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3 years ago

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What is a controlled experiment? none of the variables are changed all variables are kept constant two variables are changed at

a time one variable is changed at a time

1 answer:

Ostrovityanka [42]3 years ago

7 0

One variable is changed at a time

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A 0.5-kilogram apple falls from a height of 2 meters to 1.50 meters. Ignoring frictional effects, what is the kinetic energy of

spin [16.1K]

The final kinetic energy of the ball is 2.45 J

Explanation:

We can solve this problem by using the law of conservation of energy.

In absence of frictional effect, the mechanical energy of the apple must be conserved during the fall. So we can write:

$U_i +K_i = U_f + K_f$

where :

$U_i$ is the initial potential energy, at the top

$K_i$ is the initial kinetic energy, at the top

$U_f$ is the final potential energy, at the bottom

$K_f$ is the final kinetic energy, at the bottom

By explicing the potential energy, we can rewrite the equation as:

$mgh_i + K_i = mgh_f + K_f$

where:

m = 0.5 kg is the mass of the apple

$g=9.8 m/s^2$ is the acceleration of gravity

$h_i = 2 m$ is the initial height

$h_f=1.50 m$ is the final height

The initial kinetic energy is zero, since the ball starts from rest:

$K_i = 0$

Therefore we can solve the equation for $K_f$ , the final kinetic energy of the ball:

$K_f = mg(h_i-h_f)=(0.5)(9.8)(2-1.50)=2.45 J$

Learn more about kinetic energy and potential energy:

brainly.com/question/6536722

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3 0

3 years ago

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At the county fair, Chris throws a 0.12kg baseball at a 2.4kg wooden milk bottle, hoping to knock it off its stand and win a pri

viva [34]

Answer:

$v_{f2}$ =6.5% $v_{i1}$

Explanation:

Mass of the ball: $m_{1} =0.12kg$ ]

Initial velocity of the ball: $v_{i1}$

final velocity of the ball: $v_{f1}$ which is -30/100 of $v_{i1}$ = $-0.3v_{i1}$

Mass of the bottle: $m_{2} =2.4kg$

Initial velocity of the bottle: $v_{i2}=0m/s$

final velocity of the bottle: $v_{f2}$ is unknown (to find)

<em>by using conservation momentum, which stated that the initial momentum is equal to the final momentum.</em>

<em /> $m_{1} v_{i1} +m_{2} v_{i2} =m_{1} v_{f1} +m_{2} v_{f2}$ <em />

<em>so since the bottle is at rest firstly, therefore </em> $v_{i2} =0$ <em />

<em /> $m_{1} v_{i1} +m_{2} (0) =m_{1} v_{f1} +m_{2} v_{f2}$ <em />

<em /> $m_{1} v_{i1} =m_{1} v_{f1} +m_{2} v_{f2}$ <em> </em><em>equation 1</em>

so now substitute $v_{f1}$ into equation 1

$m_{1} v_{i1} =m_{1} (-0.3v_{i1} ) +m_{2} v_{f2}$

<em /> $m_{1} v_{i1} = -0.3m_{1}v_{i1} +m_{2} v_{f2}$ <em />

<em>collect the like terms</em>

$m_{1} v_{i1} +0.3m_{1}v_{i1} =m_{2} v_{f2}$

$1.3m_{1} v_{i1} =m_{2} v_{f2}$

divide both side by $m_{2}$

$v_{f2}=\frac{1.3m_{1} v_{i1}}{m_{2} }$

Now substitute

$v_{f2} =\frac{1.3*0.12*v_{i1}}{2.4}\\v_{f2} =\frac{0.156v_{i1} }{2.4} \\v_{f2} =0.065v_{i1}$

$v_{f2} =$ 6.5% $v_{i1}$

<em />

6 0

3 years ago

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A block (mass = 61.2 kg) is hanging from a massless cord that is wrapped around a pulley (moment of inertia = 1/2MR2 kg · m2, wh

kolezko [41]

Answer:

The angular velocity is $w = 53.35 \ rounds /minute$

Explanation:

From the question we are told that

The mass of the block is $m = 61.2kg$

The of the pulley is $M = 14.2 kg$

The radius of the pulley is $R = 1.5m$

The radius of the cord around the pulley is $r = 1.5 m$

The distance of the block to the floor is $d = 8.0 m$

From the question we are told that the moment of inertia of the pulley is

$I = \frac{1}{2} MR^2 kg \cdot m^2$

Substituting value

$I = \frac{1}{2} * 14.2 * (1.5)^2$

$I = 15.975 kg \cdot m^2$

Using the Newtons law we can express the force acting on the vertical axis as

$ma = mg -T$

=> $T = mg -ma$

Now when the pulley is rotated that torque generated on the massless cord as a r result of the tension T and the radius of the cord around the pulley is mathematically represented as

$\tau = I \alpha$

Here $\alpha$ is the angular acceleration

Here $\tau$ is the torque which can be equivalent to

$\tau = T r$

Substituting this above

$Tr = I \alpha$

Substituting for T

$(mg - ma ) r = I\ r \alpha$

Here $a$ is the linear acceleration which is mathematically represented as

$a = r\alpha$

$(mg - m(r\alpha ) ) r = I\ r \alpha$

$mgr = I\alpha + m(r\alpha ) r$

$mgr = \alpha [ I + mr^2]$

making $\alpha$ the subject

$\alpha = \frac{mgr}{I -mr ^2}$

Substituting values

$\alpha = \frac{61.2 * 1.5 * 9.8}{15.975 + (61.2 ) * (1.5)^2}$

$\alpha =5.854 rad /s^2$

Now substituting into the equation above to obtain the acceleration

$a = 5.854 * 1.5$

$a=8.78 m/s^2$

This acceleration is $a = \frac{v}{t}$

and v is the linear velocity with is mathematically represented as

$v = \frac{d}{t}$

Substituting this into the formula acceleration

$a = \frac{d}{t^2}$

making t the subject

$t = \sqrt{\frac{d}{a} }$

substituting value

$t = \sqrt{\frac{8}{8.78}}$

$t = 0.9545 \ s$

Now the linear velocity is

$v = \frac{8}{0.9545}$

$v = 8.38 m/s$

The angular velocity is

$w = \frac{v}{r}$

So

$w = \frac{8.38}{1.5}$

$w = 5.59 rad/s$

Generally 1 radian is equal to 0.159155 rounds or turns

So 5.59 radian is equal to x

Now x is mathematically obtained as

$x = \frac{5.59 * 0.159155}{1}$

$= 0.8892 \ rounds$

Also

60 second = 1 minute

So 1 second = z

Now z is mathematically obtained as

$z = \frac{ 1}{60}$

z $= 0.01667 \ minute$

Therefore

$w = \frac{0.8892}{0.01667}$

$w = 53.35 \ rounds /minute$

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3 years ago

How to understand that an inductor behaves like short circuit and capacitor like an open circuit in steady state in a network?

garik1379 [7]

It depends on the steady-state frequency. At zero frequency an inductor behaves like an open circuit. As the frequency increases, the inductor acts more like an open circuit and a capacitator acts more like a short circuit

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C is the answer for this problem

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