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lianna [129]
3 years ago
9

Hello please help i’ll give brainliest

Physics
1 answer:
fenix001 [56]3 years ago
5 0

Answer:

B. Large mountain ranges

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(4A) The mass of Earth is 5.972 * 10^24 kg, and the radius of Earth is 6,371 km.
faltersainse [42]

Answer:

x₁ = 345100 km

Explanation:

The direction of the attraction forces between the earth and the object, and between the moon and the object, are in opposite direction and  (along the straight line between the centers of earth and moon) and as gravity is always attractive, the net force will become zero when both forces are equal. According to this:

Let  call "x₁"  distance between center of the earth and the object, and

"x₂" the distance between center of the moon and the object, Mt mass of the earth, Ml mass of the moon, m₀ mass of the object

we can express:

F₁  ( force between earth and the object )

F₁ = K *  Mt * m₀/ ( x₁)²        K is a gravitational constant

F₂  (force between mn and the object)

F₂ = K * Ml * m₀ / (x₂)²

Then:

F₁ = F₂               K*Mt*m₀ / x₁²   =  K*Ml*m₀ /x₂²

Or  simplifying the expression

Mt/ x₁²  =  Ml/ x₂²

We know that   x₁   +  x₂  = 384000 Km then

x₁ =  384000 - x₂

Mt/( 384000 - x₂)²  =  Ml / x₂²

Mt *  x₂²  =  Ml *( 384000 - x₂)²

We need to solve for x₂

Mt *  x₂²  =  Ml *[ ( 384000)² + x₂² - 768000*x₂]

By substitution:

5.972*10∧24*x₂² = 7.348*10∧22 * [ 1.47*10∧11 ] + 7.348*10∧22*x₂² -

                                7.348*10∧22*768000*x₂

Simplifying by 10∧22

5.972*10²*x₂²  = 7.348* [ 1.47*10∧11 ] + 7.348*x₂²- 7.348*768000*x₂

Sorting out

5.972*10²*x₂²- 7.348*x₂² = 10.80*10∧11 - 56,43* 10∧5*x₂

(597,2 - 7,348 )* x₂²  = 10.80*10∧11 - 56.43*10∧5*x₂

590x₂²  + 56.43*10∧5*x₂ - 10.80*10∧11 = 0

Is a second degree equation

x₂  =  -56.43*10∧5 ± √3184*10∧10 + 25488*10∧11  / 1160

x₂ ₁  = -56.43*10∧5 + √3184*10∧10 + 25488*10∧11  / 1160

x₂ ₁  =  -56.43*10∧5 + √3184*10∧10 + 254880*10∧10  / 1160

x₂ ₁  = -56.43*10∧5 + 10∧5 [ √3184 + 254880 ] /1160

x₂ ₁  =  -56.43*10∧5 + 508* 10∧5  / 1160

x₂ ₁  =  451.27*10∧5/1160

x₂ ₁  =  4512.7*10∧4 /1160

x₂ ₁  = 3.89*10∧4  km (distance between the moon  and the object)

x₂ ₁  = 38900 km

x₂ = 38900 km

We dismiss the other solution because is negative and there is not a negative distance

Then the distance between the earth and the object is:

x₁  = 384000 - x₂

x₁ = 384000 - 38900

x₁ = 345100 km

5 0
3 years ago
The diagram shows the movement of air as a result of convection currents. At which point is the air at its highest density?
stellarik [79]

When you bring two objects of different temperature together, energy will always be transferred from the hotter to the cooler object.  The objects will exchange thermal energy, until thermal equilibrium is reached, i.e. until their temperatures are equal.  We say that heat flows from the hotter to the cooler object.  Heat is energy on the move.  

Units of heat are units of energy.  The SI unit of energy is Joule.  Other often encountered units of energy are 1 Cal = 1 kcal = 4186 J, 1 cal = 4.186 J, 1 Btu = 1054 J.

