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The Lambda-CDM model contains a cosmological constant, denoted by a lambda (λ) which is associated with dark energy and ___________.

Answer:

a)  y_total = 575.4 m , b)   t_total = 22.56 s , c)  v = 106.20 m

Explanation:

a) We can solve this exercise using the scientific relations

Let's find the height reached while the engines are running

         v₁ = v₀ + a t

         v₁ = 62.3 + 4.30 3.73

         v₁ = 78.339 m / s

in this time it rose

        y₁ = v₀ t₁ + ½ a t₁²

        y₁ = 62.3 3.73 + ½ 4.30 3.73²

         y₁ = 262.29 m

At this moment the engines are turned off and the rocket continues in a vertical launch, suppose that the acceleration of gravity is constant in this path

        v₂ = v₁ 2 - 2 g y₂

at maximum height the velocity is zero (v₂ = 0)

        y₂ = v₁ 2 / 2g

        y₂ = 78.339 2/2 9.8

        y₂ = 313.11 m

the total height is

       y_total = y₁ + y₂

       y_total = 262.29 + 313.11

       y_total = 575.4 m

b) the time the rocket is in the air is

       t_total = t₁ + t₂

where t₂ is the time after the engines have shut down

       y = y₁ + v₁ t₂ - ½ g t₂²

      y = 0

      0 = 262.29 + 78.339 t₂ - ½ 9.8 t₂²

      t₂² - 15.99 t₂ - 53.53 = 0

      t₂ = [15.99 ±√ (15.99² +4  53.53)] / 2

      t₂ = [15.99 ± 21.67] / 2

      t₂´ = 18.83 s

      t₂´´ = -2.84 s

since the time must be a positive quantity the correct result is t2 = 18.83 s, the total time in the air is

      t_total = 3.73 + 18.83

      t_total = 22.56 s

c) the speed when it hits the ground

We can perform this calculation starting with the maximum height y = 575.4 m where it has zero initial velocity (vo = 0)

      v² = v₀² + 2 g y

      v = √ 2gy

       v = √ (2 9.8 575.4)

       v = 106.20 m

Mass if a substance is the product of its volume and density. The mass of gold of 0.02 m³ with a density of 19300 kg/m³ is 386 kg.

Density of a substance is the measure of its mass per unit volume. Thus, it says how much denser the object is in a given volume. Density of a substance is dependent on its bond type, temperature and pressure beside the volume and mass.

Volume can be defined as the space occupied by the substance. Larger the volume , less dense the substance is. However as the mass increases volume also increases.

Mass of an object is the product of its volume and density.

Given the volume of gold = 0.02 m³

density = 19300 kg/m³.

mass = volume ×  density

         = 0.02 m³ × 19300 kg/m³

         = 386 kg

Therefore, the mass of gold is 386 Kg.

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1. Heat transfer from a hot burner on the stove into a pot or pan

2. A metal spoon that gets hot from the boiling water

3. Chocolate candy in your hand that melts because of the heat

4. The wires in your house that conduct electricity

5. Roasting marshmallows, the heat will soften the marshmallow

Answer:

the engine heat from a car you just turned on.

a heater

a pad warmer

a hot pot

using a campfire to roast hot dogs

Answer:

9 - 10N to the left

10 - There is no change on the object

Explanation:

Can I have brainliest answer pls?

Answer:

3.766 kg or 3.7 kg

Explanation:

F= M*A equation

Answer:

Digestion begins in the mouth, when you chew.

Answer:

reflection

Explanation:

Answer:

reflection

Explanation:

A p e x

Answer:

The shortest possible distance is  [tex]|s| = 91.9 \ m[/tex]

Explanation:

From the question we are told that

   The mass of the car is   [tex]m = 1800 \ kg[/tex]

    The speed along the horizontal road is  [tex]v_o =u = 18.4 \ m/s[/tex]

     The static friction coefficient is  [tex]\mu_s = 0.188[/tex]

      The kinetic friction coefficient is  [tex]\mu_k = 0.1316[/tex]

Generally the static frictional force acting on the car is  mathematically represented as

          [tex]F_f = m * g * \mu_s[/tex]

Generally the force propelling the car is mathematically represented as

        [tex]F = m * a[/tex]

Here a is the maximum acceleration

at the point which the car stops ,

       [tex]F = F_f[/tex]  

=> [tex]m * g * \mu_s = ma[/tex]

=> [tex]g * \mu_s =a[/tex]

