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3 years ago

In an experiment, the independent variable __________.

Physics

2 answers:

alexira [117]3 years ago

7 0

Answer:The correct answer is option D.

Explanation:

Independent variables are those variables which do not dependent on any other variable.These are those variables which can be controlled or varied by researchers to test their effect on the dependent variables.

Dependent variables are those variables which depends upon independent and other dependent variables.

Hence, from the given the option the correct answer is option D.

joja [24]3 years ago

4 0

D) is manipulated or varied

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An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.300 rev/s . The magnitude

Salsk061 [2.6K]

1) 1.2 m/s

First of all, we need to find the angular velocity of the blade at time t = 0.200 s. This is given by

$\omega_f = \omega_i + \alpha t$

where

$\omega_i = 0.300 rev/s$ is the initial angular velocity

$\alpha = 0.895 rev/s^2$ is the angular acceleration

Substituting t = 0.200 s, we find

$\omega_f = 0.300 + (0.895)(0.200)=0.479 rev/s$

Let's now convert it into rad/s:

$\omega_f = 2\pi \cdot 0.479 rev/s=3.01 rad/s$

The distance of a point on the tip of the blade is equal to the radius of the blade, so half the diameter:

$r=\frac{0.800}{2}=0.400 m$

And so now we can find the tangential speed at t = 0.200 s:

$v=\omega_f r =(3.01)(0.400)=1.2 m/s$

2) $2.25 m/s^2$

The tangential acceleration of a point rotating at a distance r from the centre of the circle is

$a_t = \alpha r$

where $\alpha$ is the angular acceleration.

First of all, we need to convert the angular acceleration into rad/s^2:

$\alpha = 0.895 rev/s^ \cdot 2 \pi =5.62 rad/s^2$

A point on the tip of the blade has a distance of

r = 0.400 m

From the centre; so, the tangential acceleration is

$a_t = (5.62)(0.400)=2.25 m/s^2$

3) $3.6 m/s^2$

The centripetal acceleration is given by

$a=\frac{v^2}{r}$

where

v is the tangential speed

r is the distance from the centre of the circle

We already calculate the tangential speed at point a):

v = 1.2 m/s

while the distance of a point at the end of the blade from the centre is

r = 0.400 m

Therefore, the centripetal acceleration is

$a=\frac{1.2^2}{0.400}=3.6 m/s^2$

7 0

3 years ago

Which of the following is/are ALWAYS true concerning collision and transition state theory?

Scrat [10]

Answer:

1) Transition states are short-lived

Explanation:

Transition state theory explains the rates of elementary chemical reactions. It assumes a quasi-equilibrium between reactants and activated transition state complexes.

The following are the characteristics of transition states

Instability
Ill-defined
High energy
short-lived

The species that must collide for the reaction to occur are shown by the mechanism of reaction and not the balanced reaction itself

Intermediates are consumed in each step of the overall reaction, they are not short lived

6 0

3 years ago

Give an example of a normal force. type below.

Oksi-84 [34.3K]

Normal force is the force exerted when an object is on an surface. So an example could be a pile of books on top of a table.

8 0

3 years ago

Water enters a shower head

zalisa [80]

Answer:

13.1

Explanation:

thats what i put in for acellus and its right

4 0

3 years ago

A mass of 1 slug, when attached to a spring, stretches it 2 feet and then comes to rest in the equilibrium position. Starting at

Vesna [10]

Answer:

$Y=(\dfrac{3}{16}+t \dfrac{3}{8})e^{-2t}-\dfrac{3}{16}cos 4t$

Explanation:

Given that m= 1 slug and given that spring stretches by 2 feet so we can find the spring constant K

mg=k x

1 x 32= k x 2

K=16

And also give that damping force is 8 times the velocity so damping constant C=8.

We know that equation for spring mass system

my''+Cy'+Ky=F

Now by putting the values

1 y"+8 y'+ 16y=6 cos 4 t ----(1)

The general solution of equation Y=CF+IP

Lets assume that at steady state the equation of y will be

y(IP)=A cos 4t+ B sin 4t

To find the constant A and B we have to compare this equation with equation 1.

Now find y' and y" (by differentiate with respect to t)

y'= -4A sin 4t+4B cos 4t

y"=-16A cos 4t-16B sin 4t

Now put the values of y" , y' and y in equation 1

1 (-16A cos 4t-16B sin 4t)+8( -4A sin 4t+4B cos 4t)+16(A cos 4t+ B sin 4t)=6sin4 t

So by comparing the coefficient both sides

-16A+32B+16A=0 So B=0

-16 B-32 A+16B=6 So A=-3/16

y=-3/16 cos 4t

Now to find the CF of differential equation 1

y"+8 y'+ 16y=6 cos 4 t

Homogeneous version of above equation

$m^2+8m+16=0$

So $CF =(C_1+tC_2)e^{-2t}$

So the general equation

$Y=(C_1+tC_2)e^{-2t}-3/16 cos 4t$

Given that t=0 Y=0 So

$C_1=\dfrac{3}{16}$

t=0 Y'=0 So

$C_2 =\dfrac{3}{8}$

$Y=(\dfrac{3}{16}+t \dfrac{3}{8})e^{-2t}-\dfrac{3}{16}cos 4t$

The above equation is the general equation for motion.

3 0

3 years ago