Without an external agent doing work, heat will always flow from a hotter to a cooler object.  Two objects of different temperature always interact.  There are three different ways for heat to flow from one object to another.  They are conduction, convection, and radiation.

 

6 0
3 years ago
Read 2 more answers
A mass of 3.6 kg oscillate on a horizontal spring with a spring constant of 160 N/m.
Darya [45]

Answer:

48.7 J

Explanation:

For a mass-spring system, there is a continuous conversion of energy between elastic potential energy and kinetic energy.

In particular:

- The elastic potential energy is maximum when the system is at its maximum displacement

- The kinetic energy is maximum when the system passes through the equilibrium position

Therefore, the maximum kinetic energy of the system is given by:

KE=\frac{1}{2}mv^2

where

m is the mass

v is the speed at equilibrium position

In this problem:

m = 3.6 kg

v = 5.2 m/s

Therefore, the maximum kinetic energy is:

KE=\frac{1}{2}(3.6)(5.2)^2=48.7 J

6 0
3 years ago
When a person wearing a helmet rides a bicycle, the helmet experiences which type of friction?
vodomira [7]

Answer:

B

Explanation:

3 0
3 years ago
Read 2 more answers
A projectile is fired with an initial speed of 65.2 m/s at an angle of 34.5º above the horizontal on a long flat firing range. (
Margarita [4]

Answer:

A) h = 69.58 m

D) v = 58.12 \frac{m}{s}  (Speed magnitude)

α = 22.49° (Speed direction above the horizontal)

Explanation:

Conceptual analysis:

To solve this problem we consider the following concepts:

1) The projectile in its movement describes a curved line called a parabola, therefore two coordinates are required to fix the position at each instant of time, since the movement is performed in the X-Y plane.

The initial velocity (Vo) is tangent to the trajectory at the initial point and can be broken down into two components, one vertical (Voy) and one horizontal(Vox):

V_{ox} =  V_{o}Cos\alpha _{o} Formula (1)

V_{oy} =  V_{o}Sin\alpha _{o} Formula (2)

Where:

Vo: Initial velocity in m/s

\alpha_{o}: Initial angle above the horizontal in grades

2) The formula to calculate its velocity at any vertical position(y) is as follows:

v_{y}^{2} = v_{oy}^{2} -2gy Formula (3)

Where:

v_{f}^{2}: Final speed component in vertical direction in m/s

v_{oy}^{2}: Initial speed component in vertical direction in m/s

g: acceleration due to gravity in m/s2

3) The formulas to calculate the projectile velocity components at any time (t) are:

v_{x} = v_{ox} Because the movement is uniform in the x direction (constant speed)

v_{y} = v_{oy}-g*t Formula (4)  Because the movement is uniformly accelerated in the y direction

Known information:

We know the following data:

v_{o} = 65.2 \frac{m}{s}

\alpha _{o} = 34.5º above the horizontal

g=9.8 \frac{m}{s^{2}}

Development of the problem:

Initial speed components(Vox, Voy), (Formula (1), Formula (2)

v_{ox} =  65.2*Cos34.5 = 53.7\frac{m}{s}

v_{oy} =  65.2*Sin34.5 = 36.93\frac{m}{s}

A) Maximum height (h):

When the projectile reaches its maximum height (h) ,the speed component Vy = 0, then, we replace this value in the Formula (3):

0=(36.93)^2-2*9.8*h

h=\frac{ (36.93)^2}{2*9.8} = 69.58 m

D) Speed of the projectile 1.50s after firing

We replace t=1.5 s in the formula(4)

v_{y} = 36.93-9.8*1.5 = 22.23 \frac{m}{s}

v_{x} = v_{ox} = 53.7 \frac{m}{s}

v = \sqrt{53.7^{2}+22.23^{2}} = 58.12 \frac{m}{s}  (Speed magnitude)

\alpha = tan^{-1} (\frac{v_{y}}{v_{x}}) =  tan^{-1} (\frac{22.23}{53.7})

α = 22.49° (Speed direction above the horizontal)

6 0
3 years ago
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