=> [tex]a = 9.8 * 0.188[/tex]

=> [tex]a = 1.8424 \ m/s^2[/tex]

Generally from kinematic equation

    [tex]v^2 = u^2 + 2as[/tex]

Here v  is the final velocity of the car and the value is zero given that the car comes to rest

So

        [tex]0^2 = 18.4^2 + 2* 1.8424 s[/tex]

=>     [tex]s = - \frac{18.4^2}{2 * 1.8424}[/tex]

=>     [tex]|s| = 91.9 \ m[/tex]

Answer: the answer is c

Explanation:

Answer:

men

Explanation:

Answer:

Explanation:

The car will fall with acceleration due to gravity which is equal to 9.8 m /s²

For downward fall ,

initial velocity u = o

acceleration due to gravity = g = 9.8 m /s

final velocity v = ?

displacement h = 5.2 m

v² = u² + 2 gh

v² = 0 + 2 x 9.8 x 5.2

= 101.92

v = 10.1 m /s

Answer:

c = 894.90 m/s

Explanation:

Given data:

Frequency of wave = 471 Hz

Wavelength of wave = 1.9 m

Speed of wave = ?

Solution:

Formula:

Speed of wave = frequency × wavelength

c = f×λ

c = 471 Hz × 1.9 m

  Hz = s⁻¹

c = 471s⁻¹ × 1.9 m

c = 894.90 m/s

The speed of wave is 894.90 m/s.

Answer:

c

Explanation:

c. the ability to do work or to produce heat

Answer:

Explanation:

For vertical displacement of food-supply

if time taken to reach the ground be t

235 = 1/2 g t²  , initial velocity downwards is zero.

t = 6.92 s

Horizontal displacement of food during this period

= 6.92 x 69.4 m

= 480.25 m .

b )

time required to cover horizontal distance of 425 m

= 425 / 69.4 = 6.124 s  

This time period will be the time of vertical fall of 235 m . Let initial vertical velocity required be u

h = ut + 1/2 gt²

235 = u x 6.124 + .5 x 9.8 x 6.124²

235 = u x 6.124 + 183.76

u = 8.36 m /s

c )

v = u + gt

= 8.36 + 9.8 x 6.124

= 68.37 m /s

This will be vertical component of velocity .

horizontal velocity = 69.4 m /s

resultant velocity = √ ( 68.37² + 69.4²)

= √(4674.45 +4816.36)

= 97.42 m /s

Answer:

His third law states that for every action (force) in nature there is an equal and opposite reaction. Explanation:

Answer:

His third law states that for every action (force) in nature there is an equal and opposite reaction

Explanation:

In other words, if object A exerts a force on object B, then object B also exerts an equal and opposite force on object A. Notice that the forces are exerted on different objects.

Answer:

The force needed to launch the projectile is 150000 N.

Explanation:  

We can find the force using the following equation:

[tex] F = ma [/tex]

Where:

m: is the mass = 15 kg

a: is the acceleration

First, we need to find the acceleration of the projectile:

[tex] v_{f}^{2} = v_{0}^{2} + 2ax [/tex]                          

Where:

[tex]v_{f}[/tex]: is the final speed

[tex]v_{0}[/tex]: is the initial speed = 0

x: is the distance = 2 m

The final speed is:

[tex]v_{f} = \frac{1 km}{5 s}*\frac{1000 m}{1 km} = 200 m/s[/tex]

Then, the acceleration is:

[tex]a = \frac{v_{f}^{2}}{2x} = \frac{(200 m/s)^{2}}{2*2 m} = 10000 m/s^{2}[/tex]

Finally, the force is:

[tex]F = ma = 15 kg*10000 m/s^{2} = 150000 N[/tex]

Therefore, the force needed to launch the projectile is 150000 N.

I hope it helps you!  

Answer:

I think it's part c . but sorry if its wrong

Explanation:

... in every interaction, there is a pair of forces acting on the two interacting objects. The size of the force on the first object equals the size of the force on the second object. The direction of the force on the first object is opposite to the direction of the force on the second object. Forces always come in pairs - equal and opposite action-reaction force pairs.

Answer:

Its A

Explanation:

To reduce the strength of an electromagnet, Juan must  reduced the number of loops in the wire.

The strength of an electromagnetic is the effect or force that an electromagnet exerts in a given field.

The strength of an induced emf in a given magnetic field is given by;

emf = NBA/t

where;

Decrease in number of turns of the wire, reduces the strength of the magnetic field.

