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3 years ago

Please help me,its urgent!!

Physics

1 answer:

GREYUIT [131]3 years ago

4 0

Test:

Performing a Litmus Test

Result:

Litmus paper gives the user a general indication of acidity or alkalinity as it correlates to the shade of red or blue that the paper turns.

To test the pH of a substance, dip a strip of litmus paper into the solution or use a dropper or pipette to drip a small amount of solution onto the litmus paper.
Blue litmus paper can indicate an acid with a pH between 4 and 5 or lower.
Red litmus paper can show a base with a pH greater than 8.
If a solution has a pH between 5 and 8, it will show little color change on the litmus paper.
A base tested with blue litmus paper will not show any color change, nor will an acid tested with red litmus paper register a change in color.

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An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.220 rev/s . The magnitude

matrenka [14]

Answer:

1) The fan's angular velocity after 0.208 seconds is approximately 2.585 rad/s

2) The number of revolutions the blade has travelled in 0.208 s is approximately 0.066 revolutions

3) The tangential speed of a point on the tip of the blade at time t = 0.208 s is approximately 1.034 m/s

4) The magnitude of the tangential acceleration of a point on the tip of the blade at time t = 0.208 seconds is approximately 2.312 m/s²

Explanation:

The given parameters are;

The initial velocity of the fan, n = 0.220 rev/s

The magnitude of the angular acceleration = 0.920 rev/s²

The direction of the angular acceleration and the angular velocity = Clockwise

The diameter of the circle formed by the electric ceiling fan blades, D = 0.800 m

1) The initial angular velocity of the fan, ω₀ = 2·π × n = 2·π × 0.220 rev/s = 1.38230076758 rad/s

The angular acceleration of the fan, α = 2·π×0.920 rad/s² = 5.78053048261 rad/s²

The fan's angular velocity, 'ω', after a time t = 0.208 seconds has passed is given as follows;

ω = ω₀ + α·t

From which we have;

ω = 1.38230076758 rad/s + 5.78053048261 rad/s × 0.208 s = 2.58465110796 rad/s

The fan's angular velocity after 0.208 seconds is ω ≈ 2.585 rad/s

2) The number of revolutions the blade has travelled in the given time interval is given from the angle turned, 'θ', in the given time as follows;

θ = ω₀·t + 1/2·α·t²

θ = 1.38230076758 × 0.208 + 1/2 × 5.78053048261 × 0.208² = 0.41256299505 radians

2·π radians = 1 revolution

∴ 0.41256299505 radians = 0.41256299505 radian× 1 revolution/(2·π radian) = 0.06566144 revolution

The number of revolutions the blade has travelled in 0.208 s ≈ 0.066 revolutions

3) The tangential speed of a point on the tip of the blade at time t = 0.208 s is given as follows;

The tangential speed, $v_t$ = ω × r = ω × D/2

At t = 0.208 s, ω = 2.58465110796 rad/s, therefore, we have;

$v_t$ = ω × D/2 = 2.58465110796 × 0.800/2 = 1.0338604413

The tangential speed, $v_t$ = 1.0338604413 m/s

The tangential speed ≈ 1.034 m/s

4) The magnitude of the tangential acceleration of a point on the tip of the blade at time t = 0.208 seconds, 'a' is given as follows;

a = α × r = α × D/2

a = 5.78053048261 × 0.800/2 = 2.31221219304

The tangential acceleration, a ≈ 2.312 m/s²

4 0

3 years ago

Find the three longest wavelengths (call them λ1, λ2, and λ3) that "fit" on the string, that is, those that satisfy the boundary

zepelin [54]

If you have a string that is fixed on both ends the amplitude of the oscillation must be zero at the beginning and the end of the string. Take a look at the pictures I have attached. It is clear that our fundamental harmonic will have the wavelength of:
$\lambda_1=\frac{L}{2}$
All the higher harmonics are just multiples of the fundamental:
$\lambda_n=n\lambda_1\\ \lambda_n=n\frac{L}{2}$
Three longest wavelengths are:
$n=3; \lambda_3=\frac{3L}{2}\\
n=2; \lambda_2=L\\
n=1; \lambda_=\frac{L}{2}\\
$

6 0

4 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Karolina [17]

Answer:

1.00 m/s^2

Explanation:

The problem can be solved by using Newton's second law:

$F=ma$

where

F is the net force acting on an object

m is the mass of the object

a is its acceleration

In this problem, F=35 N (the force applied by Amber) and

$m=32 kg+3 kg=35 kg$

is the mass of the wagon + the brother, so the acceleration of the wagon and the brother is

$a=\frac{F}{m}=\frac{35 N}{35 kg}=1.00 m/s^2$

5 0

4 years ago

A rugby player passes the ball 6.8 m across the field, where the ball is caught at the same height as it left his hand. show ans

kherson [118]

Answer:

11.86°

Explanation:

Projectile motion is a form of motion where an object moves in parabolic path (trajectory). Projectile motion only occurs when there is one force applied at the beginning on the trajectory, after which the only interference is from gravity.

The range of an object experiencing projectile motion is given by:

R = u²sin(2θ) / g

where u is the initial velocity, θ is the angle with horizontal and g is the acceleration due to gravity

Given that R = 6.8 m, g = 10 m/s², u = 12 m/s.

R = u²sin(2θ) / g

substituting:

6.8 = 13²sin(2θ) / 10

13²sin(2θ) = 68

sin(2θ) = 0.4024

2θ = sin⁻¹(0.4024)

2θ = 23.73

θ = 11.86°

3 0

3 years ago

Como interactúan los Objetos cargados con otros otros odjetos cargados

almond37 [142]

Interactúan dos objetos cargados dependiendo de la carga domínate en el objeto, ya que cargas iguales se repelen y cargas opuestas se atraen. ... Al frotar un cuerpo cargado con otro objeto de carga neutra, este se cargara pero adquiriendo la carga opuesta al objeto cargado inicialmente.

5 0

3 years ago