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Andrej [43]
2 years ago
8

A(n) __________ refers to a substance or treatment that is often used in an experiment but does not have an actual medicinal or

experimental benefit. A. Extraneous variable B. Placebo C. Confounding variable D. Independent control Please select the best answer from the choices provided A B C D.
Physics
2 answers:
dimaraw [331]2 years ago
6 0

Answer:

A __placebo (B)___ refers to a substance or treatment that is often used in an experiment but does not have an actual medicinal or experimental benefit.

Explanation:

Got it right on EDGE2022

kipiarov [429]2 years ago
3 0

Answer:

A placebo refers to a substance or treatment that is often used in an experiment but does not have an actual medicinal or experimental benefit.

Explanation:

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Balance this chemical equation Fe(s) + 02(g) --- Fe0(s)
Softa [21]
2Fe(s) + O2 -> 2FeO(s) 

<span>2 'Fe' atoms on both sides </span>
<span>2 'O' atoms on both sides</span>
6 0
3 years ago
How much work, in N*m, is done when a 10.0 N force moves an object 2.5 m?
Alik [6]
W = F * d
W = 10N * 2.5 m
W = 25 N m
So the answer you want is the third one down.
8 0
3 years ago
Protons, neutrons, electrons, and a nucleus are
tatuchka [14]
It would be Atoms, they’re all made up of these tiny particles
8 0
3 years ago
The spring of a spring gun has force constant k = 400 N/m and negligible mass. The spring is compressed 6.00 cm and a ball with
nikdorinn [45]

Answer:

A) v = 6.93 m/s

B) v = 4.9 m/s

C) x_m = 0.015m

D) v_max = 5.2 m/s

Explanation:

We are given;

x = 6 cm = 0.06 m

k = 400 N

m = 0.03 kg

F = 6N

A) from work energy law, work dome by the spring on ball which now became a kinetic energy is;

Ws = K.E = ½kx²

Similarly, kinetic energy of ball is;

K.E = ½mv²

So, equating both equations, we have;

½kx² = ½mv²

Making v the subject gives;

v = √(kx²/m)

Plugging in the relevant values to give;

v = √((400 × 0.06²)/0.03)

v = √48

v = 6.93 m/s

B) If there is friction, the total work is;

Ws = ½kx² - - - (1)

Work of the ball is;

Wb = KE + Wf

So, Wb = ½mv² + fx - - - (2)

Combining both equations, we have;

½mv² + fx = ½kx²

Plugging in the relevant values, we have;

(½ × 0.03 × v²) + (6 × 0.06) = ½ × 400 × 0.06²

0.015v² + 0.36 = 0.72

0.015v² = 0.72 - 0.36

v² = 0.36/0.015

v = √24

v = 4.9 m/s

C) The speed is greatest where the acceleration stops i.e. where the net force on the ball is zero. (ie spring force matches 6.0N friction)

So, from F = Kx;

(x is measured into barrel from end where F = 0)

Thus; 6.0 = 400x

x_m = 6/400

x_m = 0.015m from the end after traveling 0.045m

D) Initial force on ball = (Kx - F) =

[(400 x 0.06) - 6.0] = 18N

Final force on ball = 0N

Mean Net force on ball = ½(18 + 0)

Mean met force, F_m = 9N

Net Work Done on ball = KE = 9N x 0.045m = 0.405 J

Thus;

½m(v_max)² = 0.405J

(v_max)² = 2 x 0.405/0.03

(v_max)² = 27

v(max) = √27

v_max = 5.2 m/s

6 0
2 years ago
What is the period of a wave with a frequency of 100 Hz and a wavelength of 2.0 m
spin [16.1K]

The answer for the following answer is answered below.

  • <u><em>Therefore the time period of the wave is 0.01 seconds.</em></u>
  • <u><em>Therefore the option for the answer is "B".</em></u>

Explanation:

Frequency (f):

The number of  waves that pass a fixed place in a given amount of time.

The SI unit of frequency is Hertz (Hz)

Time period (T):

The time taken for one complete cycle of vibration to pass a given point.

The SI unit of time period is seconds (s)

Given:

frequency (f) = 100 Hz

wavelength (λ) = 2.0 m

To calculate:

Time period (T)

We know;

According to the formula;

<u>f =</u>\frac{1}{T}<u></u>

Where,

f represents the frequency

T represents the time period

from the formula;

  T = \frac{1}{f}

 T = \frac{1}{100}

  T = 0.01 seconds

<u><em>Therefore the time period of the wave is 0.01 seconds.</em></u>

5 0
2 years ago
Read 2 more answers
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