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lapo4ka [179]
2 years ago
11

Suppose we replace both hover pucks with pucks that are the same size as the originals but twice as massive. otherwise, we keep

the experiment the same. compared to the pucks in the video, this pair of pucks will rotate
Physics
2 answers:
Mrac [35]2 years ago
6 0

The pucks will rotate with the same angular rate even after changing the masses of the puck.

Further Explanation:

The two hover pucks collide with each other and stick to one another after the collision. Use the concept of momentum conservation in order to obtain the final rate of rotation of the pucks.

The conservation of momentum means that the momentum of the two bodies remains equal to the momentum of the combined two bodies after they collide if there is no other means of energy loss.

Concept:

Let the two hover pucks have masses {m_1} and {m_2} which are being rotated at the rate of {v_1} and {v_2} respectively.

We can write the expression for the momentum conservation of the pucks as.

\fbox{\begin\\{m_1}{v_1}+{m_2}{v_2}=\left({{m_1}+{m_2}}\right)v\end{minispace}}                        …… (1)

Here, {m_1} is the mass of first puck, {m_2} is the mass of the second puck, {v_1} is the linear rate of the first puck, {v_2} is the linear rate of the second puck and v is the final rate of the two hover pucks.

The rate of the rotation of the pucks after collision will be,

v = \dfrac{{{m_1}{v_1} + {m_2}{v_2}}}{{{m_1} + {m_2}}}

Now, when the masses of the two pucks is doubled and the other conditions are kept same.

Substitute 2{m_1} for {m_1} and 2{m_2} for {m_2} in above expression.

\begin{aligned}v'&=\frac{{2{m_1}{v_1}+2{m_2}{v_2}}}{{2{m_1}+2{m_2}}}\\&=\frac{{{m_1}{v_1}+{m_2}{v_2}}}{{{m_1}+{m_2}}}\\&=v\\\end{gathered}

Thus, the above expression shows that the momentum of the hover pucks remains conserved and the hover pucks rotate with the same angular rate.

Learn More:

1.  Collision of a car with the wall brainly.com/question/9484203

2.  A ball falling under the acceleration due to gravity brainly.com/question/10934170

3. Net force on a body brainly.com/question/4033012

Answer Details:

Grade: High School

Subject: Physics

Chapter: Conservation of Momentum.

Keywords:

Hover, pucks, rotate, momentum, conservation, angular rate, rate of rotation, collide, masses, collision, m1v1, M1V1, (m1+m2)v, m2v2, first puck, second puck, same rate.

kvv77 [185]2 years ago
5 0

This pair of pucks will rotate at the same rate.

Further Explanation:

Hover Puck:  

The Hover Puck coasts effectively on a self-produced pad of air until followed up on by an outside power.  

Employments of hover puck:  

The Hover Puck coasts effectively on a self-produced pad of air until followed up on by an outside power.  

• Use one Hover Puck to show themes from Newton's First Law to impacts to reflection.  

• Use at least two pucks to examine the protection laws in two measurements.  

• Set up a bowling alley utilizing plastic beverage bottles.  

Innovation of hover puck:  

The First Rubber Hockey Pucks Were Made From Sliced-Up Lacrosse Balls. At the point when the game moved inside, entire balls were initially utilized, yet arena proprietors before long thought that it was desirable over cut them into thirds and keep the center segment. This fundamental plan was the standard by 1885.  

hockey puck was developed:  

The hockey puck appeared in 1875. It's hazy who really imagined it. Specialists accept the main hockey puck was likely only an elastic ball cut down the middle. This gave players an article with a level side that would slide over the ice.

Subject: physics

Level: High School

Keywords: Hover Puck, Employments of hover puck, Innovation of hover puck, hockey puck was developed.  

Related links:  

Learn more about evolution on

brainly.com/question/727976

brainly.com/question/6953278

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The banking angle in a turn on the Olympic bobsled track is not constant, but increases upward from the horizontal. Coming aroun
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Answer:

the curve inclination is increased so that a weight component helps keep the car on track

Explanation:

In the sledging competition these devices go at quite high speeds over 100 km/h, so when reaching the curves the friction force is not enough to keep the car on the track. For this reason,  the curve inclination is increased so that a weight component helps keep the car on track.

In general we can solve Newton's second law for this case, with the condition of no friction, it is found that

   V² = r g tan θ

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8 0
2 years ago
Calculate the change in the kinetic energy (KE) of the bottle when the mass is increased. Use the formula KE = mv2, where m is t
Aliun [14]

kinetic energy is given as

KE = (0.5) m v²

given that : v = speed of the bottle in each case =  4 m/s

when m = 0.125 kg

KE = (0.5) m v² =  (0.5) (0.125) (4)² = 1 J

when m = 0.250 kg

KE = (0.5) m v² =  (0.5) (0.250) (4)² = 2 J

when m = 0.375 kg

KE = (0.5) m v² =  (0.5) (0.375) (4)² = 3 J

when m = 0.0.500 kg

KE = (0.5) m v² =  (0.5) (0.500) (4)² = 4 J

6 0
2 years ago
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The position of a particle moving along the x axis may be determined from the expression x(t) = btu + ctv, where x will be in me
KIM [24]

