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3 years ago

Which of the following does Bernoulli's principle help to explain?

Physics

2 answers:

Kryger [21]3 years ago

7 0

B. Surface tension

Hope this helps

Anon25 [30]3 years ago

6 0

I would say B surface tension

You might be interested in

For the circuit shown, calculate

RSB [31]

For the circuit shown

a.the total resistance = R = 35.9 Ω

b. when total current = 2 A, then total voltage = V = 71.8 V

c.the current through resistor of resistance 56Ω = I₁ = 1.28 A

the current through resistor of resistance 100Ω = I₂ = 0.72 A

Explanation:

Resistors can be connected in series or in parallel. For series combination Resultant resistance of the circuit is given by

R = R₁ + R₂ + R₃ +...........+Rₙ

But is our case as shown in the picture, Both the resistors are connected in parallel and for parallel combination resultant, resultant resistance of the circuit is given by

(1 / R) = (1 / R₁) + (1 / R₂) + (1 / R₃) +.......+ (1 / Rₙ)

1 / R = (1 / R₁) + (1 / R₂)

Let

R₁ = 56Ω and R₂ = 100Ω

1 / R = 1/56 + 1/100

1 / R = 39 / 1400

R = 1400 / 39

R = 35.9 Ω

Part b can be solved by Ohm's Law Which is stated as "The current flowing through a circuit is directly proportional to the potential difference across its ends provided the physical state such as temperature of the conductor (circuit) does not change."

Mathematically

V ∝ I

V = IR

Where V is the potential difference across the ends of the conductor and I is the Current flowing through the circuit. R is the resistance of the circuit.

Given data:

Total current = I = 2 A

Resistance = R = 35.9 Ω

Voltage = ?

By Ohm's Law

V = IR

V = 2*35.9

V = 71.8 V

To solve current through each resistor, We can use current division formula which is given as

I₁ = I[ R₂ / (R₁ + R₂) ]

I₂ = I[ R₁ / (R₁ + R₂) ]

Also we can take Voltage across each resistor equal to V = 71.8 V because Potential difference across each resistors connected in parallel remain same . And by using Ohm's law divide the value of potential difference with the value of respective resistor to find the current through each resistors.

By Ohm's Law

V = IR

I = V / R

I₁ = 71.8 / 56

I₁ = 1.28 A

I₂ = V / R

I₂ = 71.8 / 100

I₂ = 0.72 A

Learn more about Resistors and Ohm's law from

https://brainly.in/question/9744300

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7 0

3 years ago

In a charging process, 4 × 1013 electrons are removed from one small metal sphere and placed on a second identical sphere. Initi

Alina [70]

Answer:

The distance between the two spheres is 914.41 X 10³ m

Explanation:

Given;

4 X 10¹³ electrons, and its equivalent in coulomb's is calculated as follows;

1 e = 1.602 X 10⁻¹⁹ C

4 X 10¹³ e = 4 X 10¹³ X 1.602 X 10⁻¹⁹ C = 6.408 X 10⁻⁶ C

V = Ed

where;

V is the electrical potential energy between two spheres, J

E is the electric field potential between the two spheres N/C

d is the distance between two charged bodies, m

$V = \frac{K*q}{d^2}*d = \frac{K*q}{d}$

$d = \frac{K*q}{V}$

where;

K is coulomb's constant = 8.99 X 10⁹ Nm²/C²

d = (8.99 X 10⁹ X 6.408 X 10⁻⁶)/0.063

d = 914.41 X 10³ m

Therefore, the distance between the two spheres is 914.41 X 10³ m

3 0

3 years ago

An infant's toy has a 120 g wooden animal hanging from a spring. If pulled down gently, the animal oscillates up and down with a

Morgarella [4.7K]

Answer:

0.37 m

Explanation:

The angular frequency, ω, of a loaded spring is related to the period, T, by

$\omega = \dfrac{2\pi}{T}$

The maximum velocity of the oscillation occurs at the equilibrium point and is given by

$v = \omega A$

A is the amplitude or maximum displacement from the equilibrium.

$v = \dfrac{2\pi A}{T}$

From the the question, T = 0.58 and A = 25 cm = 0.25 m. Taking π as 3.142,

$v = \dfrac{2\times3.142\times0.25\text{ m}}{0.58\text{ s}} = 2.71 \text{ m/s}$

To determine the height we reached, we consider the beginning of the vertical motion as the equilibrium point with velocity, v. Since it is against gravity, acceleration of gravity is negative. At maximum height, the final velocity is 0 m/s. We use the equation

$v_f^2 = v_i^2+2ah$

$v_f$ is the final velocity, $v_i$ is the initial velocity (same as v above), a is acceleration of gravity and h is the height.

