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3 years ago

8.

Physics

1 answer:

vladimir1956 [14]3 years ago

3 0

Answer:

a. petrification

Explanation:

tar seeps = natural trade that, because of its close proximity to the ground surface, seeps from the cracks in the Earth or between rocks forming pits or pools (tar pits)

amber = fossilized resin produced by extinct coniferous trees, typically yellow in color

mummification = a process in which the skin and flesh of a corpse can be preserved by embalming and drying

You might be interested in

A que profundidad esta nadando una persona dentro de una alberca si la presión absoluta sobre ésta es de 156kPa?

lara31 [8.8K]

Answer:

La persona está nadando en la alberca a una profundida de 5.575 metros.

Explanation:

La presión absoluta ( $P_{tot}$ ) experimentada por la persona es la suma de la presión atmosférica ( $P_{atm}$ ) y la presión hidrostática de la columna de agua de la alberca ( $P_{h}$ ), medidas en kilopascales. Es decir,

$P_{tot} = P_{atm}+P_{h}$ (1)

$P_{tot} = P_{atm} + \frac{\rho\cdot g \cdot z}{1000}$ (2)

Donde:

$\rho$ - Densidad del fluido de la alberca, medida en kilogramos por metro cúbico.

$g$ - Aceleración gravitacional, medida en metros por segundo al cuadrado.

$z$ - Profundidad de la persona en la alberca, medida en metros.

Si sabemos que $P_{atm} = 101.325\,kPa$ , $\rho = 1000\,\frac{kg}{m^{3}}$ , $g = 9.807\,\frac{m}{s^{2}}$ y $P_{tot} = 156\,kPa$ , entonces la profundidad de la persona en la alberca es:

$156 = 101.325 +\frac{(1000)\cdot (9.807)\cdot z}{1000}$

$54.675 = 9.807\cdot z$

$z = 5.575\,m$

La persona está nadando en la alberca a una profundida de 5.575 metros.

5 0

3 years ago

An object initially at rest experiences an acceleration of 0.281 m/s2 to the South for a time of 5.44 seconds. It then increases

andre [41]

Answer:

12.0 meters

Explanation:

Given:

v₀ = 0 m/s

a₁ = 0.281 m/s²

t₁ = 5.44 s

a₂ = 1.43 m/s²

t₂ = 2.42 s

Find: x

First, find the velocity reached at the end of the first acceleration.

v = at + v₀

v = (0.281 m/s²) (5.44 s) + 0 m/s

v = 1.53 m/s

Next, find the position reached at the end of the first acceleration.

x = x₀ + v₀ t + ½ at²

x = 0 m + (0 m/s) (5.44 s) + ½ (0.281 m/s²) (5.44 s)²

x = 4.16 m

Finally, find the position reached at the end of the second acceleration.

x = x₀ + v₀ t + ½ at²

x = 4.16 m + (1.53 m/s) (2.42 s) + ½ (1.43 m/s²) (2.42 s)²

x = 12.0 m

5 0

3 years ago

Formula for work done

Alenkinab [10]

Answer:

Work = Force × Distance

-----------------------------------

hope this helps :)

6 0

3 years ago

Assuming this is a distance time graph( ignore the speed time title) assume metres on vertical scale. describe in as much detail

riadik2000 [5.3K]

Answer:

The journey started from point A, with a speed of 1 m/s for 10 seconds after which the it became stationary at 10 meters from the start point, for 10 seconds. The journey continued at a higher speed of 2 m/s for 5 seconds and halted (became stationary) again at 20 meters which was the maximum distance reached from the start point

The return journey to the start point started at the 20 meter mark and lasted for 5 seconds, at a speed of 4 m/s

The total distance travelled during the journey from the start point A to the final point B is 40 m

Explanation:

From the start point A to point B, we have;

The speed from A to B = 10 m/(10 s) = 1 m/s

The distance traveled from A to B = 10 m

The time it takes to move from A to B = 10 seconds

From the point B to point C, we have;

The distance traveled from B to C = 0 m, (stationary)

The time it remains at point B distance from the start point = 10 seconds

The speed between point B to C = 0 m/(10 s) = 0 m/s

From the point C to point D, we have;

The distance traveled from C to D = 10 m

The time it takes to move from C to D = 5 seconds

The speed between point C and D = 10 m/(5 s) = 2 m/s

From the point D to point E, we have;

The distance traveled from D to E = 0 m, (stationary)

The time it remains at point D distance from the start point = 10 seconds

The speed between point D to E = 0 m/(10 s) = 0 m/s

From the point E to point F, we have;

The distance traveled from E to F = 20 m (return journey starts at point E)

The time it takes to move from E to F = 5 seconds

The speed between point E to F = 20 m/(5 s) = 4 m/s (Return journey)

Therefore, the journey started from point A, with a speed of 1 m/s for 10 seconds after which the it became stationary at 10 meters from the start point, for 10 seconds. The journey continued at a higher speed of 2 m/s for 5 seconds and halted (became stationary) again at 20 meters which was the maximum distance reached from the start point

The return journey to the start point started at the 20 meter mark and lasted for 5 seconds, at a speed of 4 m/s

The total distance moved, 'd', to and from the start point with reference to the graph is given as follows;

d = (From A to B) 10 m + (From B to C) 0 m + (From C to D) 10 m + (From D to E) 0 m + (From E to F) 20 m = 40 m

The total distance travelled in the journey is 40 m

The total displacement, $\underset{d}{\rightarrow}$ = 10 m + 10 m - 20 m = 0 m

7 0

3 years ago

A 22-turn circular coil of radius 3.00 cm and resistance 1.00 Ω is placed in a magnetic field directed perpendicular to the plan

zloy xaker [14]

Answer:

23.5 mV

Explanation:

number of turn coil 'N' =22

radius 'r' =3.00 cm=> 0.03m

resistance = 1.00 Ω

B= 0.0100t + 0.0400t²

Time 't'= 4.60s

Note that Area'A' = πr²

The magnitude of induced EMF is given by,

lƩl =ΔφB/Δt = N (dB/dt)A

=N[d/dt (0.0100t + 0.0400 t²)A

=22(0.0100 + 0.0800(4.60))[π(0.03)²]

=0.0235

=23.5 mV

Thus, the induced emf in the coil at t = 4.60 s is 23.5 mV

8 0

3 years ago