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3 years ago

As a person squeezes and applies pressure to the outside of a balloon, air particles inside

1 answer:

7 0

The air particle inside the balloon will collide more with each other and the temperature inside the balloon will increase.

As a person squeezed and applies the pressure to the outside of a balloon, the air particle inside the balloon gains energy and collide with each other, the particle of the air also try leave the balloon surface will implies equal pressure on the wall of the balloon, as the pressure outside the balloon increase, the inside pressure will also increase.

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Why is learning by assimilation considered to be so effective.

emmainna [20.7K]

<span>Learning by assimilation is so effective for it involves a concise and logical approach. Most people who require clear understanding and explanation of ideas, information and abstract concepts must use this approach of learning. The reason is that, they are more attracted to logically sound theories rather than more focus on people approach and based on practical value. </span>

In assimilating learning, people prefer readings, listening to lectures, exploring analytical models, and having time to think things through on their own. With this, all the important concept and data is clearly inculcated and stored in mind.

6 0

4 years ago

A thin, uniformly charged spherical shell has a potential of 832 V on its surface. Outside the sphere, at a radial distance of 2

Zolol [24]

Answer:

The radius is $r_1 = 0.315m$

The total charge is $Q= 2.912*10^{-8}C$

The electric potential inside sphere is the same with that outside the sphere which implies that the electric potential is 832 V

The magnitude of the electric field is $E= 2641.3 V/m$

The velocity is $v= 1.76 *10^{14} m/s$

Explanation:

From the question we are told that

The potential is $V_1 = 832 V$

The radial distance from the sphere is $d = 21.0cm = \frac{21}{100} = 0.21m$

The potential at the radial distance is $V_2 = 499V$

The potential at the surface of the sphere is mathematically represented as

$V = \frac{kQ}{r}$

$Vr = kQ$

Where kQ is a constant what this means that the the charge Q and the coulomb constant do not change

This means that

$V_1 r_1 = V_2 r_2$

Where $r_1$ is the radius of the sphere

and $r_2$ is the distance from that point where the second potential was measured to the center of the sphere which is mathematically represented as

$r_2 = r_1 + d$

Substituting this into the equation

$v_1 r_1 = V_2 (r_1 +d)$

Now substituting value

$832 * r_1 = 499 * (r_1 + 0.21)$

$832r_1 - 499r_1 = 104.79$

$333r_1 = 104.79$

$r_1 = \frac{104.79}{333}$

$r_1 = 0.315m$

From the equation above

$V = \frac{kQ}{r_1}$

making Q the subject

$Q = \frac{V r_1 }{k}$

k has a values of $k = 9*10^9 \ kg\cdot m^3 \cdot s^{-4} \cdot A^{-2}$

Substituting into the equation

$Q =\frac{832 * 0.315}{9*10^9}$

$Q= 2.912*10^{-8}C$

According to Gauss law the electric field from outside a sphere is taken to be an electric field from a point charge this mean that the potential outside a sphere is also taken as electric potential inside a sphere

The magnitude of a electric field from a sphere (point charge ) is mathematically represented as

$E = \frac{kQ}{r_1^2}$

Substituting values

$E = \frac{9*10^{9} * 2.912*10^{-8}}{0.315^2}$

$E= 2641.3 V/m$

According the the law of energy conservation

The electric potential energy at the point outside the surface where the second potential was measured(21 cm from the sphere surface) = The electric potential energy at the surface + The kinetic energy of the charge (electron) at that the surface

Generally Electric potential energy is mathematically represented as

$EPE = V * e$

Where is e is an electron

And Kinetic energy is mathematically represented as

$KE = \frac{1}{2} m v^2$

From the statement above

$V_2 e = V_1 e + \frac{mv^2}{2}$

But from the question we can deduce that the potential at the surface is zero

So the equation becomes

$V_2 e = \frac{mv^2}{2}$

The charge an electron has a value $e = 1.602*10^{-19}C$

And the mass of an electron is $m = 9.109 *10^{-31}kg$

Making v the subject

$v = \sqrt{\frac{2V_2 e}{m} }$

Substituting value

$v = \sqrt{\frac{2 * 499 * 1.602 *10^{-19}}{9.109*10^{-31}} }$

$v= 1.76 *10^{14} m/s$

7 0

3 years ago

What is instantaneous acceleration?

KengaRu [80]

Answer:

how fast the car goes

Explanation:

3 0

4 years ago

Read 2 more answers

Â (a) How long can you rapidly climb stairs (116/min) on the 166.0 kcal of energy in a 200g apple? (b) How many flights is this

adoni [48]

Answer:

a) It will take you 16.94 min to climb the stairs.

b) The numbers of flights are 123 flights.

Explanation:

a) Let the energy consumption while climbing the stairs be 685 W in kcal/min is given by:

rate of energy consumption = (685 W)×[(0.01433kcal/min)/1 W]

= 9.81605 kcal/min

the time it takes to climb the stairs is given by:

t = E/P

= (166.0 kcal)/(9.8 kcal/min)

= 16.94 min

Therefore, it will take you 16.94 min to climb the stairs.

b) Given t is the time it takes to climb stairs and N is the number of stairs to climb per min, then the number of stairs is:

n = t×N

= (16.94)×(116)

= 1965 stairs

given the number of stairs each flight as N1, then the number of flights is given by:

flights = n/N1

= (1965)/(16)

= 123 flights

Therefore, the numbers of flights are 123 flights.

7 0

4 years ago

Lions can run at speeds up to approximately 80.0 km / h. A hungry 109 kg lion running northward at top speed attacks and holds o

sukhopar [10]

Answer:

17.34 m/s

Explanation:

Given:

Mass of lion (m₁) = 109 kg

Initial speed of lion (v₁) = 80.0 km/h (Northward direction)

Mass of gazelle (m₂) = 39.0 kg

Initial speed of gazelle (v₂) = 78.5 km/h (Eastward direction)

Final velocity of both lion and gazelle is, $v_f=?$

First, let us convert the speeds from km/h to m/s using the conversion factor.

We know that, 1 km/h = 5/18 m/s

Therefore,

$v_1= 80.0\ km/h=80\times \frac{5}{18}=22.22\ m/s\\\\v_2=78.5\ km/h=78.5\times \frac{5}{18}=21.81\ m/s$

Now, the concept of conservation of total momentum is used here as this is a case of perfectly inelastic collision. In inelastic collision, the masses move together with same velocity after collision.

Here, as the lion and gazelle are moving in directions at right angles to each other, the vector sum of their momentums will give the net initial momentum of the system.

So, initial momentum is given as:

$P_i=\sqrt{P_1^2+P_2^2}\\\\Where,\\\\P_1\to initial\ momentum\ of\ lion\\P_2\to initial\ momentum\ of\ gazelle$