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sweet [91]
3 years ago
6

When a sacred item or symbol is removed from its special place or is duplicated in mass quantities, then it becomes profane as a

result of ________.
Physics
1 answer:
Galina-37 [17]3 years ago
7 0

Answer:

It becomes profane as a result of desacralization

Explanation:

Because desacralization means when a dedicated religious structure is no longer used for its intended purpose both rather used for another purpose other than the original purpose

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A quantity of gasoline injected into car engine has a potential energy of 100J. All of this energy is not converted into useable
neonofarm [45]

Answer:because some energy are lost due to friction

Explanation:some energy are lost due to friction

5 0
3 years ago
A fighter plane is flying overhead at mach 1.20. What angle does the wave front of the shock wave produced make relative to the
Sever21 [200]

Answer: 56.44°

Explanation:

<u>Given:</u>

  • Let u represent the current speed of the plane, <u>1.2 Mach</u>

<em>Converting to SI Units (m/s):</em>

= (1.2 mach)(340 ms^-1 / 1 Mach)

u  = 408 m/s

  • Speed of sound in air, v = 340 m/s

<u>Find:</u>

  • Angle the wave front of the shock wave relative to the plane's direction of motion, θ

We have, sinθ = speed of sound / speed of object

               sinθ = v / u

                   θ = sin^-1 (v / u)  

                      = sin^-1 (340 / 408)

                   θ = 56.44°

7 0
2 years ago
Your car burns gasoline as you drive up a large mountain. What energy transformation is the car performing?
Sonja [21]

Your car is performing a transformation of energy of:

Chemical energy to Mechanical energy

The chemical is the gasoline which is then converted to fire as the car runs thus creating the movement of the car which is mechanical energy.

7 0
3 years ago
Read 2 more answers
Two stones are launched from the top of a tall building. One stoneis thrown in a direction 30.0^\circ above the horizontal with
Butoxors [25]

Answer:

Part A)

t(1) > t(2), the stone thrown 30 above the horizontal spends more time in the air.

Part B)

x(f1) > x(f2), the first stone will land farther away from the building.

Explanation:

<u>Part A)</u>

Let's use the parabolic motion equation to solve it. Let's define the variables:

  • y(i) is the initial height, it is a constant.
  • y(f) is the final height, in our case is 0
  • v(i) is the initial velocity (v(i)=16 m/s)
  • θ1 is the first angle, 30°
  • θ2 is the first angle, -30°

For the first stone

y_{f1}=y_{i1}+v*sin(\theta_{1})t_{1}-0.5gt_{1}^{2}              

0=y_{i1}+16*sin(30)t_{1}-0.5*9.81*t_{1}^{2}

0=y_{i1}+8t_{1}-4.905*t_{1}^{2} (1)  

For the second stone  

0=y_{i2}+16*sin(-30)t_{2}-4.905t_{2}^{2}    

0=y_{i2}-8t_{2}-4.905t_{2}^{2} (2)            

 

If we solve the equation (1) we will have:

t_{1}=\frac{-8\pm \sqrt{64+19.62*y_{i}}}{-9.81}  

We can do the same procedure for the equation (2)

t_{1}=\frac{8\pm \sqrt{64+19.62*y_{i}}}{-9.81}

We can analyze each solution to see which one spends more time in the air.

It is easy to see that the value inside the square root of each equation is always greater than 8, assuming that the height of the building is > 0. Now, to get positive values of t(1) and t(2) we need to take the negative option of the square root.

Therefore, t(1) > t(2), it means that the stone thrown 30 above the horizontal spends more time in the air.

<u>Part B)</u>

We can use the equation of the horizontal position here.

<u>First stone</u>

x_{f1}=x_{i1}+vcos(30)t_{1}

x_{f1}=0+13.86*t_{1}

x_{f1}=13.86*t_{1}

<u>Second stone</u>

x_{2}=x_{i2}+vcos(-30)t_{2}

x_{1}=0+13.86*t_{1}

x_{1}=13.86*t_{2}

Knowing that t(1) > t(2) then x(f1) > x(f2)

Therefore, the first stone will land farther away from the building.

They land at different points at different times.

I hope it helps you!

3 0
3 years ago
according to newton's third law, when a horse pulls on a cart, the cart pulls back on the horse with an equal force on the horse
Debora [2.8K]
Based on Newton's principle, whenever objects A and B interact with each other, they exert forces upon each other.

When a horse pulls on a cart, t<span>he horse exerts a force only to the cart. But that force applies only to the cart, not to the horse.
 
The cart in turn exerts a force on the horse. But that force applies only to the horse, not the cart also.
</span>
There are two forces resulting from this interaction - a force on the horse and a force on the cart. T<span>he net force on the cart remains as it was --- a positive force in the direction of the horse's movement. Therefore, the cart begins to accelerate and move.</span><span>


</span>
8 0
3 years ago
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