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4 years ago

11

What is the moment of inertia of an object that rolls without slipping down a 2.00-m-high incline starting from rest, and has a

final velocity of 6.00 m/s?

1 answer:

Paha777 [63]4 years ago

7 0

$E = mgh + \frac{1}2} m v^{2} + \frac{1}{2} I \omega^{2} = mgh + \frac{1}2} m r ^{2} \omega ^{2} + \frac{1}{2} I \omega^{2}$

for a solid cylinder: $I = \frac{1}{2} m r^{2}$
for a hollow cylinder: $I = mr^{2}$

I will look at the case of a hollow cylinder:

$E = mgh + I \omega ^{2} = constant \\ \\ I = \frac{mgh}{ \omega^{2} }$

That is as far as i get.

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A new prototype cup has been designed to keep liquids, such as hot coffee or cold annk, near their original temperature for long

san4es73 [151]

Answer:

hope this was good for u and I believe it would be solid

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3 years ago

A 4.9-MeV (kinetic energy) proton enters a 0.28-T field, in a plane perpendicular to the field. Part APart complete What is the

BartSMP [9]

Answer:

$r=1.14m$

Explanation:

$\theta$ is the angle between the velocity and the magnetic field. So, the magnetic force on the proton is:

$F_m=qvBsen\theta\\F_m=qvBsen(90^\circ)\\F_m=qvB$

A charged particle describes a semicircle in a uniform magnetic field. Therefore, applying Newton's second law to uniform circular motion:

$F_m=F_c\\qvB=F_c(1)$

$F_c$ is the centripetal force and is defined as:

$F_c=m\frac{v^2}{r}$

Here $v$ is the proton's speed and $r$ is the radius of the circular motion. Replacing this in (1) and solving for r:

$qvB=\frac{mv^2}{r}\\r=\frac{mv^2}{qvB}\\r=\frac{mv}{qB}$

Recall that 1 J is equal to $6.242*10^{12}MeV$ , so:

$4.9MeV*\frac{1J}{6.242*10^{12}MeV}=7.85*10^{-13}J$

We can calculate $v$ from the kinetic energy of the proton:

$K=\frac{mv^2}{2}\\\\v=\sqrt{\frac{2K}{m}}\\v=\sqrt{\frac{2(7.85*10^{-13}J)}{1.67*10^{-27}kg}}\\v=3.06*10^{7}\frac{m}{s}$

Finally, we calculate the radius of the proton path:

$r=\frac{mv}{qB}\\r=\frac{1.67*10^{-27}kg(3.06*10^{7}\frac{m}{s})}{1.6*10^{-19}C(0.28T)}\\r=1.14m$

8 0

3 years ago

A seagull flies at a velocity of 9.00 m/s straight into the wind.

RideAnS [48]

a)If it takes the bird 18.0 minutes to fly 6 km away from the earth, the wind's speed will be 4 m/s.

b) The bird would need 7 minutes and 42 seconds to fly back 6 kilometers if he turned around and flew with the wind.

c)Compared to the 133.33 seconds it would take without the wind, the overall round-trip time is affected by the wind.

<h3>What is velocity?</h3>

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The given data in the problem is

A seagull flies at a velocity, $\rm v_{SA} = 9 \ m/sec$

The time the bird takes,t=18.0 min

The distance traveled relative to the earth = 6.00 km

a)

The seagull's relative velocity with reference to the ground as;

$\rm v_{sg} = \frac{6.00 \times 10^3 \ m }{(20 min) \times \frac{60 s }{1 \ min}} \\\\ v_{sg}= 5.00 \ m/sec$

Air velocity with reference to the ground is;

$\rm v_{AG}= v_{SG}-v_{SA} \\\\ v_{AG} = 5.00 \ m/sec - 9.00 \ m/sec \\\\ v_{AG} = -4.00 \ m/sec$

b)

If the bird turns around and flies with the wind, The time will he take to return 6.00 km is;

$\rm v_{SG}=v_{SA}+v_{AG} \\\\ v_{SG}=-900 \ m/sec +(-4.00 \ m/sec) \\\\ v_{SG}= -13.00 \ m/sec$

The time the bird takes;

$\rm t = \frac{x_{SG}}{v_{SG}} \\\\ t = \frac{6.00 \times 10^3 \ m }{13.00 \ m/sec } \\\\ t = 462 m/sec \\\\ t = 7 \ min \ and \ 42 \ sec$

c)\

The total round-trip time compared to what it would be with no wind. is;

$\rm t = 20 \ min( \frac{60 \ sec }{1 \ min} )+ 462 \ sec \\\\ t = 1200 \ sec +6 462 \ ec \\\\ t= 1662 \ sec$

The time for the round trip is;

$\rm t = \frac{12 \times 10^ 3 }{ 9 \ m/sec } \\\\ t = 1333.33 \ sec$

Hence the wind's speed, the time bird would need to fly back the total round-trip time will be 4 m/s, 7 minutes and 42 seconds and 1333.33 sec.

To learn more about the velocity, refer to the link: brainly.com/question/862972.

#SPJ1

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2 years ago

An 80.0 kg skier slides down a hill shaped as shown. Assume

umka21 [38]

The height above the ground from where the skier start is 11.5 m.

<h3>Conservation of energy</h3>

The height above the ground from where the skier start is determined by applying the principle of conservation of energy as shown below;

P.E = K.E

mgh = ¹/₂mv²

gh = ¹/₂v²

$h = \frac{v^2}{2g} \\\\h = \frac{15^2}{2 \times 9.8} \\\\h = 11.5 \ m$

Thus, the height above the ground from where the skier start is 11.5 m.

Learn more about conservation of energy here: brainly.com/question/166559

8 0

3 years ago

Lena helps a child who is choking on his food. Which reliable method prepared her for helping with this injury?

Ivanshal [37]

knowledge of first aid ... eg St John Ambulance, Red Cross etc. I think that everyone in a school should be taught First Aid.

7 0

3 years ago

Read 2 more answers