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a_sh-v [17]
2 years ago
12

A 58.7-g sample of nickel (s = 0.443 J/(g ∙ °C)), initially at 147.4°C, is placed in an insulated vessel containing 184.0 g of w

ater (s = 4.18 J/(g ∙ °C)), initially at 29.9°C. Once equilibrium is reached, what is the final temperature of the metal–water mixture?
Physics
1 answer:
kari74 [83]2 years ago
3 0

Answer:

Explanation:

Let t be the final temp

temp. drop of nickel=147.4-t

temp. gain by water=t-29.9

in equilibrium

Heat lost=Heat gained

58.7×0.443×(147.4-t)=184×4.18×(t-29.9)

58.7×0.443×147.4-58.7×0.443×t=184×4.18×t-184×4.18×29.9

     3833.00434 -26.0041 t=769.12t-22996.688

(769.12 +26.0041)t= ?

find t

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A champion athlete can produce one horsepower (746 W) for a short period of time. The number of 16-cm-high steps a 70-kg athlete
erastovalidia [21]

Answer:

407 steps

Explanation:

From the question,

P = mgh/t........... Equation 1

Where P = power, m = mass, g = acceleration due to gravity, h = height, t = time.

Make h the subject of the equation

h = Pt/mg............. Equation 2

Given: P = 746 W, t = 1 minute = 60 seconds, m = 70 kg.

Constant: g = 9.8 m/s²

Substitute into equation 2

h = 746(60)/(70×9.8)

h = 44760/686

h = 65.25 m

h = 6525 cm

number of steps = 6525/16

number of steps = 407 steps

6 0
1 year ago
A child of mass m is at the edge of a merry-go-round of diameter d. When the merry-go-round is rotating with angular acceleratio
dem82 [27]

Answer:

The torque on the child is now the same, τ.

Explanation:

  • It can be showed that the external torque applied by a net force on a rigid body, is equal to the product of the moment of inertia of the body with respect to the axis of rotation, times the angular acceleration.
  • In this case, as the movement of the child doesn't create an external torque, the torque must remain the same.
  • The moment of inertia is the sum of the moment of inertia of the merry-go-round (the same that for a solid disk) plus the product of  the mass of the child times the square of the distance to the center.
  • When the child is standing at the edge of the merry-go-round, the moment of inertia is as follows:

       I_{to} = I_{d} + m*r^{2}  = m*\frac{r^{2}}{2} +  m*r^{2} = \frac{3}{2}*  m*r^{2} (1)

  • So, τ = 3/2*m*r²*α (2)
  • When the child moves to a position half way between the center and the edge of the merry-go-round, the moment of inertia of the child decreases, as the distance to the center is less than before, as follows:

       I_{t} = I_{d} + m*\frac{r^{2}}{4}   = m*\frac{r^{2}}{2} + m*\frac{r^{2}}{4}  = \frac{3}{4}*  m*r^{2} (3)

  • Since the angular acceleration increases from α to 2*α, we can write the torque expression as follows:

       τ = 3/4*m*r² * (2α) = 3/2*m*r²

        same result than in (2), so the torque remains the same.

7 0
2 years ago
THE RELATIVE ANGLE AT THE KNEE CHANGES FROM 0O TO 85O DURING THE KNEE FLEXION PHASE OF A SQUAT EXERCISE. IF 10 COMPLETE SQUATS A
Mice21 [21]

Answer: angular distance = 1700° and 29.7 rad

      also the angular displacement = 0

Explanation:

To explain this, i will give a breakdown of this works.

we are asked to find both the angular distance and displacement the knee undergo.

Ok to get the distance of the knee, we would first take note that for one to squat down and get back up, the knee would travel through 85° of flexion to gpo down, and also through another 85° of extension to return standing (upright). So, the actual angular distance of the squat is 170°.

taking ten squats, the knee would have to go through 170° motion times 10 i.e;

10 * 170° = 1700°

Therefore the angulaar distance is 1700°

now converting this distance to radians since we will be required to have our answer in both degree and rad.

Given that 2pi = 360°, it means that one degree will give 57.3°;

∅ (rad) = ∅ (deg) * 2π/360°

∅ (rad) = 1170° * 2π/360°  = 29.7 rad

∅ (rad) = 29.7 rad

b. For the other part, let us remember that angular displacement is equal to angular distance divided by time, so the angular displacement displacement of the knee will be zero, because the knee's position at the final third will be the same as the initial position.

cheers i hope this helps!!!!

π

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Answer:

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Explanation:

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