
Because,

In left image pin is not touch to the wire.

In right image pin is touch to the wire.

Hope it helps you.....

Plz...Plz...Plz...Plz…Plz…

Mark be Brainliest.....

Please.....

And..

Please thanks me.....

Plz.....Plz.....
8 0
3 years ago
Read 2 more answers
A 40W lamp wastes 34 J of energy every second by heating its surroundings.
Artemon [7]

Answer:

15\%.

Explanation:

The efficiency of a machine is the percentage of energy input that was turned into useful energy.

The power rating of this lamp is 40\; \rm W (same as 40\; \rm J \cdot s^{-1},) meaning that 40\; \rm J of energy is supplied to this lamp every second.

The question states that 34\; \rm J out of that 40\; \rm J of energy input would be turned into heat, which is not useful energy output in this scenario. Assuming that all other forms of energy loss is negligible. The rest of the (40\; \rm J - 34\; \rm J) = 6\; \rm J of energy supplied to this lamp would be turned into useful energy output.

Thus, every second, this lamp would receive 40\; \rm J of energy input and would outputs 6\; \rm J of useful work. The efficiency of this lamp would be:

\begin{aligned}& \text{Efficiency} \\ =\; & \frac{\text{Useful energy out}}{\text{Total energy in}} \times 100\% \\ =\; & \frac{6\; \rm J}{40\; \rm J} \times 100\%\\ =\; &15\% \end{aligned}.

4 0
2 years ago
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