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Gemiola [76]
1 year ago
11

A solid disk is rolling without slipping on a level surface at a constant speed of 2.50 m/s. (a) if the disk rolls up a 30.0o ra

mp, how far along the ramp will it move before it stops? (b) explain why your answer in part (a) does not depend on either the mass or the radius of the disk.
Physics
1 answer:
AnnyKZ [126]1 year ago
8 0

(a) By using the law of conservation of energy, the distance moved by the solid disk along the ramp is 0.96 m.

(b) The answer does not depend on the mass and radius of the disc as these values are canceled when the law of conservation of energy is applied.

What is the law of conservation of energy?

The law of conservation of energy states that the total energy of an isolated system is conserved.

The total initial energy Ei of the disc rolling down an inclined plane is given by the formula,

Ei=1/2*mv^2 + 1/2*Iω^2

where m is the mass of the disc, v is the velocity of the disc, I is the moment of inertia and ω is the angular velocity of the disc.

For a solid disc, I=1/2mr^2, and since it is rolling without slipping, the rolling velocity of the disc will be equal to its translational velocity, that is,

v=ωr or ω=v/r

So using ω=v/r and I=1/2mr^2, it can be written,

Ei=1/2*m(v)^2 + 1/2*(1/2mr^2)(v/r)^2

Ei=1/2*m(v)^2*(1+1/2*)

Ei=3/4*mv^2

If the height covered by the disc is h before stopping, then its final total  energy Ef will be equal to the potential energy, that is,

Ef=m*g*h

From the law of conservation of energy, it can be written,

Ei=Ef

3/4*mv^2=m*g*h

h=3v^2/(4g)

The length of the ramp is then given by the formula,

l=hcosecθ

where θ is the inclination angle. So

l= 3v^2/(4g)*cosecθ

Here g=9.8 m/s^2, v=2.5 m/s  and θ=30.0 degree. Using these values,

l= 3*(2.5 )^2/(4*9.8)*cosec( 30)

l= 0.96 m

Learn more about the law of conservation of energy,

brainly.com/question/14312374

#SPJ4

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3 years ago
An archer shoots an arrow at a 75.0 m distant target; the bull’s-eye of the target is at same height as the release height of th
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Answer:

the shooting angle ia 18.4º

Explanation:

For resolution of this exercise we use projectile launch expressions, let's see the scope

      R = Vo² sin (2θ) / g

      sin 2θ = g R / Vo²

      sin 2θ = 9.8 75/35²

      2θ = sin⁻¹ (0.6)

      θ = 18.4º

To know how for the arrow the tree branch we calculate the height of the arrow at this point

       X2 = 75/2 = 37.5 m

We calculate the time to reach this point since the speed is constant on the X axis

       X = Vox t

       t2 = X2 / Vox = X2 / (Vo cosθ)

        t2 = 37.5 / (35 cos 18.4)

        t2 = 1.13 s

With this time we calculate the height at this point

        Y = Voy t - ½ g t²

        Y = 35 sin 18.4   1.13 - ½ 9.8 1,13²

        Y = 6.23 m

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8 0
3 years ago
Read 2 more answers
2.5 g of helium at an initial temperature of 300 K interacts thermally with 9.0 g of oxygen at an initial temperature of 620 K .
muminat

Answer:

Explanation:

2.5 g of He = 2.5 / 4  mole

= .625 moles

9 g of oxygen = 9/32

= .28 mole of oxygen

C_p of He = 3/2 R

C_p of O₂ = 5/2 R

A ) Initial thermal energy of He = 3/2 n R T

= 1.5 x .625 x 8.32 x 300

= 2340 J

Initial thermal energy of O₂ = 5/2 n R T

= 2.5 x .28 x 8.32 x 620

= 3610.88 J

B ) If T be the equilibrium temperature after mixing

gain of heat by helium

= n C_p Δ T

= .625 x 3/2 R x ( T - 300 )

Loss of heat by oxygen

n C_p Δ T

= .28 x 5/2 R x ( 620 - T )

Loss of heat = gain of heat

.625 x 3/2 R x ( T - 300 ) = .28 x 5/2 R x ( 620 - T )

1.875 T- 562.5 = 868- 1.4 T

3.275 T = 1430,5

T = 436.8 K

Thermal energy of He

= 1.5 x .625 x 8.32 x 436.8

= 3407 J

thermal energy of O₂

= 2.5 x .28  x 8.32 x 436.8

= 2543.92 J

C )

Heat energy transferred

=  .28 x 5/2 R x ( 620 - T )

=  .28 x 5/2 x  8.32 x ( 620 - 436.8 )

1066.95 J

Heat will flow from O₂ to He

Final temperature is 436.8 K

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