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timofeeve [1]
3 years ago
10

40 POINTS!!

Physics
2 answers:
Mazyrski [523]3 years ago
5 0

Answer:

1-put the black can in the temperature thing.

2-put the silver can well put water in it and leave it out side for a bit.

3-judge of which one will have a larger increase.

Explanation:

Mademuasel [1]3 years ago
3 0

Answer:

1 put the black can in the temperature thing.

2.put the silver can well put water in it and leave it out side for a while.

3.Then you can be the judge of witch one will have a larger increase.

hope this helps :)

happy to help any time:)

Explanation:

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A vertical spring stretches 4.0 cm when a 12-g object is hung from it. The object is replaced with a block of mass 28 g that osc
marin [14]

Answer:

Time period of the osculation will be 0.0671 sec  

Explanation:

It is given a vertical spring is stretched by 4 cm

So change in length of the spring x = 4 cm = 0.04 m

Mass which is hung from it m = 12 gram = 0.012 kg

Sprig force will be equal to weight of the mass

So kx=mg

k\times 0.04=0.012\times 9.8

k = 244.7 N/m

Now new mass is m = 28 gram = 0.028 kg

So time period with new mass will be

T=2\pi \sqrt{\frac{m}{k}}

=2\times 3.14 \sqrt{\frac{0.028}{244.7}}=0.0671sec

4 0
3 years ago
If a rock is thrown upward on the planet Mars with a velocity of 10 ms, its height in meters seconds later is given by (a) Find
Delicious77 [7]

Answer:

1) X(t)=Xo+So(t)+\frac{1}{2} g(t)^{2}

2) S(Δt)=So+g(Δt)

Explanation:

I think the equation they gave you for the Height upon time in seconds it's (1), where if you see, you will find the gravity, that you should multiply by 2 because, its divided by two in (1) (that should be your (a)), then, once you find your gravity, you can use the equation (2) to know the Final Speed replacing g , at the time asked, remember that g is gravity, and Δt is the: final time- initial time.

so in the [1,2] interval of time, your Δt=1s, and in [1,1.5] is Δt=0,5s.

i hoped it helped you even though i cant give you the exact answer right now.

6 0
4 years ago
The air in a tire pump has a volume of 1.5 L at a temperature of 5 ℃. If the temperature is increased to 25 ℃ and the pressure r
tankabanditka [31]

Answer:

Explanation:

5 C = 278 K

25 C = 298 K

V1 / T1 = V2 / T2

1.5L / 278 K = V2 / 298 K

V2 = (1.5L * 298) / 278

V2 = 1.61 L

5 0
3 years ago
What is the weight of a feather (mass = 0.0001 kg) that floats through earth's and the moon's atmospheres?
olya-2409 [2.1K]

Weight = (mass) x (acceleration of gravity)

Acceleration of gravity = 9.81 m/s² on Earth, 1.62 m/s² on the Moon.

The feather's weight is . . .

On Earth:  (0.0001 kg) x (9.81 m/s²) = <em>0.000981 Newton </em>

On the Moon:  (0.0001 kg) x (1.62 m/s²) = <em>0.000162 N</em>

The presence or absence of atmosphere makes no difference.  In fact, the numbers would be the same if the feather were sealed in a jar, or spinning wildly in a tornado, or hanging by a thread, or floating in a bowl of water or chicken soup.  Weight is just the force of gravity between the feather and the Earth.  It's not affected by what's around the feather, or what's happening to it.

6 0
3 years ago
If the Earth's ozone (O3) layer has a total volume of 1.00 × 1020 km3, a partial pressure of 1.6 × 10-9 atm, and an average temp
choli [55]

To solve this problem we will apply the ideal gas equations for which the product of pressure and volume is defined, as the equivalent between the ideal gas constant by the amount of matter and the temperature, mathematically this equation is described as

PV = nRT

Here,

P = Pressure

V = Volume

R = Ideal gas Constant

T = Temperature

n = Number of molecules

The pressure is in atmospheres, and considering the units of the other values we have finally that,

R = 0.08206 \cdot atm\cdot  L\cdot  mol^{-1}\cdot  K^{-1}

V = 1*10^{20}km^3 (\frac{1*10^{12}L}{1km^3 })

V = 1*10^{32}L

P = 1.6*10^{-9}atm

T = 230K

Replacing,

(1.6*10^{-9})(1*10^{32}) = n(0.08206)(230)

n = \frac{(1.6*10^{-9})(1*10^{32})}{(0.08206)(230)}

n = 8.47736*10^{-21}

Multiplying the number of moles by Avogadro's number we have,

M =(8.47736*10^{-21})(6.022*10^{23})

M = 5.1*10^{45}

Therefore the number of ozone molecules in the Earth's ozone layer are 5.1*10^{45}

4 0
4 years ago
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