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2 years ago

15. Some people have observed that animal behavior changes just before an earthquake. What might explain this?

Physics

1 answer:

Ber [7]2 years ago

3 0

Answer:

15: Animals are able to detect the first of an earthquake's seismic waves—the P-wave, or pressure wave, that arrives in advance of the S-wave, or secondary, shaking wave. This likely explains why animals have been seen snapping to attention, acting confused or running right before the ground starts to shake.

16: No, it is not even remotely true. Microwaves cook food from the outside in, just like a regular oven. In fact, most of the cooking on the inside of the food, depending on its thickness, is done by heat conduction from the outside surfaces inwards, as the microwaves do not actually penetrate that far into the food.

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A well lagged copper calorimeter of mass 120g contains 70g of water and 10g of ice both at 0 degrees Celsius. Dry steam at 100 d

shtirl [24]

Answer:

7.6 g

Explanation:

"Well lagged" means insulated, so there's no heat transfer between the calorimeter and the surroundings.

The heat gained by the copper, water, and ice = the heat lost by the steam

Heat gained by the copper:

q = mCΔT

q = (120 g) (0.40 J/g/K) (40°C − 0°C)

q = 1920 J

Heat gained by the water:

q = mCΔT

q = (70 g) (4.2 J/g/K) (40°C − 0°C)

q = 11760 J

Heat gained by the ice:

q = mL + mCΔT

q = (10 g) (320 J/g) + (10 g) (4.2 J/g/K) (40°C − 0°C)

q = 4880 J

Heat lost by the steam:

q = mL + mCΔT

q = m (2200 J/g) + m (4.2 J/g/K) (100°C − 40°C)

q = 2452 J/g m

Plugging the values into the equation:

1920 J + 11760 J + 4880 J = 2452 J/g m

18560 J = 2452 J/g m

m = 7.6 g

7 0

3 years ago

HELP FAST!!!!!! How does thermal energy affect different materials?

Sergeeva-Olga [200]

Thermal energy affects doffrwnt materials by the movement of the tiny particles, which transfers from one object to another, aka heat.

7 0

3 years ago

Read 2 more answers

What is the net electric force on the charge located at the lower right-hand corner of the triangle.

Inga [223]

Answer:

$\sqrt{3}\frac{kq^{2}}{a^{2} }$

Explanation:

none sadly

6 0

3 years ago

Four electrons and one proton are at rest, all at an approximate infiitne distance away from each other. This original arrangmen

murzikaleks [220]

Answer:

a) W = 1.63 10⁻²⁸ J, b) W = 1.407 10⁻²⁷ J, c) W = 1.68 10⁻²⁸ J,

d) W = - 4.93 10⁻²⁸ J

Explanation:

a) In this problem we have an electron at the origin, work is requested to carry another electron from infinity to the point x₂ = 0, y₂ = 2.00m

If we use the law of conservation of energy, work is the change in energy of the system

W = ΔU = U_∞ -U

the potential energy for point charges is

U =k $\sum \frac{q_i q_j}{r_{ij} }$

in this case we only have two particles

U = k $\frac{q_1q_2}{r_{12} }$

the distance is

r₁₂ = $\sqrt{(x_2-x_1)^2 + ( y_2-y_1)^2 }$

r₁₂ = $\sqrt{ 0 + ( 2-0)^2}$ Ra 0 + (2-0)

r₁₂ = √2= 1.4142 m

we substitute

W = k \sum \frac{q_i q_j}{r_{ij} }

let's calculate

W = $\frac{ 9 \ 10^9 (1.6 \ 10^{-19})^2 }{1.4142}$ 9 109 1.6 10-19 1.6 10-19 / 1.4142

