Which city did King David make the royal capital of the Kingdom of Israel?
Canaan
Jerusalem
Philistine
Cairo

Answers

Answer 1

Answer:

Jerusalem

Explanation:

because now a days jerusalem is the royal capital of the kingdom of israel so i conclude that the answer is jerusalem


Related Questions

Acceleration is sometimes expressed in multiples of g, where g = 9.8 m/s^2 is the magnitude of the acceleration due to the earth's gravity. In a test crash, a car's velocity goes from 26 m/s to 0 m/s in 0.15 s. How many g's would be experienced by a driver under the same conditions?

Answers

Answer:

Acceleration = 18g

Explanation:

Given the following data;

Initial velocity, u = 26m/s

Final velocity, v = 0

Time = 0.15 secs

To find the acceleration;

In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.

This simply means that, acceleration is given by the subtraction of initial velocity from the final velocity all over time.

Hence, if we subtract the initial velocity from the final velocity and divide that by the time, we can calculate an object’s acceleration.

Mathematically, acceleration is given by the equation;

[tex]Acceleration (a) = \frac{final \; velocity - initial \; velocity}{time}[/tex]

Substituting into the equation, we have;

[tex]a = \frac{0 - 26}{0.15}[/tex]

[tex]a = \frac{26}{0.15}[/tex]

Acceleration = 173.33m/s2

To express it in magnitude of g;

Acceleration = 173.33/9.8

Acceleration = 17.7 ≈ 18g

Acceleration = 18g

A car moves forward up a hill at 12 m/s with a uniform backward acceleration of 1.6 m/s2. What is its displacement after 6 s?

Answers

Answer:

The displacement of the car after 6s is 43.2 m

Explanation:

Given;

velocity of the car, v = 12 m/s

acceleration of the car, a = -1.6 m/s² (backward acceleration)

time of motion, t = 6 s

The displacement of the car after 6s is given by the following kinematic equation;

d = ut + ¹/₂at²

d = (12 x 6) + ¹/₂(-1.6)(6)²

d = 72 - 28.8

d = 43.2 m

Therefore, the displacement of the car after 6s is 43.2 m

An object with a mass of 3.0 kg has a
force of 9.0 newtons applied to it. What
is the resulting acceleration of the
object?

Answers

[tex] \LARGE{ \underline{ \tt{Required \: answer:}}}[/tex]

We have:

Mass of the object = 3 kgForce on the object = 9 N

We need to find:

Resulting accleration of the object?

Solution:

According to Newton's 2nd law of motion, or quantitative measure of Force:

Force = Mass × Accleration

Using this,

➝ F = ma

➝ 9N = 3 kg × a

➝ a = 9/3 m/s²

➝ a = 3 m/s²

Hence,

The resulting accleration of the object is 3 m/s². And we are done! :D

⛱️ [tex] \large{ \blue{ \bf{FadedElla}}}[/tex]

If mass (3.0 kg) multiplying (*) acceleration gives you force (newtons), force dividing mass gives you acceleration; 9/3 =


3 m/s^2

A rolling ball moves from x1 = 8.0 cm to x2 = -4.1 cm during the time from t1 = 2.9 s to t2 = 6.0 s .

Answers

Complete Question

A rolling ball moves from [tex]x_1 = 8.0 \ cm[/tex] to [tex]x_2 = - 4.1 \ cm[/tex] during the time from [tex]t_1 = 2.9 s[/tex]  to  [tex]t_2 = 6.0s[/tex]

What is its average velocity over this time interval?

Answer:

The velocity is  [tex]v = 3.903 \ m/s[/tex]

Explanation:

From the question we are told that

    The first position of the ball is  [tex]x_1 = 8.0 \ cm[/tex]

    The second position of the ball is  [tex]x_2 = - 4.1 \ cm[/tex]

Generally the average velocity is mathematically represented as

       [tex]v = \frac{ x_1 - x_2}{t_2 - t_1}[/tex]

=>    [tex]v = \frac{ 8 - -4.1 }{ 6 - 2.9 }[/tex]

=>    [tex]v = 3.903 \ m/s[/tex]

A woman standing before a cliff claps her hands, and 2.8s later she hears the echo. How far away is the cliff? The speed of sound in air a ordinary temperature is 343 m/s.

Answers

Answer:

480.2 m

Explanation:

The following data were obtained from the question:

Speed of sound (v) = 343 m/s.

Time (t) = 2.8 s

Distance (x) of the cliff =?

