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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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Water exits straight down from a faucet with a 1.96-cm diameter at a speed of 0.55 m/s. The volume flow rate of the water as it

d1i1m1o1n [39]

Answer:

Q = 165.95 cm³ / s, 1) v = $\sqrt{0.55^2 + 19.6 y}$ , 2) v = 2.05 m / s,

3) d₂ = 1.014 cm

Explanation:

This is a fluid mechanics exercise

1) the continuity equation is

Q = v A

where Q is the flow rate, A is area and v is the velocity

the area of a circle is

A = π r²

radius and diameter are related

r = d / 2

substituting

A = π d²/4

Q = π/4 v d²

let's reduce the magnitudes

v = 0.55 m / s = 55 cm / s

let's calculate

Q = π/4 55 1.96²

Q = 165.95 cm³ / s

If we focus on a water particle and apply the zimematics equations

v² = v₀² + 2 g y

where the initial velocity is v₀ = 0.55 m / s

v = $\sqrt{0.55^2 + 2 \ 9.8\ y}$

v = $\sqrt{0.55^2 + 19.6 y}$

2) ask to calculate the velocity for y = 0.2 m

v = $\sqrt{0.55^2 + 19.6 \ 0.2}$

v = 2.05 m / s

3) We write the continuous equation for this point 2

Q = v₂ A₂

A₂ = Q / v₂

let us reduce to the same units of the SI system

Q = 165.95 cm³ s (1 m / 10² cm) ³ = 165.95 10⁻⁶ m³ / s

A₂ = 165.95 10⁻⁶ / 2.05

A₂ = 80,759 10⁻⁶ m²

area is

A₂ = π/4 d₂²

d₂ = $\sqrt{4 A_2 / \pi }$

d₂ = $\sqrt{ \frac{4 \ 80.759 \ 10^{-6} }{\pi } }$

d₂ = 10.14 10⁻³ m

d₂ = 1.014 cm

4 0

3 years ago

A 4 kg textbook sits on a desk. It is pushed horizontally with a 50 N applied force against a 15 N frictional force.

GarryVolchara [31]

a) See free-body diagram in attachment

b) The book is stationary in the vertical direction

c) The net horizontal force is 35 N in the forward direction

d) The net force on the book is 35 N in the forward horizontal direction

e) The acceleration is $8.75 m/s^2$ in the forward direction

Explanation:

The free-body diagram of a body represents all the forces acting on the body using arrows, where the length of each arrow is proportional to the magnitude of the force and points in the same direction.

From the diagram of this book, we see there are 4 forces acting on the book:

- The applied force, F = 50 N, pushing forward in the horizontal direction

- The frictional force, $F_f = 15 N$ , pulling backward in the horizontal direction (the frictional force always acts in the direction opposite to the motion)

- The weight of the book, $W=mg$ , where m is the mass of the book and $g=9.8 m/s^2$ is the acceleration of gravity, acting downward. We can calculate its magnitude using the mass of the book, m = 4 kg:

$W=(4)(9.8)=39.2 N$

- The normal reaction exerted by the desk on the book, N, acting upward, and balancing the weight of the book

The book is in equilibrium in the vertical direction, therefore there is no motion.

In fact, the magnitude of the normal reaction (N) exerted by the desk on the book is exactly equal to the weight of the book (W), so the equation of motion along the vertical direction is

$N-W=ma$