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lukranit [14]
3 years ago
13

Rubbing your hands together warms them by converting work into thermal energy. If a woman rubs her hands back and forth for a to

tal of 12 rubs, at a distance of 7.50 cm per rub, and with a frictional force of 45.0 N, what is the temperature increase in degrees Celsius?
Physics
1 answer:
Feliz [49]3 years ago
5 0

Answer:

Explanation:

Total heat = Work done = Force × distance

distance = 0.075 × 12 = 0.9 m

W = 45 × 0.9 = 40.5 joules

Specific heat of the human hand = 3.5 kj/kg = 3.5 j/g

Q = MCΔT

ΔT = (Q) ÷ (MC)

ΔT = 40.5 ÷ (3.5 × 1) = 11.57°C

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3 years ago
Read 2 more answers
How many electrons will constitute 2A current in unit time
soldi70 [24.7K]

Answer:

2 charges of electron (2C)

Explanation:

I = Q/t

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Q = 2 charge of electron

4 0
3 years ago
According to the text, there is no energy shortage now, nor will there ever be. what reason (s) is given to support this stateme
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The reason why there is no energy shortage nor will there ever be is because energy is being preserved and conserved and only changes form. It never gets lost or increased.
8 0
3 years ago
Two blocks with masses 1 and 2 are connected by a massless string that passes over a massless pulley as shown. 1 has a mass of 2
Bess [88]

Answer:

The acceleration of M_2 is  a =  0.7156 m/s^2

Explanation:

From the question we are told that

    The mass of first block is  M_1 =  2.25 \ kg

    The angle of inclination of first block is  \theta _1 =  43.5^o

    The coefficient of kinetic friction of the first block is  \mu_1  = 0.205

      The mass of the second block is  M_2 = 5.45 \ kg

     The angle of inclination of the second block is  \theta _2 =  32.5^o

      The coefficient of kinetic friction of the second block is \mu _2 = 0.105

The acceleration of M_1 \ and\  M_2 are same

The force acting on the mass M_1 is mathematically represented as

     F_1 = T -  M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1

=> M_1 a = T -  M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1

Where T is the tension on the rope

The force acting on the mass M_2 is mathematically represented as    

  F_2 =  M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2

   M_2 a =  M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2

At equilibrium

  F_1 =  F_2

So

 T -  M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1 =M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2

making a the subject of the formula

    a =  \frac{M_2 g sin \theta_2 - M_1 g sin \theta_1 - \mu_1 M_1g cos \theta - \mu_2 M_2 g cos \theta_2 }{M_1 +M_2}

substituting values a =  \frac{(5.45) (9.8) sin (32.5) - (2.25) (9.8) sin (43.5) - (0.205)*(2.25) *9.8cos (43.5) - (0.105)*(5.45) *(9.8) cos(32.5) }{2.25 +5.45}

    => a =  0.7156 m/s^2

     

3 0
4 years ago
Please help on this one?
Nonamiya [84]

30 or c because 60-30=30

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4 years ago
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