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Vsevolod [243]
3 years ago
15

Anyone know how to solve this?

Physics
1 answer:
AnnZ [28]3 years ago
7 0
Beni humnexexec4dCallCalllollolI7
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Which of the following states that all matter tends to "warp" space in its vicinity and that objects react to this warping by ch
Natasha2012 [34]
There are no choices on the list you provided that make such a statement,
and it's difficult to understand what is meant by "the following".

That statement is one way to describe the approach to 'forces of gravity'
taken by the theory of Relativity. 
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A motorcycle running on gasoline wastes a large amount of energy mainly as A) heat energy and sound energy. B) light energy and
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The metal ions in iron are held together by
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Metals are giant structures of atoms held together by metallic bonds. “Giant” implies that large but variable numbers of Atoms are involved - depending on the size of the bits of metal. most metals are close packed - that is, they fit as many items as possible into the available volume.
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4 years ago
A 400g sample of water absorbs 500j of energy. how did the water temperature change if the specific heat of water is 4.18j/g©. S
Mashutka [201]

In general, the quantity of heat energy, Q, required to raise a mass m kg of a substance with a specific heat capacity of <span>c </span>J/(kg °C), from temperature t1 °C to t2 °C is given by:

<span>Q </span>= <span>mc(t</span><span>2 </span><span>– t</span>1<span>) joules</span>

<span>So:</span>

(t2-t1) =Q / mc

<span>As we know:
Q = 500 J </span>

<span>m = 0.4 kg</span>

<span>c = 4180 J/Kg </span>°c

<span>We can take t1 to be 0</span>°c

t2 - 0 = 500 / ( 0.4 * 4180 )

t2 - 0 = 0.30°c

6 0
3 years ago
The initial height of the water in a sealed container of diameter 100.0 cm is 5.00 m. The air pressure inside the container is 0
katen-ka-za [31]

Answer:

a)  F = 2.66 10⁴ N, b)   h = 1.55 m

Explanation:

For this fluid exercise we use that the pressure at the tap point is

Exterior

          P₂ = P₀ = 1.01 105 Pa

inside

         P₁ = P₀ + ρ g h

the liquid is water with a density of ρ=1000 km / m³

         P₁ = 0.85   1.01 10⁵ + 1000   9.8  5

         P₁ = 85850 + 49000

         P₁ = 1.3485 10⁵ Pa

the net force is

         ΔP = P₁- P₂

         Δp = 1.3485 10⁵ - 1.01 10⁵

         ΔP = 3.385 10⁴ Pa

Let's use the definition of pressure

         P = Fe / A

         F = P A

the area of ​​a circle is

         A = pi r² = [i d ^ 2/4

let's reduce the units to the SI system

         d = 100 cm (1 m / 100 cm) = 1 m

         F = 3.385 104 pi / 4 (1) ²

         F = 2.66 10⁴ N

b) the height for which the pressures are in equilibrium is

        P₁ = P₂

        0.85 P₀ + ρ g h = P₀

        h = \frac{P_o ( 1-0.850)}{\rho \ g}

        h = \frac{1.01 \ 10^5 ( 1 -0.85)}{1000 \ 9.8}

        h = 1.55 m

4 0
3 years ago
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