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makkiz [27]
3 years ago
7

What may thermal energy stores also be called? PLEASE REALLY NEED THE HELP!

Physics
1 answer:
ollegr [7]3 years ago
3 0

Answer:

The thermal energy is the sum of the potential energy and kinetic energy that is known to make all the particle in an object it is the energy that is responsible for its temperature heat is the flow of thermal energy thermal energy is the result of something that has some internal temperature which can be measured

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Negative peer influence
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Rutherford proposed that...
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It is B because when he made the rays go through the atoms, a very small little of it created a repulsion which resulted in him proposing that there is a dense, positive nucleus in the atom while electrons are spinning around. 
7 0
3 years ago
A fathom is a unit of length about 6 ft long commonly used in measuring depths of water. Because a fathom is about the length of
Naddik [55]

Distance to the moon = 4×10^{8}m.

1 m = 3.28 ft

Distance to the moon in ft = 4×10^{8}×3.28 ft

= 13.12 ×10^{8}ft

1 fathom = 6 ft

Hence, distance to the moon in fathom

= \frac{13.2}{6}×10^{8}

≈ 2× 10^{8}fathom



4 0
4 years ago
Calculate the approximate force on a square meter (1.00 m3) of sail, given the horizontal velocity of the wind is 6.00 m/s paral
KiRa [710]

Answer:

The force exerted on square meter of cubic sail is F = 15.3 N  

Explanation:

Given:-

- The face area of cubic sail, A = 1 m^2

- The velocity at frontal face, v1 = 6.0 m/s

- The velocity at back face, v2 = 3.5 m/s

- The density of air. ρ = 1.29 kg/m^3

Find:-

Calculate the approximate force on a square meter (1.00 m3) of sail

Solution:-

- We will apply the Bernoulli's equation to the flow of air around the cubic sail. Assuming that elevation changes are negligible. The constant elevation Bernoulli's equation is:

                           P1 + ρ*v1^2 / 2 = P2 + ρ*v2^2 / 2  

- The force (F) applied by any fluid is given by:

                           F = ( P2 - P1 )*A

- Re-arranging bernoulli's expression:

                          P2 - P1 = ρ/ 2 [ v2^2 - v1^2 ]

- Multiple the equation by area A:

                          A*[P2 - P1] = A*ρ/ 2 [ v1^2 - v2^2 ]

                          F = A*ρ/ 2 [ v1^2 - v2^2 ]

- Plug in the values:

                         F = (1)*(1.29/2)*[ 6^2 - 3.5^2 ]

                         F = 15.3 N  

7 0
4 years ago
A mass on a spring vibrates in simple harmonic motion at a frequency of 3.26 Hz and an amplitude of 5.76 cm. If the mass of the
V125BC [204]

Answer:

91.48N/m

Explanation:

In a spring-mass system undergoing a simple harmonic motion, the inverse of the frequency f, of oscillation is proportional to the square root of the mass m, and inversely proportional to the square root of the spring constant, k. This can be expressed mathematically as follows;

\frac{1}{f} = 2\pi\sqrt{\frac{m}{k} }            -----------(i)

From the question;

f = 3.26 Hz

m = 0.218kg

Substitute these values into equation (i) as follows;

\frac{1}{3.26} = 2\pi\sqrt{\frac{0.218}{k} }                            [<em>Square both sides</em>]

(\frac{1}{3.26})² = (2\pi)²(\frac{0.218}{k})    

(\frac{1}{10.6276}) = 4\pi²(\frac{0.218}{k})                      [<em>Take </em>\pi<em> to be 3.142</em>]

(\frac{1}{10.6276}) = 4(3.142)²(\frac{0.218}{k})

(\frac{1}{10.6276}) = 39.488(\frac{0.218}{k})

(\frac{1}{10.6276}) = (\frac{8.608}{k})                            [<em>Switch sides</em>]

(\frac{8.608}{k}) = (\frac{1}{10.6276})                            [<em>Re-arrange</em>]

(\frac{k}{8.608}) = (\frac{10.6276}{1})                            [<em>Cross-multiply</em>]

k = 8.608 x 10.6276

k = 91.48N/m

Therefore, the spring constant of the spring is 91.48N/m

4 0
4 years ago
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