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kondor19780726 [428]
3 years ago
11

An underwater air bubble has an excess inside pressure of 13 pa. what is the excess pressure inside an air bubble with twice the

radius?
Physics
1 answer:
Evgesh-ka [11]3 years ago
6 0
Pressure = 2 * pi * sigma * radius where sigma is the pressure constant of the medium or the material.

Thus, Pressure/Radius = constant

Based on this:
(pressure*radius) in case 1 = (pressure*radius) in case 2
(13 * R) = (P * 2R)
P = 13/2 = 6.5 pa
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Only if a force acts upon it, it can move.
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3 years ago
it is determined that a certain light wave has a wavelength of 3.012x10^-12 m. the light travels at 2.99x10^8 m/s. what is the f
Dahasolnce [82]

Answer:

9.93\times 10^{19}Hz

Explanation:

Speed of light is the product of its wavelength and frequency, expressed as

S=fw

Where s represent speed, f is frequency while w is wavelength

Making f the subject of the formula then

f=s/w

Substituting 2.99x10^8 m/s for s and 3.012x10^-12 m for w then

f=\frac {2.99\times 10^{8}}{3.012\times 10^{-12}}=9.926958831341\times 10^{19}\\f\approx 9.93\times 10^{19}Hz

Therefore, the frequency equals to 9.93\times 10^{19}Hz

4 0
3 years ago
The front 1.20 m of a 1,600-kg car is designed as a "crumple zone" that collapses to absorb the shock of a collision. (a) If a c
eimsori [14]

To develop the problem it is necessary to apply the kinematic equations for the description of the position, speed and acceleration.

In turn, we will resort to the application of Newton's second law.

PART A) For the first part we look for the time, in a constant acceleration, knowing the speeds and the displacement therefore we know that,

X_f = X_i +\frac{1}{2}(V_i+V_f)t

Where,

X = Desplazamiento

V = Velocity

t = Time

In this case there is no initial displacement or initial velocity, therefore

X_f = \frac{1}{2} (V_i+V_f)t

Clearing for time,

t = \frac{2X_f}{(V_i+V_f)}

t = \frac{2*1.2}{24+0}

t = 0.1s

PART B) This is a question about the impulse of bodies, where we turn to Newton's second law, because:

F = ma

Where,

m=mass

a = acceleration

Acceleration can also be written as,

a= \frac{\Delta V}{t}

Then

F = m\frac{\Delta V}{t}

F = m\frac{V_f-V_i}{t}

F = m\frac{-V_i}{t}

F = \frac{(1600kg)(-24m/s)}{(0.1s)}

F = -384000N

Negative symbol is because the force is opposite of the direction of moton.

PART C) Acceleration through kinematics equation is defined as

V_f^2=V_i^2-2ax

0 = (24m/s)^2-2*a(1.2m)

a = \frac{(24m/s)^2}{1.2m}

a=480m/s^2

The gravity is equal to 0.8, then the acceleration is

a = 480*\frac{g}{9.8}

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3 0
3 years ago
A spaceship ferrying workers to Moon Base I takes a straight-line path from the earth to the moon, a distance of 384,000 km. Sup
Tema [17]

Answer:

a) v = 19,149.6 m/s

b) f = 95%

c) t = 346.5min

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The equation of motion you need is:

v_f = a*t+v_0

where v_f is the final velocity, a is acceleration and t is time in hours.

Since the spaceship starts from 0 velocity:

v_f = a*t = 20.2*948 = 19,149.6 m/s

Next, you need to calculate the distances traveled on each interval, considering that both starting and final intervals travel the same distance because the acceleration and time are equal. For this part you need the next motion equation:

x=\frac{v_0+v_f}{2}t

solving for first and last interval:

Since the spaceship starts and finish with 0 velocity:

x=\frac{v}{2}t=\frac{19,149.6}{2}948=9,076,910.4m=9,076.9104km

Then the ship traveled 384,000-9,076.9104*2 = 361,846.1792km at constant speed, which means that it traveled:

f_{constant_speed} =\frac{ x_{constant_speed}}{x_total} =\frac{361,846.1792}{380,000} =0.95

Which in percentage is 95% of the trip.

to calculate total time you need to calculate the time used during constant speed:

t = \frac{361,846,179.2}{19,149.6} = 18,895.75s = 314min

That added to the other interval times:

t_{total} = t_1+t_2+t_3=15.8+314.93+15.8=346.5min

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A magnet is a substance which attracts or repels another substance. In a magnet, the atoms are aligned in a particular direction in domains. A magnet has two poles: North pole and South pole. The domains are oppositely aligned in unlike poles. Like poles repel each other where as unlike poles attract each other. Hence, when we bring like poles closer, repulsion would be experienced. In case of unlike poles, they would stick together.

6 0
3 years ago
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