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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

You might be interested in
The mass of Jupiter is 1.9 × 1027 and that of the sun is 1.99 × 1030. The
n200080 [17]

Answer:

F = 4.147 × 10^23

v = 1.31 × 10^4

Explanation:

Given the following :

mass of Jupiter (m1) = 1.9 × 10^27

Mass of sun (m2) = 1.99 × 10^30

Distance between sun and jupiter (r) = 7.8 × 10^11m

Gravitational force (F) :

(Gm1m2) / r^2

Where ; G = 6.673×10^-11 ( Gravitational constant)

F = [(6.673×10^-11) × (1.9 × 10^27) × (1.99 × 10^30)] / (7.8 × 10^11)^2

F = [25.231 × 10^(-11+27+30)] / (60.84 × 10^22)

F = (25.231 × 10^46) / (60.84 × 10^22)

F = 3.235 × 10^(46 - 22)

F = 0.4147 × 10^24

F = 4.147 × 10^23

Speed of Jupiter (v) :

v = √(Fr) / m1

v = √[(4.147 × 10^23) × (7.8 × 10^11) / (1.9 × 10^27)

v = √32.3466 × 10^(23+11) / 1.9 × 10^27

v = √32.3466× 10^34 / 1.9 × 10^27

v = √17. 023 × 10^34-27

v = √17.023 × 10^7

v = 13047.221

v = 1.31 × 10^4

4 0
4 years ago
Which best describes the scientific method?
Mrrafil [7]

Your answer would be, The process of Hypothesis, and Testing through which scientific inquiry occurs.

Hope that helps!!!! : )

3 0
3 years ago
Read 2 more answers
And 8 kg bowling ball is rolling along the frictionless alley
VLD [36.1K]
It will stop eventually
8 0
3 years ago
A mass attached to a 50.0 cm long string starts from rest and is rotated 40 times in one minute before reaching a final angular
ch4aika [34]

To solve this problem it is only necessary to apply the kinematic equations of angular motion description, for this purpose we know by definition that,

\theta = \frac{1}{2}\alpha t^2 +\omega_0 t + \theta_0

Where,

\theta = Angular Displacement

\alpha =Angular Acceleration

\omega_0 = Angular velocity

\theta_0 =Initial angular displacement

For this case we have neither angular velocity nor initial angular displacement, then

\theta = \frac{1}{2}\alpha t^2

Re-arrange for \alpha,

\alpha = \frac{2\theta}{t^2}

Replacing our values,

\alpha = \frac{2(40rev*\frac{2\pi rad}{1rev})}{60^2}

\alpha = 0.139rad/s

Therefore the ANgular acceleration of the mass is 0.139rad/s^2

4 0
3 years ago
Part D
anygoal [31]

Answer: I didn't see a difference because the large ball's vertical displacement and velocity are the same as the small one's.

Explanation:

5 0
3 years ago
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