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Rudiy27
2 years ago
13

Climate is the pattern of ________________ in an area over the years.

Physics
2 answers:
pav-90 [236]2 years ago
8 0

Answer:

Weather conditions

Explanation:

Maru [420]2 years ago
3 0

Answer:

Weather

Explanation:

Climate is the the long-term pattern of weather in a specific area.

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Give the correct balanced equation for the single replacement reaction of magnesium with aluminum oxide
diamong [38]

Answer:

Al2O3 + 3Mg ===》 3MgO + 2Al

Explanation:

Reaction of aluminium oxide with magnesium metal would form magnesium oxide and aluminium metal.

Balancing the stoichiometric equation, the number of atoms at the reactant must be equal to the number of atoms at the product so that law of conservation of matter must hold.

Al203 + 3Mg ====》 3MgO + 2Al

2 aluminium at reactant = 2 aluminium at product

3 oxygen at reactant = 3 oxygen at product

3 magnesium at reactant = 3 magnesium at product.

8 0
3 years ago
An object weighs 15N in air and 13N when completely
anastassius [24]

Answer:

The volume of water displaced = 0.204 m³

Explanation:

<em>From Archimedes' principle,</em>

<em>Upthrust = lost in weight = weight of water displaced.</em>

<em>U = W₁ - W₂..................... Equation 1</em>

<em>Where U = upthrust, W₁ = weight in air, W₂ = weight when completely submerged in water.</em>

<em>Given: W₁ = 15 N, W₂ = 13 N</em>

<em>Substituting these values into equation 1</em>

<em>U = 15 - 13 </em>

<em>U = 2 N</em>

<em>But,</em>

<em>Weight of water displaced = mass of water displaced. × acceleration due to gravity.</em>

<em>Mass of water displaced = weight of water displaced/ acceleration due to gravity.................... Equation 2</em>

Where: acceleration due to gravity = 9.8 m/s², weight of water displaced = 2 N

Substituting these values into equation 2

Mass of water displaced = 2/9.8

Mass of water displaced = 0.204 kg.

Also,

Volume of water displaced = mass of water displaced/ Density of water.............. Equation 3

Where Density of water = 1.0 kg/m³, mass of water displaced = 0.204 kg

Substituting these values into equation 3 above,

Volume of water displaced = 0.204/1

volume of water displaced = 0.204 m³

Therefore the volume of water displaced = 0.204 m³

<em />

8 0
3 years ago
A lion has a mass of 45 kg. Answer the following questions about it, using correct units. a. The lion runs at a speed of 14.2 m/
Eva8 [605]
A) The kinetic energy of an object is given by:
K= \frac{1}{2}mv^2
where m is the mass of the object, and v its speed. For the lion in our problem, m=45 kg and v=14.2 m/s, so its kinetic energy is
K= \frac{1}{2}mv^2= \frac{1}{2}(45 kg)(14.2 m/s)^2=4537 J

b) the increase in gravitational potential energy of the lion is given by:
\Delta U = mg \Delta h
where g is the gravitational acceleration, and \Delta h is the increase in altitude of the lion. In this problem, \Delta h=28 m, so the increase in gravitational potential energy is
\Delta U=mg \Delta h=(45 kg)(9.81 m/s^2)(28 m)=12361 J

c) When the fox reaches the top of the tree, its gravitational potential energy is
U=mgh=(1.8 kg)(9.81 m/s^2)(3.8 m)=67 J
As it jumps, its kinetic energy is
K= \frac{1}{2}mv^2= \frac{1}{2}(1.8 kg)(8.1 m/s)^2=59 J
So the total mechanical energy of the fox as it jumps is
E=U+K=67 J + 59 J =126 J
6 0
3 years ago
4<br> Explica las características principales de las fuer-<br> zas de acción y reacción.
Naya [18.7K]
-Surgen de una interacción.
-Nunca aparece una sola: son dos y simultáneas.
-Actúan sobre cuerpos diferentes: una en cada cuerpo.
-Nunca forman un par de fuerzas: tienen la misma línea de acción.
-Un cuerpo que experimenta una única interacción no está en equilibrio, pues sobre el aparece una fuerza unica que lo acelera. Para estar en equilibrio se requieren por lo menos dos interacciones.




Las mas importantes son la 2,3,4 característica
5 0
3 years ago
A car goes round a curve of radius 48m, the road is banked at an angle of 15 with the horizontal,at what maximum speed may the c
Marizza181 [45]

Answer:

11 m/s

Explanation:

Draw a free body diagram.  There are two forces acting on the car:

Weigh force mg pulling down

Normal force N pushing perpendicular to the incline

Sum the forces in the +y direction:

∑F = ma

N cos θ − mg = 0

N = mg / cos θ

Sum the forces in the radial (+x) direction:

∑F = ma

N sin θ = m v² / r

Substitute and solve for v:

(mg / cos θ) sin θ = m v² / r

g tan θ = v² / r

v = √(gr tan θ)

Plug in values:

v = √(9.8 m/s² × 48 m × tan 15°)

v = 11.2 m/s

Rounded to 2 significant figures, the maximum speed is 11 m/s.

3 0
3 years ago
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