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Dvinal [7]
3 years ago
9

The manipulated variable is plotted on the horizontal axis

Physics
1 answer:
tatuchka [14]3 years ago
8 0
<span>True.

Manipulated variable or also called the controlled variables are variables in which you regulate. Manipulate or as said control. By this you want to certain the outcome of a certain experiment. Making it close to which you want it or desired it to be, possibly. Characteristics could be: </span>
<span><span>1.       </span>You are able to have govern or a certain variable is controllable.</span> <span><span>
2.       </span>The outcome or effect of a particular variable was what you were determining or were testing with high probability</span> 
<span><span>3.       </span>You can either increase or decrease the level of which the variable could take effect on a dependent variable.<span>
</span></span>
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9. If the frequency of a certain light is 3.8 x 1024 Hz, what is the energy of this light?
german

Answer:

E=hf

Were, h = Planck constant 6.67*10^11

E=3.8*10^24 * 6.67*10^11= 2.508*q0^36j

4 0
2 years ago
What are examples of convection currents?
hram777 [196]

Answer:

I would say all of the above.

Explanation:

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5 0
3 years ago
If the frequencies of two component waves are 24 Hz and 20 Hz, they should produce _______ beats per second.
horrorfan [7]
This can be answered using the beat frequency formula, which is simply the difference between 2 frequencies.

Let: <span>fᵇ = beat frequency
</span>f₁ = first frequency
f₂ = second frequency

fᵇ = |f₁ - f₂|

substituting the values:
fᵇ = |24Hz - 20Hz|
fᵇ = 4Hz

The unit Hz also means beats per second, therefore:
<span>fᵇ = 4 beats per second
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Therefore, the answer is C. 4
8 0
3 years ago
Read 2 more answers
A book sits on a bookshelf without moving until a student picks it up. Which law best explains why the book remains at rest unti
svetoff [14.1K]

Answer:

Newtons first law

Explanation:

object in rest stays at rest

object in motion stays in motion

8 0
3 years ago
A hunter aims at a deer which is 40 yards away. Her cross- bow is at a height of 5ft, and she aims for a spot on the deer 4ft ab
shutvik [7]

Answer:

a)  θ₁ = 0.487º , b)   t = 0.400 s ,        x = 11.73 ft

Explanation:

For this exercise let's use the projectile launch relationships.

The initial height is I = 5 ft and the final height y = 4 ft

            y = y₀ + v_{oy} t - ½ g t²

The distance to the band is x = 40 yard (3 ft / 1 yard) = 120 ft

            x = v₀ₓ t

            t = x / v₀ₓ

We replace

             y –y₀ = v_{oy} x / v₀ₓ - ½ g x² / v₀ₓ²

             v_{oy} = v₀ sin θ

             v₀ₓ = vo cos θ

             

             y –y₀ = x tan θ - ½ g x² / v₀² cos² θ

                5-4 = 120 tan θ - ½ 32 120 / (300 2 cos2 θ)

                1 = 120 tan θ - 0.0213 sec² θ

Let's use the trigonometry relationship

               Sec² θ = 1 - tan² θ

                 1 = 120 tan θ - 0.0213 (1 –tan²θ)

                 0.0213 tan²θ + 120 tanθ -1.0213 = 0

                 

We change variables

          u = tan θ

          u² + 5633.8 u - 48.03 = 0

We solve the second degree equation

          u = [-5633.8 ±√(5633.8 2 + 4 48.03)] / 2

          u = [- 5633.8 ± 5633.82] / 2

           u₁ = 0.0085

           u₂= -5633.81

           u = tan θ

           θ = tan⁻¹ u

For u₁

           θ₁ = tan⁻¹ 0.0085

           θ₁ = 0.487º

For u₂

           θ₂ = -89.99º

The launch angle must be 0.487º

b) let's look for the time it takes for the arrow to arrive

         x = v₀ₓ t

         t = x / v₀ cos θ

         

         t = 120 / (300 cos 0.487)

         t = 0.400 s

The deer must be at a distance of

           v = 20 mph (5280 ft / 1 mi) (1 h / 3600s) = 29.33 ft / s

           x = v t

           x = 29.33 0.4

           x = 11.73 ft

3 0
3 years ago
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