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shepuryov [24]
3 years ago
9

Water boiling Which one is shown? (Look at pic)

Physics
1 answer:
Mkey [24]3 years ago
4 0

Answer:

conduction.

Explanation:

Hoped I helped! Im Eve btw have a great day and consider marking this brainliest if you do thank you in advanced!

You might be interested in
Un movil viaja a 40km/h y comienza a reducir su velocidad a partir del instante t=0. Al cabo de 6 segundo se detiene completamen
aleksklad [387]

Answer:

1,85 m / s²

Explanation:

De la pregunta anterior, se obtuvieron los siguientes datos:

Velocidad inicial (u) = 40 km / h

Hora inicial (t₁) = 0

Tiempo final (t₂) = 6 s

Velocidad final (v) = 0

Aceleración (a) =?

A continuación, convertiremos 40 km / ha m / s. Esto se puede obtener de la siguiente manera:

1 km / h = 0,2778 m / s

Por lo tanto,

40 km / h = 40 km / h × 0,2778 m / s / 1 km / h

40 km / h = 11,11 m / s

Por tanto, 40 km / h equivalen a 11,11 m / s.

Finalmente, determinaremos la aceleración del móvil durante el período en el que desaceleró. Esto se puede obtener de la siguiente manera:

Velocidad inicial (u) = 11,11 m / s

Hora inicial (t₁) = 0

Tiempo final (t₂) = 6 s

Velocidad final (v) = 0

Aceleración (a) =?

a = (v - u) / (t₂ - t₁)

a = (0 - 11,11) / (6 - 0)

a = - 11,11 / 6

a = –1,85 m / s²

Por tanto, la aceleración del móvil durante el período en el que se ralentizó es de –1,85 m / s²

6 0
3 years ago
n deep space, sphere A of mass 47 kg is located at the origin of an x axis and sphere B of mass 110 kg is located on the axis at
Volgvan

Answer:

a)-1.014x 10^{-7J

b)3.296 x  10^{-7J

Explanation:

For Sphere A:

mass 'Ma'= 47kg

xa= 0

For sphere B:

mass 'Mb'= 110kg

xb=3.4m

a)the gravitational potential energy is given by

U_{i = -GMaMb/ d

U_{i= - 6.67 x 10^{-11} x 47 x 110/ 3.4 => -1.014x 10^{-7J

b) at d= 0.8m (3.4-2.6) and U_{i=-1.014x 10^{-7J

The sum of potential and kinetic energies must be conserved as the energy is conserved.

K_{i + U_{i= K_{f + U_{f

As sphere starts from rest and sphere A is fixed at its place, therefore K_{i is zero

U_{i= K_{f + U_{f

The final potential energy is

U_{f= - GMaMb/d

Solving for 'K_{f '

K_{f = U_{i + GMaMb/d => -1.014x 10^{-7 + 6.67 x 10^{-11} x 47 x 110/ 0.8

K_{f = 3.296 x  10^{-7J

6 0
3 years ago
One of the world's largest Ferris wheels, the Cosmo Clock 21 with a radius of 50.0 m is located in Yokohama City, Japan. Each of
STatiana [176]

Answer:

a = 0.55 m / s²

Explanation:

The centripetal acceleration is given by the relation

         a = v² / r

angular and linear velocities are related

         v = w r

we substitute

          a = w² r

In the exercise they indicate the angular velocity w = 1 rev/min, let's reduce to the SI system

          w = 1 rev / min (2pi rad / 1rev) (1min / 60s) = 0.105 rad/ s

let's calculate

          a = 0.105² 50.0

          a = 0.55 m / s²

4 0
3 years ago
Blood in a carotid artery carrying blood to the head is moving at 0.15 m/s when it reaches a section where plaque has narrowed t
sp2606 [1]

Answer:

26.9 Pa

Explanation:

We can answer this question by using the continuity equation, which states that the volume flow rate of a fluid in a pipe must be constant; mathematically:

A_1 v_1 = A_2 v_2 (1)

where

A_1 is the cross-sectional area of the 1st section of the pipe

A_2 is the cross-sectional area of the 2nd section of the pipe

v_1 is the velocity of the 1st section of the pipe

v_2 is the velocity of the 2nd section of the pipe

In this problem we have:

v_1=0.15 m/s is the velocity of blood in the 1st section

The diameter of the 2nd section is 74% of that of the 1st section, so

d_2=0.74d_1

The cross-sectional area is proportional to the square of the diameter, so:

A_2=(0.74)^2 A_1=0.548 A_1

And solving eq.(1) for v2, we find the final velocity:

v_2=\frac{A_1 v_1}{A_2}=\frac{A_1 (0.15)}{0.548 A_1}=0.274 m/s

Now we can use Bernoulli's equation to find the pressure drop:

p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2

where

\rho=1025 kg/m^3 is the blood density

p_1,p_2 are the initial and final pressure

So the pressure drop is:

p_1 - p_2 = \frac{1}{2}\rho (v_2^2-v_1^2)=\frac{1}{2}(1025)(0.274^2-0.15^2)=26.9 Pa

8 0
3 years ago
Determine the angular speed, in rad/s of:
ryzh [129]

Answer:

Explanation:

A. The earth about its axis:

The earth makes one revolution in 24 hours. to know the number of revolutions per second it makes, we need to convert hours to seconds and the revolution to rad.

\frac{1 rev}{24hours}\times \frac{1 hr}{3600s}\times \frac{2\pi}{1}= 7.27 \times10^-5 rad/s

B. The minute hand of the clock makes one revolution in 60 minutes

To convert this to rad per second, we have

\frac{1 rev}{60mins}\times \frac{1 min}{60s}\times \frac{2\pi}{1}= 1.745 \times10^-3rad/s

C. The hour hand of  a clock completes one revolution in 12 hours

\frac{1 rev}{12hours}\times \frac{1 hr}{3600s}\times \frac{2\pi}{1}= 1.454 \times10^-4 rad/s

D. an egg beater turning at 300 rpm.

\frac{300 rev}{1minute}\times \frac{1 min}{60s}\times \frac{2\pi}{1}= 7.27 \times10^-5 rad/s=0.0218rad/s

6 0
3 years ago
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