W = 1.63 10⁻²⁸ J

b) the two electrons are fixed, what is the work to bring another electron to x₃ = 3.00 m y₃ = 0

in this case we have two fixed electrons

U = k $( \frac{q_1q_3}{r_{13} } + \frac{q_2q_3}{r_{23} } )$

in this case all charges are electrons

q₁ = q₂ = q₃ = q

W = U = k q² $( \frac{1}{r_{13} } + \frac{1}{r_{23} } )$

the distances are

r₁₃ = $\sqrt{(3-0)^2 + 0}$ RA (3.00 -0) 2 + 0

r₁₃ = 3

r₂₃ = $\sqrt{ 3^2 + 2^2}$ Ra (3 0) 2 + (2 0) 2

r₂₃ = √13

r₂₃ = 3.606 m

let's look for the job

W = U

let's calculate

W = ${9 \ 10^3 ( 1.6 10^{-19})^2 }({\frac{1}{3} + \frac{1}{3.606} } )$

W = 1.407 10⁻²⁷ J

c) the three electrons are fixed, we bring the four electron to x₄ = 3.00m,

y₄ = 4.00 m

W = U = k $( \frac{q_1q_4}{r_{14 }} + \frac{q_2q_4}{r_{24} } + \frac{q_3q_4}{r_{34} } )$

all charges are equal q₁ = q₂ = q₃ = q₄ = q

W = k q² $(\frac{1}{r_{14} } + \frac{1}{r_{24} } + \frac{1}{r_{34} } )$

let's look for the distances

r₁₄ = $\sqrt{3^2 +4^2}$

r₁₄ = 5 m

r₂₄ = $\sqrt{3^2 + ( 4-2)^2}$

r₂₄ = √13 = 3.606 m

r₃₄ = $\sqrt{(3-3)^2 + (4-0)^2}$

r₃₄ = 4 m

we calculate

W = 9 10⁹ (1.6 10⁻¹⁹)² $( \frac{1}{5} + \frac{1}{3.606} + \frac{1}{4} )$

W = 1.68 10⁻²⁸ J

d) we take the proton to the location x5 = 1m y5 = 1m

W = U = k $( \frac{q_1q_5}{r_{15} } + \frac{q_2q_5}{r_{25} } + \frac{q_3q_5}{r_{35} } + \frac{q_4q_5}{r_{45} } )$

in this case the charges have the same values but charge 5 is positive and the others negative, so the products of the charges give a negative value

W = - k q² $( \frac{1}{r_{15} } + \frac{1}{r_{25} } + \frac{1}{r_{35} } + \frac{1}{r_{45} } )$

we look for distances

r₁₅ = $\sqrt{ 1^2 +1^2}$ Ra (1-0) 2 + (1-0) 2

r₁₅ = √ 2 = 1.4142 m

r₂₅ = $\sqrt{ (2-1)^2 +1^2}$

r₂₅ = √2 = 1.4142 m

r₃₅ = $\sqrt{ ( 3-1)^2 +1^2}$

r₃₅ = √5 = 2.236 m

r₄₅ = $\sqrt{ (3-1)^2 + (4-1)^2}$

r₄₅ = √13 = 3.606 m

we calculate

W = - 9 10⁹ (1.6 10⁻¹⁹)² $( \frac{1}{1.4142} +\frac{1}{1.4142} + \frac{1}{2.236} + \frac{1}{3.606} )$

W = - 4.93 10⁻²⁸ J

3 0

3 years ago

Pls help i begg youuuuu

Anna [14]

Pitch is the sensation of certain frequencies to the ear. High frequency = high pitch, low frequency = low pitch.

f = c(speed of the wave) / λ (wavelength)

1. 343m/s / 0.77955m = 439.99 Hz
This corresponds to pitch A

2. 343m/s / 0.52028m = 659.26 Hz
 This corresponds to pitch E

3. 343m/s / 0.65552m = 523.349 Hz
 This corresponds to pitch C

4. using f = c / λ
λ = c / f
= 343m/s / 587.33 = 0.583999 m = 0.584 m

3 0

3 years ago