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3 years ago

Which statement describes a controlled experiment?

1 answer:

8 0

Answer:The answers is c, a group exposed to the factor being tested, and a group that isn’t exposed to the factor tested.

Explanation:

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A man pushed the box with a force of 20 N. He doubled his force to 40 N. The box did not

Alex787 [66]

Answer:

Explanation:

if the man doubles his force to 40 and the box was yet to move that means acceleration also doubled so your answer would be A

8 0

2 years ago

How are dwarf planets different from true planets?

morpeh [17]

Answer:

The only difference between a planet and a dwarf planet is the area surrounding each celestial body. A dwarf planet has not cleared the area around its orbit, while a planet has.

Explanation:

the three criteria of the IAU for a full-sized planet are: It is in orbit around the Sun. It has sufficient mass to assume hydrostatic equilibrium (a nearly round shape). It has "cleared the neighborhood" around its orbit .

3 0

2 years ago

Diagnostic ultrasound of frequency 3.82 MHz is used to examine tumors in soft tissue. (a) What is the wavelength in air of such

maxonik [38]

Explanation:

It is given that,

Frequency of diagnostic ultrasound, f = 3.82 MHz = 3820 Hz

The speed of the sound in air, v = 343 m/s

(a) We need to find the wavelength in air of such a sound wave. Let it is given by λ₁

i.e. $\lambda=\dfrac{v}{\nu}$

$\lambda_1=\dfrac{343\ m/s}{3820\ Hz}$

$\lambda_1=0.089\ m$

(b) If the speed of sound in tissue is 1650 m/s .

$\lambda_2=\dfrac{v}{\nu}$

$\lambda_2=\dfrac{1650\ m/s}{3820\ Hz}$

$\lambda_2=0.43\ m$

Hence, this is the required solution.

7 0

2 years ago

Ray Of Light [21]

Acceleration = Force \ mass

0,375N/0,60kg=0.6ms-2

4 0

1 year ago

New 5G networks utilize millimeter-wave radiation. Millimeter-wave radiation refers to electromagnetic waves with frequencies in

seraphim [82]

Answer:

It corresponds to 1mm-10 mm range.

Explanation:

Electromagnetic waves (such as the millimeter-wave radiation) travel at the speed of light, which is 3*10⁸ m/s in free space.
As in any wave, there exists a fixed relationship between speed, frequency and wavelength, as follows:

$v = \lambda * f (1)$

Replacing v= c=3*10⁸ m/s, and the extreme values of f (which are givens), in (1) and solving for λ, we can get the free-space wavelengths that correspond to the 30-300 GHz range, as follows:

$\lambda_{low} = \frac{c}{f_{high}} = \frac{3e8m/s}{300e9Hz} = 1 mm (2)$

$\lambda_{high} = \frac{c}{f_{low}} = \frac{3e8m/s}{30e9Hz} = 10 mm (3)$

4 0

2 years ago