PHYSICS HIGH SCHOOL

Stable system where forces cancel eachother out

Answers

Answer 1
Answer: Equilibrium - stable system where forces cancel each other out

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Within a single fiber, the tension developed during a twitch depends upon the
An arrow is fired horizontally at a target that is 40 m away (horizontal d). The arrow drops 0.5 m before striking the target. With what velocity is the arrow fired (what is the horizontal vi)?

The flywheel of a steam engine runs with a constant angular velocity of 140 rev/min. When steam is shut off, the friction of the bearings and of the air stops the wheel in 1.9 h. (a) What is the constant angular acceleration, in revolutions per minute-squared, of the wheel during the slowdown? (b) How many revolutions does the wheel make before stopping? (c) At the instant the flywheel is turning at 70.0 rev/min, what is the tangential component of the linear acceleration of a flywheel particle that is 40 cm from the axis of rotation? (d) What is the magnitude of the net linear acceleration of the particle in (c)?

Answers

Answer:

A) α = -1.228 rev/min²

B) 7980 revolutions

C) α_t = -8.57 x 10^(-4) m/s²

D) α = 21.5 m/s²

Explanation:

A) Using first equation of motion, we have;

ω = ω_o + αt

Where,

ω_o is initial angular velocity

α is angular acceleration

t is time the flywheel take to slow down to rest.

We are given, ω_o = 140 rev/min ; t = 1.9 hours = 1.9 x 60 seconds = 114 s ; ω = 0 rev/min

Thus,

0 = 140 + 114α

α = -140/114

α = -1.228 rev/min²

B) the number of revolutions would be given by the equation of motion;

S = (ω_o)t + (1/2)αt²

S = 140(114) - (1/2)(1.228)(114)²

S ≈ 7980 revolutions

C) we want to find tangential component of the velocity with r = 40cm = 0.4m

We will need to convert the angular acceleration to rad/s²

Thus,

α = -1.228 x (2π/60²) = - 0.0021433 rad/s²

Now, formula for tangential acceleration is;

α_t = α x r

α_t = - 0.0021433 x 0.4

α_t = -8.57 x 10^(-4) m/s²

D) we are told that the angular velocity is now 70 rev/min.

Let's convert it to rad/s;

ω = 70 x (2π/60) = 7.33 rad/s

So, radial angular acceleration is;

α_r = ω²r = 7.33² x 0.4

α_r = 21.49 m/s²

Thus, magnitude of total linear acceleration is;

α = √((α_t)² + (α_r)²)

α = √((-8.57 x 10^(-4))² + (21.49)²)

α = √461.82

α = 21.5 m/s²

If an alloyed guitar B string (147 Hz resonance) has an average diameter of .6 mm over an effective oscillating length of 65 cm, what is the tension in Newtons if the alloy is 10% aluminum and 90% copper? (assume the specific gravities of the metals are 2.6 and 8.9 grams per cubic centimeter, respectively)

Answers

Answer:

73.8 N

Explanation:

The total volume is,

V = (m_Al)/(P_Al) = (m_copper)/(P_copper)

=  (10m)/((100)(2.6)) = (90m)/((100)(8.9))

= 0.1396 m

The average  density is,

p = (m)/(V)

= (m)/(0.1396)

= 7.169 g/cm³

The linear mass density is,

μ = pπr²

= (7.169 x 10⁹) (π (0.3 x 10⁻³)²)

= 2.026 x 10⁻³ Kg/m

The fundamental mode of length is,

L =  λ/2

λ=2L

= 2 x 0.65

= 1.3 m

The speed of the wave is,

v = λf

= 1.3 m x 147 Hz

= 1.91 m/s

The tension is,

v =  √T/ц

T = ц v²

= 2.026 x 10⁻³)(1.91 m/s)²

= 73.769N

73.8N

Two objects are moving at equal speed along a level, frictionless surface. the second object has twice the mass of the first object. they both slide up the same frictionless incline plane. which object rises to a greater height?

Answers

Answer:

They both rises to same height.

Explanation:

When an object is sliding up in friction less surface than according to conservation of energy its potential energy will be converted into kinetic energy.

mgH=(1)/(2)mv^(2)\n v=√(2gH)

Here, m is the mass, v is the velocity, g is the acceleration due to gravity and H is the height.

Here the height is independent on the mass of an object and its only depend on velocity.

Now according to the question, two objects have same velocity but they have different masses.

Therefore, they rises to the same height because  height will not change with mass.

Both objects rise to the same height

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Further explanation

Let's recall Kinetic Energy Formula as follows:

\large {\boxed{Ek = (1)/(2)mv^2} }

Ek = Kinetic Energy ( Joule )

m = mass of the object ( kg )

v = speed of the object ( m/s )

Let us now tackle the problem !

\texttt{ }

Given:

initial speed of first object = initial speed of second object = v

final speed of first object = final speed of second object = 0

mass of first object = m

mass of second object = 2m

Asked:

height = H = ?

Solution:

We will use Conservation of Energy to solve this problem:

Ep_1 + Ek_1 = Ep_2 + Ek_2

0 + (1)/(2)mv^2 = mgH + 0

(1)/(2)v^2 = gH

v^2 = 2gH

\boxed {H = (v^2)/(2g)}

\texttt{ }

H_1 : H_2 = (v_1^2)/(2g) : (v_2^2)/(2g)

H_1 : H_2 = v_1^2 : v_2^2

H_1 : H_2 = v^2 : v^2

H_1 : H_2 = 1 : 1

\boxed{ H_1 = H_2 }

\texttt{ }

Conclusion:

Both objects rise to the same height

\texttt{ }

Learn more

\texttt{ }

Answer details

Grade: High School

Subject: Physics

Chapter: Energy

PLEASE HELP*******use the table to find the acceleration of the object for the time interval 1.0 to 4.0 seconds ​

Answers

Answer:

acceleration for time interval from 1 sec to 4 second is 1.5 metre/second^2

mark it as brainliest!!

What is velocity of a particle?

Answers

Velocity of a particle is its speed and the direction in which it is moving.

It can also be described as the instantaneous rate of change of the particle's
distance traveled, and the direction in which the distance is changing.
velocity is generally defined as the distance travelled by a particle in a particular time interval ...........its formula is given as
v=d/t
where v is velocity
         d is distance travelled and
         t is time interval.....
    its units are - kmph , m/s read as kilometers per second and meters per second

BAlpha Centauri has an apparent magnitude of -0.27, whereas the apparent
magnitude of Alpha Crucis is 0.77. Identify which star appears brighter when observed from
Earth. Explain your answer.

Answers

Answer:

Alpha centauri will be brighter than Alpha Crucis .

Explanation:

Apparent magnitude of a star measures how bright a star is .

This scale is reverse logarithmic ie , the brighter  the  star , the lower is its magnitude . A magnitude equal to 5 scale higher represents less magnitude by a factor of 1/ 100 . In this way a difference of 1 magnitude represents a brightness ratio of 2.512 . Hence a star of brightness magnitude of 7 is less bright by a factor 2.512  than that of a star magnitude of 6 .
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