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anygoal [31]
3 years ago
9

A scientist heated a tank containing 50 g of water. The specific heat of water is 4.18 J/gºC. The temperature of the water incre

ased from 25ºC to 37ºC. How much heat energy did the water absorb?
Your answer:


2,508 Joules

-2,508 Joules

5,225 Joules

7,733 Joules
Physics
1 answer:
Amanda [17]3 years ago
6 0
I got 5,225 by 50x4.18= 209(25)=5,225
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Gaseous helium is in thermal equilibrium with liquid helium at 6.4 K. The mass of a helium atom is 6.65 × 10−27 kg and Boltzmann
chubhunter [2.5K]

Answer:

162.78 m/s is the most probable speed of a helium atom.

Explanation:

The most probable speed:

v_{mp}=\sqrt{\frac{2K_bT}{m}}

K_b= Boltzmann’s constant =1.38066\times 10^{-23} J/K

T = temperature of the gas

m = mass of the gas particle.

Given, m = 6.65\times 10^{-27} kg

T = 6.4 K

Substituting all the given values :

v_{mp}=\sqrt{\frac{2\times 1.38066\times 10^{-23} J/K\times 6.4 K}{6.65\times 10^{-27} kg}}

v_{mp}=162.78 m/s

162.78 m/s is the most probable speed of a helium atom.

4 0
3 years ago
If a 0.15 kg ball falls and has a KE of 20 J just before striking the ground, from what height did it fall. A. 1.36m B. 3m C. 13
RUDIKE [14]
According to the conservation of mechanical energy, the kinetic energy just before the ball strikes the ground is equal to the potential energy just before it fell. 

Therefore, we can say KE = PE
We know that PE = m·g·h

Which means KE = m·g·h

We can solve for h:

h = KE / m·g
   = 20 / (0.15 · 9.8) 
   = 13.6m

The correct answer is: the ball has fallen from a height of 13.6m.

5 0
3 years ago
Heat flows into a gas in a piston and work is performed on the gas by its surroundings. The amount of work done is equal to the
inna [77]

Answer:

The Internal energy of the gas did not change

Explanation:

In this situation the Internal energy of the gas did not change and this is because according the the first law of thermodynamics

Δ U = Q - W  ------ ( 1 )

Δ U  = change in internal energy

Q = heat added

W = work done

since Q = W.  the value of ΔU  will be = zero   i.e. No change

4 0
3 years ago
Ask Your Teacher Suppose the roller coaster below(h1 = 36 m, h2 = 13 m, h3 = 30) passes point A with a speed of 1.00 m/s. If the
Oliga [24]

Answer:

The answer to the question is

The roller coaster will reach point B with a speed of 14.72 m/s

Explanation:

Considering both kinetic energy KE = 1/2×m×v² and potential energy PE = m×g×h

Where m = mass

g = acceleration due to gravity = 9.81 m/s²

h = starting height of the roller coaster

we have the given variables

h₁ = 36 m,

h₂ = 13 m,

h₃ = 30 m

v₁ = 1.00 m/s

Total energy at point 1 = 0.5·m·v₁² + m·g·h₁

= 0.5 m×1² + m×9.81×36

=353.66·m

Total energy at point 2 = 0.5·m·v₂² + m·g·h₂

= 0.5×m×v₂² + 9.81 × 13 × m = 0.5·m·v₂² + 127.53·m

The total energy at 1 and 2 are not equal due to the frictional force which must be considered

Total energy at point 2 = Total energy at point 1 + work done against friction

Friction work = F×d×cosθ = (\frac{1}{5} × mg)×60×cos 180 = -117.72m

0.5·m·v₂² + 127.53·m = 353.66·m -117.72m

0.5·m·v₂² = 108.41×m

v₂² = 216.82

v₂  =  14.72 m/s

The roller coaster will reach point B with a speed of 14.72 m/s

8 0
4 years ago
An MRI scanner is based on a solenoid magnet that produces a large magnetic field. The magnetic field doesn't stop at the soleno
GaryK [48]

Answer:

The maximum change in  flux is \Delta \o = 0.1404 \ Wb

The average  induced emf     \epsilon =0.11232 V

Explanation:

   From the question we are told that

             The speed of the technician is v = 0.80 m/s

              The distance from the scanner is d = 1.0m

              The  initial magnetic field is  B_i = 0T

               The final magnetic field is B_f = 6.0T

                 The diameter of the loop is  D = 19cm = \frac{19}{100} = 0.19 m

The area of the loop is mathematically represented as

        A  =  \pi [\frac{D}{2} ]^2

             = 3.142 \frac{0.19}{2}

             = 0.02834 m^2

At maximum the change in magnetic field is mathematically represented as

            \Delta \o = (B_f - B_i)A

  =>      \Delta  \o = (6 -0)(0.02834)

                  \Delta \o = 0.1404 \ Wb

The  average induced emf is mathematically represented as

           \epsilon =  \Delta \o v

              = 0.1404 * 0.80

             \epsilon =0.11232 V

7 0
3 years ago
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