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Mice21 [21]
3 years ago
13

What does the power of a machine measure?

Physics
2 answers:
GarryVolchara [31]3 years ago
5 0

The power of a machine is the work/time ratio for that particular machine

Its the rate of doing work.

bixtya [17]3 years ago
4 0
The power of a machine is the work/time ratio for that particular machine.



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Two 20.0 g ice cubes at − 20.0 ∘ C are placed into 285 g of water at 25.0 ∘ C. Assuming no energy is transferred to or from the
Lelechka [254]

Answer:

Ft = 17.48°C

Explanation:

Ft is the final temperature. However, ice absorbs heat during two process of melting and cooling and as such, there is no loss of heat to or from the surrounding hence by conservation of energy.

Therefore,

Heat absorbed by water of 20g = heat rejected by water of 265g.

So; M(ice)[C(ice) [(ΔT) + LH(ice) + C(water)(ΔT)] = C(water) M(water) (ΔT)

So, 20[(2.108) [0 - (-20)] + 333.5 + 4.187(Ft - 0)]] = (285)(4.187) (25 - Ft)

To get;

7513 + 83.74 Ft = 29832.4 - 1193.3 Ft

So factorizing, we get;

83.74 Ft + 1193.3 Ft = 29832.4 - 7513

So; 1277.04 Ft = 22319.4

So; Ft = 22319.4/1277.04 = 17.48°C

7 0
3 years ago
A bird carrying a fish (5kg) drops it from 107 meters in the air how fast does the fish hit the ground
notka56 [123]

Answer:

The velocity of the fish hitting the ground is , v = 45.795 m/s        

Explanation:

Given data,

The mass of the fish, m = 5 kg

The height of the bird from the surface, h = 107 m

Using the III equation of motion,

                          v² = u² + 2gs

                          <em> v = √(u² + 2gs)</em>

Substituting the values,

                           v = √(0² + 2 x 9.8 x 107)  

                              = 45.795 m/s

Hence, the velocity of the fish hitting the ground is, v = 45.795 m/s        

4 0
3 years ago
A model air rocket with a mass of 50.g is free to travel along a horizontal track. It begins from rest. After 2.0s, the rocket h
katen-ka-za [31]

Answer:

v_r=5.89\ m.s^{-1}

Explanation:

Given:

  • mass of rocket, m_r=50\ g
  • time of observation, t=2\ s
  • mass lost by the rocket by expulsion of air, m_a=10\%\ of m_r=5\ g
  • velocity of air, v_a=53\ m.s^{-1}

<u>Now the momentum of air will be equal to the momentum of rocket in the opposite direction: </u>(Using the theory of elastic collision)

m_a.v_a=(m_r-m_a)\times v_r

5\times 53=(50-5)\times v_r

v_r=5.89\ m.s^{-1}

7 0
3 years ago
A point charge with charge q1 = 4.00 μC is held stationary at the origin. A second point charge with charge q2 = -4.40 μC moves
Bezzdna [24]

Answer:

W=0.94J

Explanation:

Electrostatic potential energy is the energy that results from the position of a charge in an electric field. Therefore, the work done to move a charge from point 1 to point 2 will be the change in electrostatic potential energy between point 1 and point 2.

This energy is given by:

U=\frac{K\left |q_1 \right |\left |q_2 \right |}{r}\\

So, the work done to move the chargue is:

W=U_1-U_2\\W=\frac{K\left |q_1 \right |\left |q_2 \right |}{r_1}-\frac{K\left |q_1 \right |\left |q_2 \right |}{r_2}\\r_1=\sqrt{((0.155 m)^2+0 m)^2}=0.115m\\r_2=\sqrt{((0.245 m)^2+(0.270 m)^2}=0.365m\\W=K\left |q_1 \right |\left |q_2 \right |(\frac{1}{r_1}-\frac{1}{r_2})\\W=8.99*10^9\frac{Nm^2}{c^2}(4.00*10^{-6}C)(4.40*10^{-6}C)(\frac{1}{0.115m}-\frac{1}{0.365})\\W=0.94J

The work is positive since the potential energy in 1 is greater than 2.

5 0
4 years ago
Equipotential lines are usually shown in a manner similar to topographical contour lines, in which the difference in the value o
Elza [17]

Answer:

B or 2

Explanation:

5 0
3 years ago
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