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Marina CMI [18]

1 year ago

7

In an experiment, students roll several hoops down the same incline plane. Each hoop has the same mass but a different radius. E

ach hoop rolls down the incline without slipping. Which of the following graphs best shows the linear speed V of the hoops at the bottom of the incline as a function of the radius r of the hoop?

1 answer:

insens350 [35]1 year ago

5 0

Answer:

The graph should have velocity (v) on the y-axis and radius (r) on the x-axis. It will have a straight, horizontal line that goes across the graph.

Explanation:

$KE=\frac{1}{2} I(omega)^{2}$

Shown above is the formula for Kinetic Energy in rotational terms. I'm new to brain.ly so I couldn't insert the omega symbol, sorry about that. Omega can be replaced with $\frac{v^{2} }{r^2}$ . Moment of Inertia (I) can be replaced with $mr^2$ .

The equation becomes $KE=\frac{1}{2} mr^2(\frac{v^2}{r^2} )$ .

The r's cancel out, making the different radii negligible, causing a straight horizontal line.

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Select True or False for the following statements about Heisenberg's Uncertainty Principle. True False It is not possible to mea

sveticcg [70]

Answer:

Statement 1) False

Statement 2) False

Statement 3) True

Explanation:

The uncertainty principle states that " in a physical system certain quantities cannot be measured with random precision no matter whatever the least count of the instrument is" or we can say while measuring simultaneously the position and momentum of a particle the error involved is

$P\cdot\delta x\geq \frac{h}{4\pi }$

Thus if we measure x component of momentum of a particle with 100% precision we cannot measure it's position 100% accurately as the error will be always there.

Statement 1 is false since measurement of x and y positions has no relation to uncertainty.

Statement 2 is false as both the momentum components can be measured with 100% precision.

Statement 3 is true as as demanded by uncertainty principle since they are along same co-ordinates.

6 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

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

6 0

2 years ago

A 3.0-kg mass and a 5.0-kg mass hang vertically at the opposite ends of a very light rope that goes over an ideal pulley. If the

AleksAgata [21]

Answer:

acceleration = 2.4525‬ m/s²

Explanation:

Data: Let m1 = 3.0 Kg, m2 = 5.0 Kg, g = 9.81 m/s²

Tension in the rope = T

Sol: m2 > m1

i) for downward motion of m2:

m2 a = m2 g - T

5 a = 5 × 9.81 m/s² - T

⇒ T = 49.05‬ m/s² - 5 a Eqn (a)‬

ii) for upward motion of m1

m a = T - m1 g

3 a = T - 3 × 9.8 m/s²

⇒ T = 3 a + 29.43‬ m/s² Eqn (b)

Equating Eqn (a) and(b)

49.05‬ m/s² - 5 a = T = 3 a + 29.43‬ m/s²

49.05‬ m/s² - 29.43‬ m/s² = 3 a + 5 a

19.62 m/s² = 8 a

⇒ a = 2.4525‬ m/s²

5 0

1 year ago

Joanna has become good friends with Janna, whose name begins with the same letter as hers. They sit next to each other in three

solmaris [256]

Proximity -------------------- APEX

5 0

2 years ago

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A 60 kg student in a rowboat on a still lake decides to dive off the back of the boat. The studen'ts horizontal aceleration is 2

TiliK225 [7]

As per the third law of Newton, the force exerted by the boat over the student is equal in magnitude to the force that the student exerted on the boat.

So, calculate the force on the student using the second law of Newton, Force = mass * acceleration.

Force on the student = 60 kg * 2.0 m/s^2 = 120 N.

=> horizontal force exerted by the student on the boat = 120 N

Answer: option d. 120 N. toward the back of the boat.

Of course it is toward the back because that is where the student jumped from..

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2 years ago