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2 years ago

What is the surface area for a rectangular prism with the height of 6 inches, width of 8 inches, and the length of 4 inches

Physics

2 answers:

podryga [215]2 years ago

8 0

Answer:

208 is the surface area!

Explanation:

Hope this helps! ;-)

victus00 [196]2 years ago

8 0

Answer:

208

Explanation:

208 is the answer to this question

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A 1.55-kg object hangs in equilibrium at the end of a rope (taken as massless) while a wind pushes the object with a 13.3-n hori

yarga [219]

To answer this problem, we use balancing of forces: x and y components to determine the tension of the rope.

First, the vertical component of tension (Tsin theta) is equal to the weight of the object.
T * sin θ = mg = 1.55 * 9.81 
T * sin θ = 15.2055

Second, the horizontal component of tension (t cos theta) is equal to the force of the wind.
T * cos θ = 13.3

Tan θ = sin θ / cos θ = 15.2055/13.3 = 1.143
we can find θ that is equal to 48.82.

T then is equal to 20.20 N

4 0

2 years ago

The phosphorus cycle is the biogeochemical cycle that describes the movement of phosphorus through the lithosphere,

ludmilkaskok [199]

most common source of phosphates in the soil is the weathering of the mineral apatite.

Explanation:

Apatite is a group of phosphate minerals like hydroxylapatite, fluorapatite, and chlorapatite. When apatite rocks weather, they release the phosphate minerals mainly in the form of PO₄ ³⁻ . These minerals become dissolved in water (hydrosphere), where they are readily available to plants and other organisms in the biosphere. The phosphates are taken up and used in biosynthesis. When these organisms die and become buried with sediments, the phosphate gets back to the lithosphere as sedimentary rock.

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2 years ago

En un experimento de calorimetría, 0.50 kg de un metal a 100°C se añaden a 0.50 kg de agua a 20°C en un vaso de calorímetro de a

Maru [420]

Answer:

c=0.14J/gC

Explanation:

2) The specific heat will be the same because it is a property of the substance and does not depend on the medium.

We can use the expression for heat transmission

$Q=mc(T_2-T_1)$

In this case the heat given by the metal (which is at a higher temperature) is equal to that gained by the water, that is to say

$Q_1=-Q_2$

for water we have to

c = 4.18J / g ° C

replacing we have

$c_{metal}*(500g)(100\°C-25\°C)=-(250g)(4.18\frac{J}{g\°C})(20\°C-25\°C)\\c_{metal}=0.14\frac{J}{g\°C}$

I hope this is useful for you

2) El calor específico será igual porque es una propiedad de la sustancia y no depende del medio.

Podemos usar la expresión para la transmisión de calor

$Q=mc(T_2-T_1)$

En este caso el calor cedido por el metal (que está a mayor temperatura) es igual al ganado por el agua, es decir

$Q_1=-Q_2$

para el agua tenemos que

c=4.18J/g°C

reemplazando tenemos

$c_{metal}*(500g)(100\°C-25\°C)=-(250g)(4.18\frac{J}{g\°C})(20\°C-25\°C)\\c_{metal}=0.14\frac{J}{g\°C}$

7 0

2 years ago

All biomes don’t have the same level of biodiversity. What seems to be the optimal conditions for high biodiversity?

irinina [24]

Answer:

See the answer below

Explanation:

The optimal conditions for high biodiversity seem to be a warm temperature and wet climates.

The tropical areas of the world have the highest biodiversity and are characterized by an average annual temperature of above 18 $^oC$ and annual precipitation of 262 cm. The areas are referred to as the world's biodiversity hotspots.

Consequently, it follows logically that the optimal conditions for high biodiversity would be a warm temperature of above 18 $^oC$ and wet environment with annual precipitation of not less than 262 cm.

The variation in temperature and precipitation across biomes can thus be said to be responsible for the variation in the level of biodiversity in them.

6 0

2 years ago

If the average velocity during the athlete's walk back

goblinko [34]

Hence ,From the Guide there are other parameters which with this equation will give the exact time the athlete's walk back

$T=\frac{d}{1.50}$

From the question we are told

If the average velocity during the athlete's walk back to the starting line in Guided Example 2.5 is – 1.50 m/s,

Generally the equation Time spent is mathematically given as

T=\frac{d}{v}

Therefore

$T=\frac{d}{1.50}$

Hence ,From the Guide there are other parameters which with this equation will give the exact time the athlete's walk back

$T=\frac{d}{1.50}$

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2 years ago