The distance of the cliff from the woman can be obtained as follow:

v = 2x /t

343 = 2x /2.8

Cross multiply

2x = 343 × 2.8

2x = 960.4

Divide both side by the coefficient of x i.e 2

x = 960.4/2

x = 480.2 m

Therefore, the cliff is 480.2 m away from the woman.

The distance should be 480.2 m

The calculation is as follows:

Since A woman standing before a cliff claps her hands, and 2.8s later she hears the echo. And, there is the velocity of 343 m/s

[tex]v = 2x \div t\\\\343 = 2x \div 2.8\\\\2x = 343 \times 2.8[/tex]

2x = 960.4

x = 480.2 m

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Answer as soon as possible

Answers

Answer:

the velocity of the acorn

Explanation:

just do in in real life and see

Answer:

it is probably the velocity of the acorn

If the particles were moving with a speed much less than c, the magnitude of the momentum of the second particle would be twice that of the first. However, what is the ratio of the magnitudes of momentum for these relativistic particles?

Answers

Answer:

p₂ / p₁ = 2 (v₁ / v₂)

Explanation:

The moment is a very useful concept, since it is one of the quantities that is conserved during shocks and explosions, for which it had to be redefined to be consistent with special relativity,

         p = m v / √[1+ (v/c)² ]

for the case of speeds much lower than the speed of light this expression is close to

         p = m v

 

In this exercise they indicate that the moment of the second particle is twice the moment of the first, when their velocities are small

        p₂ = 2 p₁

       p₂/p₁ = 2

in consecuense

       m v₂ = 2 m v₁

       v₂ = 2 v₁

consider particles of equal mass.

By the time their speeds increase they enter the relativistic regime

        p₂ = mv₂ /√(1 + v₂² /c²)

        p₁ = m v₁ /√(1 + v₁² / c²)

let's look for the relationship between these two moments

       p₂ / p₁ = mv₂ / mv₁   [√ (1+ v₁² / c²) /√ (1 + v₂² / c²)

       

from the initial statement

      p₂ / p₁ = 2 √(c² + v₁²) / (c² + v₂²)

we take c from the root

      p₂ / p₁ = 2 √ [(1+ v₁²) / (1 + v₂²)]

this is the exact result, to have an approximate shape suppose that the velocities are much greater than 1

      p₂ / p₁ = 2 √ [v₁² / v₂²] = 2 √ [(v₁ / v₂)²]

      p₂ / p₁ = 2 (v₁ / v₂)

we see the value of the moment depends on the speed of the particles

True or false. when objects collide , some momentum is lost

Answers

Answer:

It is neither false nor true. When they collide some of one of the objects goes to the other object.

Explanation:

Answer: True

Explanation:

(b) In the USA, drones are not allowed to be flown too high above the ground.

Suggest one possible risk of flying a drone too high above the ground.




Plz complete thank you in advance

Answers

One possible risk is it could effect planes, radars, weather patterns.

The 400-foot altitude limit was put in place for the sake of airspace safety, and there is a risk to country security as well as the privacy of citizens.

What are drones?

Unmanned aerial vehicles (UAVs), sometimes known as drones, are used for a variety of jobs, from routine to extremely dangerous. These robotic-looking planes can be seen practically everywhere, from delivering groceries to your home to rescuing avalanche victims.

The 400-foot altitude restriction was ultimately implemented for airspace safety. Given the breadth of the airspace above 400 feet, the likelihood of a drone colliding with a human aircraft is exceedingly remote, but the consequences might be disastrous.

Any aerial vehicle that uses software to fly autonomously or that may be controlled remotely by a pilot is referred to as a drone. Numerous drones come equipped with cameras to gather visual data and propellers to stabilize flying paths. Drone technology has been incorporated into industries like videography, search and rescue, agriculture, and transportation.

When in uncontrolled (Class G) airspace, your drone must be flown 400 feet above the ground or less.

To know more about drones:

https://brainly.com/question/27753670

#SPJ2

Need help ASAP..please help

Answers

Answer:

option 3

Explanation:

can i get brainliest

A 5.3 kg block rests on a level surface. The coefficient of static friction is μ_s=0.67, and the coefficient of kinetic friction is μ_k= 0.48 A horizontal force, x is applied to the block. As x is increased, the block begins moving. Describe how the force of friction changes as x increases from the moment the block is at rest to when it begins moving. Show how you determined the force of friction at each of these times ― before the block starts moving, at the point it starts moving, and after it is moving. Show your work.

Answers

As the pushing force x increases, it would be opposed by the static frictional force. As x passes a certain threshold and overcomes the maximum static friction, the block will start moving and will require a smaller magnitude x to maintain opposition to the kinetic friction and keep the block moving at a constant speed. If x stays at the magnitude required to overcome static friction, the net force applied to the block will cause it to accelerate in the same direction.