As per given equation we have

x = bt^u + ct^v

now as per the dimensional analysis we can say that dimension of right side of equation must be equal to left side of the equation

now as per left side of equation its dimension is same as length or meter

now we can say it should be meter on right side also

bt^u = M^0L^1T^0

b*T^8 = M^0L^1T^0

b = M^0L^1T^{-8}

similarly for other term we have

ct^v = M^0L^1T^0

c*T^7 = M^0L^1T^0

c = M^0L^1T^{-7}

<em>so above are the dimensions of b and c</em>

8 0
1 year ago
A major league baseball pitcher throws a pitch that follows these parametric equations: x(t) = 142t y(t) = –16t2 + 5t + 5. The t
azamat

Answer:

(a) x'(t)= 142

(b) 142

(c) y'(t)= -32t+5

(d) 96.8 mph

(e) 0.426 s

(f) 0.061 rad

Explanation:

Velocity is a time-derivative of position.

(a) x(t) = 142t

x'(t)= 142

(b) Since x'(t)= 142 is independent of t, it follows it was constant throughout. Hence, at any point or time, the horizontal velocity is 142.

(c) y(t) = - 16t^2+5t+5

y'(t)= -32t+5

(d) When it passes the home plate, the ball has travelled 60.5 ft (from the question). This is horizontal, so it is equivalent to x(t).

x(t)= 142t = 60.5

t=\dfrac{60.5}{142}= 0.426.

In this time, the vertical velocity, y'(t) is

y(t)= -32\times0.426+5 = -8.632

The speed of the ball at thus point is s=\sqrt{142^2+(-8.632)^2}=142 ft/s

To convert this to mph, we multiply the factor 3600/5280

s=142\times\dfrac{3600}{5280}=96.8 \text{ mph}

(e) The time has been determined from (d) above.

t= 0.426

(f) This angle is given by

\theta=\tan^{-1}\dfrac{y'(t)}{x'(t)}

\theta=\tan^{-1}\dfrac{-8.632}{142}=\tan^{-1}-0.0607=3.47 (Note here we are considering the acute angle so we ignore the negative sign)

In radians, this is

\theta=3.47\times\dfrac{\pi}{180}=0.061 \text{ rad}

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2 years ago
An ideal gas is contained in a vessel at 300 K. The temperature of the gas is then increased to 900 K. (i) By what factor does t
Dahasolnce [82]

The question is missing some parts. Here is the complete question.

An ideal gas is contained in a vessel at 300K. The temperature of the gas is then increased to 900K.

(i) By what factor does the average kinetic energy of the molecules change, (a) a factor of 9, (b) a factor of 3, (c) a factor of \sqrt{3}, (d) a factor of 1, or (e) a factor of \frac{1}{3}?

Using the same choices in part (i), by what factor does each of the following change: (ii) the rms molecular speed of the molecules, (iii) the average momentum change that one molecule undergoes in a colision with one particular wall, (iv) the rate of collisions of molecules with walls, and (v) the pressure of the gas.

Answer: (i) (b) a factor of 3;

              (ii) (c) a factor of \sqrt{3};

              (iii) (c) a factor of \sqrt{3};

             (iv) (c) a factor of \sqrt{3};

              (v) (e) a factor of 3;

Explanation: (i) Kinetic energy for ideal gas is calculated as:

KE=\frac{3}{2}nRT

where

n is mols

R is constant of gas

T is temperature in Kelvin

As you can see, kinetic energy and temperature are directly proportional: when tem perature increases, so does energy.

So, as temperature of an ideal gas increased 3 times, kinetic energy will increase 3 times.

For temperature and energy, the factor of change is 3.

(ii) Rms is root mean square velocity and is defined as

V_{rms}=\sqrt{\frac{3k_{B}T}{m} }

Calculating velocity for each temperature:

For 300K:

V_{rms1}=\sqrt{\frac{3k_{B}300}{m} }

V_{rms1}=30\sqrt{\frac{k_{B}}{m} }

For 900K:

V_{rms2}=\sqrt{\frac{3k_{B}900}{m} }

V_{rms2}=30\sqrt{3}\sqrt{\frac{k_{B}}{m} }

Comparing both veolcities:

\frac{V_{rms2}}{V_{rms1}}= (30\sqrt{3}\sqrt{\frac{k_{B}}{m} }) .\frac{1}{30} \sqrt{\frac{m}{k_{B}} }

\frac{V_{rms2}}{V_{rms1}}=\sqrt{3}

For rms, factor of change is \sqrt{3}

(iii) Average momentum change of molecule depends upon velocity:

q = m.v

Since velocity has a factor of \sqrt{3} and velocity and momentum are proportional, average momentum change increase by a factor of

(iv) Collisions increase with increase in velocity, which increases with increase of temperature. So, rate of collisions also increase by a factor of \sqrt{3}.

(v) According to the Pressure-Temperature Law, also known as Gay-Lussac's Law, when the volume of an ideal gas is kept constant, pressure and temperature are directly proportional. So, when temperature increases by a factor of 3, Pressure also increases by a factor of 3.

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