Thus, to reduce the strength of an electromagnet, Juan must  reduced the number of loops in the wire.

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Answer:

Evidence for continental drift

Wegener knew that fossil plants and animals such as mesosaurs, a freshwater reptile found only South America and Africa during the Permian period, could be found on many continents. He also matched up rocks on either side of the Atlantic Ocean like puzzle pieces.

Explanation:

Answer:

Metaphase

Explanation:

During metaphase,chromosome line up in the middle of the cell.

Answer:

I think it's false

Explanation:

clouds also help regulate the Earth's energy balance by reflecting and scattering solar radiation and by absorbing the Earth's infrared radiation.

Answer:

false

Explanation:

Answer:

Acceleration = 3m/s^2

Vf= 0  Vi =35m/s   t= 13s

Explanation:

[tex]Acceleration = \frac{Change in velocity}{Change in time}\\ = \frac{35m/s}{13s}\\ a = 2.69m/s^2\\ a = 2.7m/s^2\\ a = 3m/s^2[/tex]

Answer:

a. endothermic

Explanation:

The reaction shown is an endothermic reaction. In such reaction, they require an input of energy.

Answer:

The potential is  [tex]V_A = 9600 \ V[/tex]

Explanation:

From the question we are told that  

   The  magnitude of the charge is  [tex]q_1 = 4 \mu C = 4*10^{-6} \ C[/tex]

   The position of the charge is  [tex]x = + 3.0 \ m[/tex]

   The magnitude of the second charge is  [tex]q_2 = -2.0 \mu C = -2.0 *10^{-6} \ C[/tex]

   The position is  [tex]y_1 = - 1.0 \ m[/tex]

     The position of point A is  [tex]y_2 = + 4.0 \ m[/tex]  

Generally the electric potential  at A due to the first charge is mathematically represented as

         [tex]V_a = \frac{k * q_1 }{r_1 }[/tex]

Here k is the coulombs constant with value [tex]k = 9*10^{9} \ \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}[/tex]

        [tex]r_1[/tex]  is the distance between first charge and a which is mathematically represented as

         [tex]r_1 = \sqrt{x^2 + y_2 ^2 }[/tex]

=>      [tex]r_1 = \sqrt{3^2 + 4 ^2 }[/tex]    

=>      [tex]r_1 = 5 \ m[/tex]    

So

           [tex]V_a = \frac{9*10^9 * 4*10^{-6} }{5 }[/tex]

      [tex]V_a = 7200 \ V[/tex]

Generally the electric potential  at A due to the second charge is mathematically represented as

         [tex]V_b = \frac{k * q_2 }{r_2 }[/tex]

Here k is the coulombs constant with value [tex]k = 9*10^{9} \ \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}[/tex]

        [tex]r_2[/tex]  is the distance between second charge and a which is mathematically represented as

         [tex]r_2 = y_2 - y[/tex]

=>      [tex]r _2 = 4.0 - (-1.0)[/tex]    

=>      [tex]r = 5 \ m[/tex]    

So

           [tex]V_a = \frac{9*10^9 * -2*10^{-6} }{5 }[/tex]

      [tex]V_a = -3600 \ V[/tex]

So the net potential difference at point A  due to the charges is mathematically represented as

       [tex]V_n = V_a + V_b[/tex]

=>     [tex]V_n = 7200 - 3600[/tex]

=>     [tex]V_n = 3600 V[/tex]

Generally the net potential difference at the origin due to both charges is mathematically represented as

     [tex]V_N = V_c + V_d[/tex]

Here  

      [tex]V_c = \frac{k * q_1 }{x}[/tex]

=>   [tex]V_c = \frac{9*10^9 * 4*10^{-6} }{3}[/tex]

=>   [tex]V_c = 12000 V[/tex]

and

              [tex]V_d= \frac{k * q_2 }{y}[/tex]

=>   [tex]V_c = \frac{9*10^9 * -2*10^{-6} }{1}[/tex]

=>   [tex]V_c =- 18000 V[/tex]

Generally the net potential difference at the origin is  

       [tex]V_N = 12000 - 18000[/tex]

=>     [tex]V_N = -6000[/tex]

Generally the potential difference at A relative to zero at the origin is mathematically evaluated as

         [tex]V_A = V_n - V_N[/tex]

=>      [tex]V_A = 3600 - (-6000)[/tex]

=>      [tex]V_A = 9600 \ V[/tex]