$h = \dfrac{v_f^2 - v_i^2}{2a}$

$h = \dfrac{0^2 - 2.71^2}{2\times-9.81} = 0.37 \text{ m}$

3 0

3 years ago

The engineer of a passenger train traveling at 25.0 m/s sights a freight train whose caboose is 200 m ahead on the same track. T

zaharov [31]

a) The train collide after 22.5 seconds

b) The trains collide at the location x = 537.5 m

c) See graph in attachment

d) The freight train must have a head start of 500 m

e) The deceleration must be smaller (towards negative value) than $-0.25 m/s^2$

f) The two trains avoid collision if the acceleration of the freight train is at least $0.35 m/s^2$

Explanation:

We can describe the position of the passenger train at time t with the equation

$x_p(t)=u_p t + \frac{1}{2}at^2$

where

$u_p = 25.0 m/s$ is the initial velocity of the passenger train

$a=-0.100 m/s^2$ is the deceleration of the train

On the other hand, the position of the freight train is given by

$x_f(t)=x_0 + v_f t$

where

$x_0=200 m$ is the initial position of the freight train

$v_f = 15.0 m/s$ is the constant velocity of the train

The collision occurs if the two trains meet, so

$x_p(t)=x_f(t)\\u_pt+\frac{1}{2}at^2=x_0+v_ft\\25t+\frac{1}{2}(-0.100)t^2=200+15t\\0.050t^2-10t+200=0$

This is a second-order equation that has two solutions:

t = 22.5 s

t = 177.5 s

We are interested in the 1st solution, which is the first time at which the passenger train collides with the freight train, so t = 22.5 seconds.

In order to find the location of the collision, we just need to substitute the time of the collision into one of the expression of the position of the trains.

The position of the freight train is

$x_f(t)=x_0 +v_ft$

And substituting t = 22.5 s, we find:

$x_f(22.5)=200+(15)(22.5)=537.5 m$

We can verify that the passenger train is at the same position at the time of the collision:

$x_p(22.5)=(25.0)(22.5)+\frac{1}{2}(-0.100)(22.5)^2=537.5 m$

So, the two trains collide at x = 537.5 m.

In the graph in attachment, the position-time graph of each train is represented. We have:

The freight train is moving at constant speed, therefore it is represented with a straight line with constant slope (the slope corresponds to its velocity, so 15.0 m/s)
The passenger train has a uniformly accelerated motion, so it is a parabola: at the beginning, the slope (the velocity) is higher than that of the freight train, however later it decreases due to the fact that the train is decelerating

The two trains meet at t = 22.5 s, where the position is 537.5 m.

In order to avoid the collision, the freight train must have a initial position of

$x_0'$

such that the two trains never meet.

We said that the two trains meet if:

$x_p(t)=x_f(t)\\u_p t + \frac{1}{2}at^2=x_0' + v_f t$

Re-arranging,

$\frac{1}{2}at^2+(u_p-v_f)t-x_0'=0\\-\frac{1}{2}at^2+(v_f-u_p)t+x_0'=0$

Substituting the values for the acceleration and the velocity,

$0.05t^2-10t+x_0'=0$

The solution of this equation is given by the formula

$t=\frac{+10\pm \sqrt{10^2-4\cdot 0.05 \cdot x_0'}}{2(0.05)}$

The two trains never meet if the discrimant is negative (so that there are no solutions to the equation), therefore

$10^2-4\cdot 0.05 \cdot x_0'100\\x_0'>500 m$

Therefore, the freight train must have a head start of 500 m.

In this case, we want to find the acceleration $a'$ of the passenger train such that the two trains do not collide.

We solve the problem similarly to part d):

$x_p(t)=x_f(t)\\u_p t + \frac{1}{2}a't^2=x_0 + v_f t$

Re-arranging

$\frac{1}{2}a't^2+(u_p-v_f)t-x_0=0\\-\frac{1}{2}a't^2+(v_f-u_p)t+x_0=0$

Substituting,

$-0.5at^2-10t+200=0$

The solution to this equation is

$t=\frac{+10\pm \sqrt{10^2-4\cdot (-0.5a') \cdot (200)}}{2(0.05)}$

Again, the two trains never meet if the discriminant is negative, so

$10^2-4\cdot (-0.5a') \cdot (200)$

So, the deceleration must be smaller (towards negative value) than $-0.25 m/s^2$

In this case, the motion of the freight train is also accelerated, so its position at time t is given by

$x_f(t)=x_0 + v_f t + \frac{1}{2}a_ft^2$

where $a_f$ is the acceleration of the freight train.

Then we solve the problem similarly to the previous part: the two trains collide if their position is the same,

$x_p(t)=x_f(t)\\u_p t + \frac{1}{2}at^2=x_0 + v_f t+\frac{1}{2}a_ft^2$

Re-arranging,

$\frac{1}{2}(a_f-a)t^2+(v_f-u_p)t+x_0=0\\\\\frac{1}{2}(a_f-0.100)t^2-10t+200=0$

And the solution is

$t=\frac{+10\pm \sqrt{10^2-4\cdot (0.5a_f-0.05) \cdot (200)}}{2(0.5a_f-0.05)}$

Again, the two trains avoid collision if the discriminant is negative, so

$10^2-4\cdot (0.5a_f-0.05) \cdot (200)0.35 m/s^2$

Learn more about accelerated motion:

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8 0

3 years ago

Do you think there is water on exoplanets

Ksivusya [100]

Answer:

Explanation:

Some exoplanets may depending on the climate and vicinity from the sun.

8 0

3 years ago