Let w denote the weight of the block, n the magnitude of the normal force, x the magnitude of the pushing force, and f the magnitude of the frictional force.

The block is initially at rest, so the net force on the box in the horizontal and vertical directions is 0:

n + (-w) = 0

n = w = m g = (5.3 kg) (9.80 m/s²) = 51.94 N

The frictional force is proportional to the normal force, so that f = µ n where µ is the coefficient of static or kinetic friction. Before the block starts moving, the maximum static frictional force will be

f = 0.67 (51.94 N) ≈ 35 N

so for 0 < x < 35 N, the block remains at rest and 0 < f < 35 N as well.

The block starts moving as soon as x = 35 N, at which point f = 35 N.

At any point after the block starts moving, we have

f = 0.48 (51.94 N) ≈ 25 N

so that x = 25 N is the required force to keep the block moving at a constant speed.

As x  is increasing it will be opposed by a static frictional force and for the object to start moving and maintain its acceleration, the magnitude of x must exceed the magnitude of the static frictional force and kinetic frictional force

Magnitude of normal force ( object at rest );  n = 51.94 N Required magnitude of x before the movement of object ; x = 35 NMagnitude of x  after object start moving   x = 25 N

Given data :

mass of block at rest ( m ) = 5.3 kg

Coefficient of static friction ( μ_s ) =0.67

Coefficient of kinetic friction is ( μ_k ) = 0.48

Horizontal force applied to block = x  

First step : magnitude of normal force ( n ) when object is at rest

n = w            where w = m*g

n - w = 0

n - ( 5.3 * 9.81 ) = 0     ∴  n = 51.94 N

Second step : Required magnitude of x before the movement of object

F =  μ_s * n

F = 0.67 * 51.94  = 34.79 N  ≈ 35 N

∴ The object will start moving once F and x = 35 N

Final step : Magnitude of x  after object start moving

F = μ_k  * n

  = 0.48 * 51.94 = 24.93 N  ≈ 25 N

∴ object will continue to accelerate at a constant speed once F and x = 25N

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The x component of vector A is -25.0m and the y component id +40.0m (a) what is the magnitude of A?(b) What is the angle between the direction of A and the positive direction of x?

Answers

Answer:

θ = 122°

Explanation:

Components of a Vector

A vector in the plane can be defined by its rectangular components:

[tex]\vec A =<x,y>[/tex]

Or also can be given by its polar components:

[tex]\vec A =<r,\theta>[/tex]

Where r is the magnitude of the vector and θ is the angle it forms with the positive direction of x.

The relation between them is:

[tex]r=\sqrt{x^2+y^2}[/tex]

[tex]\displaystyle \theta=\arctan\frac{y}{x}[/tex]

It's given the x-component of vector A is x=-25 m and the y-component is y=40 m

(a)

The magnitude of the vector is:

[tex]r=\sqrt{(-25)^2+40^2}[/tex]

[tex]r=\sqrt{625+1600}[/tex]

[tex]r=\sqrt{2225}[/tex]

[tex]r\approx 47.2\ m[/tex]

(b)

[tex]\displaystyle \theta=\arctan\frac{40}{-25}[/tex]

[tex]\displaystyle \theta=\arctan (-1.6)[/tex]

The calculator gives us the value

θ = -58°

But the real angle lies on the second quadrant since x is negative and y is positive, thus:

θ = -58° + 180° = 122°

θ = 122°

A plane is flying due west at 34 m/s. It encounters a wind blowing at 19 m/s south. Find the resultant veloci

Answers

Answer:

The resultant velocity has a magnitude of 38.95 m/s

Explanation:

Vector Addition

Given two vectors defined as:

[tex]\vec v_1=(x_1,y_1)[/tex]

[tex]\vec v_2=(x_2,y_2)[/tex]

The sum of the vectors is:

[tex]\vec v=(x_1+x_2,y_1+y_2)[/tex]

The magnitude of a vector can be calculated by

[tex]d=\sqrt{x^2+y^2}[/tex]

Where x and y are the rectangular components of the vector.

