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marusya05 [52]
2 years ago
11

After observing a moth that is camouflaged against dark-colored bark, a scientist asks a question. The scientist discovers that

the question has already been asked and answered by several investigations. What should the scientist do?The scientist should test the question anyway.
The scientist should broaden the scope of the question.
The scientist should come up with a new way to test the question.
The scientist should revise the question to address a gap in knowledge.
Physics
2 answers:
Rasek [7]2 years ago
6 0
Because of how it's worded the answer would most likely be number four                                                                                       

liq [111]2 years ago
6 0

Answer: The scientist should revise the question to address a gap in knowledge.

Explanation:

A research question is a scientific inquiry that is required to be answered using suitable methodology.

Among the options given, The scientist should revise the question to address a gap in knowledge. is the correct option. This is because of the fact that analyzing and finding the possibilities in the question again the scientist can add new facts in the question which have not been found out by the research conducted by the several scientists earlier. This will lead to the increment in the knowledge related to the research topic.

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For a machine with 35-cm -diameter wheels, what rotational frequency (in rpm) do the wheels need to pitch a 85 mph fastball?
Inessa05 [86]

Answer:

The rotational frequency must be 2073.56 rpm

Explanation:

Notice that we need to obtain a rotational frequency in "rpm" (revolutions per minute), so we better start by converting all the given information into the appropriate units:

The magnitude of the velocity for the pitch is given in miles per hour, while the diameter of the machine's wheels is given in cm. Let's reduce all units of length into meters(using the metric system), and the units of time into minutes.

Conversion of the 85 mph  speed into meters per minute:

Recall that 1 mile equals 1609.34 meters, and that 1 hour equals 60 minutes, so we write:

85\,\frac{miles}{hour} = 85\,\frac{1609.34\,m}{60\,min} =2279.898\,\frac{m}{min}

which can be rounded to approximately 2280 m/min.

We also convert the 35 cm diameter into meters:

diameter = 0.35 m

Now we use the equation that relates angular velocity (w) and the radius (R) of the circular movement, with tangential velocity (v_t), in order to obtain the angular velocity of the wheel:

v_t=w*R\\w=\frac{v_t}{R}

but recall that this angular velocity is given in radians per unit of time. So first find the radius of the wheel (half its diameter). R = 0.175 m

So we have:

w=\frac{2280}{0.175}\frac{radians}{min} \\w=13028.57\,\frac{radians}{min}

And now, recalling that 2\pi radians equal one revolution, we convert the angular velocity ot revolutions per minute by dividing the "w" we found by 2\pi :

rotational frequency = \frac{13028.57}{2\pi} \frac{rev}{min} = 2073.56 \frac{rev}{min}

6 0
2 years ago
A very long, straight wire has charge per unit length 3.50×10^−10 C/m . At what distance from the wire is the electricfield magn
Dafna11 [192]

Answer:

r= 2.17 m

Explanation:

Conceptual Analysis:

The electric field at a distance r from a charge line of infinite length and constant charge per unit length is calculated as follows:

E= 2k*(λ/r) Formula (1)

Where:

E: electric field .( N/C)

k: Coulomb electric constant. (N*m²/C²)

λ: linear charge density. (C/m)

r : distance from the charge line to the surface where E calculates (m)

Known data

E= 2.9  N/C

λ = 3.5*10⁻¹⁰ C/m

k= 8.99 *10⁹ N*m²/C²

Problem development

We replace data in the formula (1):

E= 2*k*(λ/r)

2.9= 2*8.99 *10⁹*(3.5*10⁻¹⁰/r)

r =( 2*8.99 *10⁹*3.5*10⁻¹⁰) / (2.9)

r= 2.17 m

5 0
2 years ago
You have a resistor and a capacitor of unknown values. First, you charge the capacitor and discharge it through the resistor. By
Fittoniya [83]

Answer:

The frequency is    f  = 0.221 \ Hz

Explanation:

From the question we are told that  

     The  time taken for it to decay to half its original size is t  =  3.40 \ ms  =  3.40 *10^{-3} \ s

Let the voltage of the capacitor when it is fully charged be  V_o

Then the voltage of the capacitor at time t is  said to be  V  =  \frac{V_o}{2}

   Now  this voltage can be  mathematical represented as

      V  =  V_o  * e ^{-\frac{t}{RC} }

Where  RC  is the time constant

   substituting values  

    \frac{V_o}{2}  =  V_o  *  e ^{-\frac{3.40 *10^{-3}}{RC} }

    0.5  =  e^{-\frac{3.40 *10^{-3}}{RC} }

    - \frac{0.5}{RC}  =  ln (0.5)

     -\frac{0.5}{RC} =  -0.6931

     RC  =  0.721

Generally the cross-over frequency for a low pass filter is mathematically represented as

          f  = \frac{1}{2 \pi  * RC  }

substituting values  

           f  = \frac{1}{2*  3.142  * 0.72  }

           f  = 0.221 \ Hz

7 0
1 year ago
When a car is 100 meters from its starting position traveling at 60.0 m/s., it starts braking and comes to a stop 350 meters fro
NISA [10]
Remember your kinematic equations for constant acceleration. One of the equations is x_{f} =  x_{i} +  v_{i}(t) + \frac{1}{2} at^{2}, where x_{f} = final position, x_{i} = initial position, v_{i} = initial velocity, t = time, and a = acceleration. 

Your initial position is where you initially were before you braked. That means x_{i} = 100m. You final position is where you ended up after t seconds passed, so x_{f} = 350m. The time it took you to go from 100m to 350m was t = 8.3s. You initial velocity at the initial position before you braked was v_{i} = 60.0 m/s. Knowing these values, plug them into the equation and solve for a, your acceleration:
350\:m = 100\:m + (60.0\:m/s)(8.3\:s) + \frac{1}{2} a(8.3\:s)^{2}\\
250\:m = (60.0\:m/s)(8.3\:s) + \frac{1}{2} a(8.3\:s)^{2}\\
250\:m = 498\:m +34.445\:s^{2}(a)\\
-248\:m = 34.445\:s^{2}(a)\\
a \approx -7.2 \: m/s^{2}

Your acceleration is approximately -7.2 \: m/s^{2}.
4 0
2 years ago
A ball was kicked upward at a speed of 64.2 m/s. how fast was the ball going 1.5 seconds later
UNO [17]

Anything that's not supported and doesn't hit anything, and
doesn't have any air resistance, gains 9.8 m/s of downward
speed every second, on account of gravity.  If it happens to
be moving up, then it loses 9.8 m/s of its upward speed every
second, on account of gravity.

                (64.2 m/s)  -  [ (9.8 m/s² ) x (1.5 sec) ] 

            =  (64.2 m/s)  -       [      14.7 m/s      ]

            =             49.5 m/s  .  (upward)

7 0
2 years ago
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