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3 years ago

What is any factor of an experiment that is not changed called? control, dependent variable ,independent variable

Physics

2 answers:

Svetllana [295]3 years ago

8 0

Control is fixed, dependent variable is controlled but changes, the independent is observed

Softa [21]3 years ago

5 0

A control is the factor which is not changed

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How much work (in J) is done against the gravitational force on a 3.9 kg briefcase when it is carried from the ground floor to t

katrin2010 [14]

Answer:

W = 14523.6 J

Explanation:

Given,

Mass = 3.9 Kg

Vertical height , h = 380 m

Work done against gravitational force

W = m g h

g is acceleration due to gravity

W = 3.9 x 9.8 x 380

W = 14523.6 J

Hence, the work done by the gravitational force is equal to W = 14523.6 J

3 0

3 years ago

A sensational 130 kg midfielder has 695 J of

antoniya [11.8K]

Answer:

v = 3.27 m/s

Explanation:

KE = 1/2 mv^2

695 J = 1/2 (130kg)(v^2)

695 J / (1/2 x 130kg) = v^2

v^2 = square root of 10.69

v = 3.27 m/s

4 0

3 years ago

How would you explain force and movement to a friend who doesn’t know much about it?

kirza4 [7]

Answer:

I can explain it using the law of force

or Newton's law of motion

6 0

3 years ago

A high speed train in japan travels a distance of 300 km in 3.60x10^3 seconds . What is the average speed of this train

Bezzdna [24]

83.33km per second.

Speed = distance ÷ time.

Speed = 300 ÷ 3.60000

Speed = 83.33km per second

hope that helps.

8 0

3 years ago

The emission of x rays can be described as an inverse photoelectric effect.

timama [110]

Answer:

The potential difference through which an electron accelerates to produce x rays is $1.24\times 10^4\ volts$ .

Explanation:

It is given that,

Wavelength of the x -rays, $\lambda=0.1\ nm=0.1\times 10^{-9}\ m$

The energy of the x- rays is given by :

$E=\dfrac{hc}{\lambda}$

The energy of an electron in terms of potential difference is given by :

$E=eV$

So,

$\dfrac{hc}{\lambda}=eV$

V is the potential difference

e is the charge on electron

$V=\dfrac{hc}{e\lambda}$

$V=\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{1.6\times 10^{-19}\times 0.1\times 10^{-9}}$

V = 12431.25 volts

$V=1.24\times 10^4\ volts$

So, the potential difference through which an electron accelerates to produce x rays is $1.24\times 10^4\ volts$ . hence, this is the required solution.

4 0

3 years ago