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2 years ago

A 0.5 kg ball moving at a speed of 3m/s rolls up a hill. How high does the ball roll before it stops

1 answer:

3 0

According to conservation of energy
Kinetic energy at the bottom of hill = potential energy at top of hill
Kinetic energy(KE)= 0.5×mass(m)×velocity(v)²
where m=0.5 kg, and v= 3 m/s, v²=9
KE=0.5×0.5×9=2.25 J = PE
PE=m×gravitational acceleration(g)×height(h)
h=2.25/(0.5×9.8)=0.459 m

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A person jogs 4.0 km in 32 minutes, then 2.0 km in 22 minutes, and finally 1.0 km in 16 minutes. What is the jogger's average sp

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3 years ago

A wheel in the shape of a flat, heavy, uniform, solid disk is initially at rest at the top of an inclined plane of height 2.00 m

olasank [31]

Answer:

Explanation:

If friction is neglected, the wheel cannot roll and can only slide frictionlessly and will have the same velocity at the bottom of the ramp as if it had been in free fall as it has converted the same amount of potential energy.

mgh = ½mv²

v = √(2gh) = √(2(9.81)(2.00)) = 6.26418... = 6.26 m/s

However if we do not ignore all friction and the wheel rolls without slipping down the slope, the potential energy becomes linear and rotational kinetic energy

mgh = ½mv² + ½Iω²

mgh = ½mv² + ½(½mR²)(v/R)²

2gh = v² + ½v²

2gh = 3v²/2

v = √(4gh/3) =√(4(9.81)(2.00)/3) = 5.11468... = 5.11 m/s

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2 years ago

A student is observing a pendulum swinging back and forth. Each time it swings, the pendulum bob reaches a lower maximum, and ev

vodomira [7]

Answer:

the pendulum loses momentum and stops because of gravity and wind resistance. it does not violate the law of conservation of energy because it is not gaining any more momentum than what it had started with

Explanation:

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1 year ago

Read 2 more answers

A 0.0450 kg bullet is accelerated from rest to a speed of 425 m/s in a 2.25 kg rifle (which is inititally at rest). The pain of

Mrac [35]

Answer:

If the rifle is held loosely away from the shoulder, the recoil velocity will be of -8.5 m/s, and the kinetic energy the rifle gains will be 81.28 J.

Explanation:

By momentum conservation, and given the bullit and the recoil are in a straight line, the momentum analysis will be unidimentional. As the initial momentum is equal to zero (the masses are at rest), we have that the final momentum equals zero, so

$0=P_{f}=m_{b} *v_{b}+m_{r}*v_{r}$

now we clear $v_{r}$ and use the given data to get that

$v_{r}=-8.5\frac{m}{s}$

But we have to keep in mind that the bullit accelerate from rest to a speed of 425 m/s, then if the rifle were against the shoulder, the recoil velocity would be a fraction of the result obtained, but, as the gun is a few centimeters away from the shoulder, it is assumed that the bullit get to its final velocity, so the kick of the gun, gets to its final velocity $\bold{v_{r}}$ too.

Finally, using $v_{r}$ we calculate the kinetic energy as

$K=\frac{1}{2}m_{r}v_{r}^{2}=81.28J$

3 0

2 years ago

Distance between Ranjan and Gomes house is 9km. Ranjan has to attend Gomes birthday party at 7 o'clock. He started from his home

Nataliya [291]

Answer:

Kindly check explanation

Explanation:

Given the following :

Distance between Ranjan and Gomez house = 9km

Start time = 6'o clock

Time he arrived at Gomes house = 7'o clock

Time used for chatting during this period = 5 minutes

Distance covered in 40minutes = 6km

Time spend for second part of the journey = (60 - (40 + 5))minutes = 15 minutes = 15/60 = 0.25 hoir

Distance covered during second part of journey = 9 - 6 = 3 km

Speed = distance / time

Speed = 3km / 0.25 hr = 12km/hr

Average speed for entire journey :

First part :

Speed = distance / time

Time = 40 minutes = (40/60) = 0.667 hour

Speed = 6km / 0.667 hour = 9km / hr

Average speed :

(First part + second part) / 2

(9km/hr + 12km/hr) / 2 = 21km/hr / 2 = 10.5 km/hr

= 10.5km/hr

5 0

2 years ago