We have a plane flying due west at 34 m/s. Its velocity vector is:

[tex]\vec v_1=(-34,0)[/tex]

The wind blows at 19 m/s south, thus:

[tex]\vec v_2=(0,-19)[/tex]

The sum of both velocities gives the resultant velocity:

[tex]\vec v =(-34,-19)[/tex]

The magnitude of this velocity is:

[tex]d=\sqrt{(-34)^2+(-19)^2}[/tex]

[tex]d=\sqrt{1156+361}=\sqrt{1517}[/tex]

d = 38.95 m/s

The resultant velocity has a magnitude of 38.95 m/s

In the winter sport of curling, players give a 20 kg stone a push across a sheet of ice. The Slone moves approximately 40 m before coming to rest. The final position of the stone, in principle, onlyndepends on the initial speed at which it is launched and the force of friction between the ice and the stone, but team members can use brooms to sweep the ice in front of the stone to adjust its speed and trajectory a bit; they must do this without touching the stone. Judicious sweeping can lengthen the travel of the stone by 3 m.1. A curler pushes a stone to a speed of 3.0 m/s over a time of 2.0 s. Ignoring the force of friction, how much force must the curler apply to the stone to bring it op to speed?A. 3.0 NB. 15 NC. 30 N
D. 150 N2The sweepers in a curling competition adjust the trajectory of the slope byA. Decreasing the coefficient of friction between the stone and the ice.
B. Increasing the coefficient of friction between the stone and the ice.C. Changing friction from kinetic to static.D. Changing friction from static to kinetic.3. Suppose the stone is launched with a speed of 3 m/s and travel s 40 m before coming to rest. What is the approximate magnitude of the friction force on the stone?A. 0 NB. 2 NC. 20 ND. 200 N4. Suppose the stone's mass is increased to 40 kg, but it is launched at the same 3 m/s. Which one of the following is true?A. The stone would now travel a longer distance before coming to rest.B. The stone would now travel a shorter distance before coming to rest.C. The coefficient of friction would now be greater.D. The force of friction would now be greater.

Answers

Answer:82. Since you have a distance and a force, then the easiest principle to use is energy, i.e. work.

The work done by friction is F * d. This work cancels out the kinetic energy of the stone (1/2)mv^2

Fd = (1/2)mv^2

F = (1/2)mv^2/d.

Plug in m = 20 kg, v = 3 m/sec, d = 40 m.

83. With more mass, the kinetic energy is higher now. The work needed is higher. W = F * d and F is the same.

Explanation:Hope I helped :)

How much work is done by the gravitational force on the block?

Answers

Answer:

Work = Mass * Gravity * Height and is measured in Joules. Imagine you find a 2 -Kg book on the floor and lift it 0.75 meters and put it on a table. Remember, that “force” is simply a push or a pull. If you lift 100 kg of mass 1-meter, you will have done 980 Joules of work.

Explanation:

A racecar accelerates from rest at 6.5 m/s2 for 4.1 s. How fast will it be going at the end of that time?

Answers

Answer:

The final velocity of the car is 26.65 m/s.

Explanation:

Given;

acceleration of the racecar, a = 6.5 m/s²

initial velocity of the car, u = 0

time of motion, t = 4.1 s

The final velocity of the car is given by;

v = u + at

where;

v is the final velocity of the car

suvstitute the givens

v = 0 + (6.5)(4.1)

v = 26.65 m/s.

Therefore, the final velocity of the car is 26.65 m/s.

What is the force of gravity for a 12 kg turkey?

Please help asap

Answers

Answer: 117.6N

Explanation:

By the second Newton's law, we know that:

F = m*a

F = force

m = mass

a = acceleration

We know that in the surface of the Earth, the gravitational acceleration is g = 9.8m/s^2.

Then we just can input that acceleration in the above equation, and also replace m by 12kg, and find that the force due the gravity is:

F = 12kg*9.8m/s^2 = 117.6N

When particles get close to the surface, they interact with atoms in
the
(Finish the sentence)

Answers

Is there anything else in the page I think it’s missing a part

In principle, when you fire a rifle, the recoil should push you backward. How big a push will it give? Let's find out by doing a calculation in a very artificial situation. Suppose a man standing on frictionless ice fires a rifle horizontally. The mass of the man together with the rifle is 70 kg, and the mass of the bullet is 10 g. If the bullet leaves the muzzle at a speed of 500 m/s, what is the final speed of the man?

Answers

Answer:

Explanation:

m1v1=m2v2

m1=70 kg

m2=10 g=0.01 kg

v2=500 m/s

m1v1=m2v2

v1=m2v2/m1

v1=0.01*500/70

v1=0.07

How many significant figures are in 0.0067?

Answers

Answer:

2

Explanation:

there are 2 significant figures in there

What would happen if there is more male hyenas than female hyenas in a population?



Choices:
Male hyenas will compete to mate with the females.

Some male hyenas will die.

Male hyenas for wait for more females to join the population.

Answers

A. Because that’s how the wild works.

Answer:

Option 1

Explanation:

I always see animals do that

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