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umka21 [38]
2 years ago
14

Which of these will NOT help reduce c02 levels in the atmosphere?

Physics
1 answer:
denpristay [2]2 years ago
4 0

Answer:

the answer would be "using more heat" btw

Explanation:

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I run around a circular track, with radius 25m, for 4 and times before stopping. It takes me 16
Sergio039 [100]

Answer:

13m/s. is the answer probably but do u have ms gallup too?

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3 years ago
Use the figure to help you find the weight (Fg) of the body if the tension in both wires is 3.7 newtons.
topjm [15]
I'm not completely sure, but I think it's 3.4 newtons. I hope you get it correct.
4 0
3 years ago
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Amanda [17]

Two, four, eight and sixteen cells are produced from first, second, third and fourth cell division respectively.

<h3>How many cells are produced? </h3>

When the body cell passes through cell division process, it divides into two daughter cells. In the second cell division, two cells divides into four cells. In the third cell division, the four cells divides into eight cells. In the fourth cell division, the eight cells divides and sixteen cells are produced.

So we can conclude that two, four, eight and sixteen cells are produced from first, second, third and fourth cell division respectively.

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6 0
2 years ago
(d) शून्य
Ede4ka [16]
The answer is a 20 or 140 u choose because u subtract or add these numbers and sorry I can’t speak sri lanka or Indian but I know how to read them
8 0
3 years ago
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what was the initial temperature is 250 calories reply .1 kg of gold the final temperature of the gold was 175°c ? the specific
weqwewe [10]

Answer:

115.2^{\circ}C

Explanation:

When an amount of energy Q is supplied to a substance of mass m, the temperature of the substance increases by \Delta T, according to the equation

Q=mC_s \Delta T

where C_s is the specific heat capacity of the substance.

In this problem, we have:

Q=250 \cdot 4.184 =1046 J is the amount of heat supplied to the sample of gold

m = 0.1 kg = 100 g is the mass of the sample

C_s = 0.175 J/gC is the specific heat capacity of gold

Solving for \Delta T, we find the change in temperature

\Delta T = \frac{Q}{m C_s}=\frac{1046}{(100)(0.175)}=59.8^{\circ}

And since the final temperature was

T_f = 175^{\circ}

The initial temperature was

T_i = T_f - \Delta T= 175 -59.8=115.2^{\circ}C

3 0
3 years ago
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