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2 years ago

To measure the coefficient of kinetic friction by sliding a block down an inclined plane the block must be in equilibrium.

Physics

1 answer:

lozanna [386]2 years ago

8 0

Answer:

Explanation:

A block sliding down an inclined plane, is subject to two external forces along the slide.
One is the component of gravity (the weight) parallel to the incline.
If the inclined plane makes an angle θ with the horizontal, this component (projection of the downward gravity along the incline, can be written as follows:

$F_{gp} = m*g* sin \theta (1)$

(taking as positive the direction of the movement of the block)

The other force, is the friction force, that adopts any value needed to meet the Newton's 2nd Law.
When θ is so large, than the block moves downward along the incline, the friction force can be expressed as follows:

$F_{f} = \mu_{k} * N (2)$

The normal force, adopts the value needed to prevent any vertical movement through the surface of the incline:

$N = m*g* cos \theta (3)$

In equilibrium, both forces, as defined in (1), (2) and (3) must be equal in magnitude, as follows:

$m*g* sin \theta = \mu_{k} * m*g* cos \theta$

As the block is moving, if the net force is 0, according to Newton's 2nd Law, the block must be moving at constant speed.
In this condition, the friction coefficient is the kinetic one (μk), which can be calculated as follows:

$\mu_{k} = tg \theta$

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two students are on a balcony 19.6 m above the street. one student throws a ball vertically downward at 14.7 m:ds. at the same i

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A. The difference in the two ball's time in the air is 3 seconds

B. The velocity of each ball as it strikes the ground is 24.5 m/s

C. The balls 0.500 s after they are thrown are 14.7 m apart

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

<em>a = acceleration ( m/s² )</em>

<em>v = final velocity ( m/s )</em>

<em>u = initial velocity ( m/s )</em>

<em>t = time taken ( s )</em>

<em>d = distance ( m )</em>

Let us now tackle the problem!

<u>Given:</u>

Initial Height = H = 19.6 m

Initial Velocity = u = 14.7 m/s

<u>Unknown:</u>

A. Δt = ?

B. v = ?

C. Δh = ?

<u>Solution:</u>

<h2>Question A:</h2><h3>First Ball</h3>

$h = H - ut - \frac{1}{2}gt^2$

$0 = 19.6 - 14.7t - \frac{1}{2}(9.8)t^2$

$0 = 19.6 - 14.7t - 4.9t^2$

$4.9t^2 + 14.7t - 19.6 = 0$

$t^2 + 3t - 4 = 0$

$(t + 4)(t - 1) = 0$

$(t - 1) = 0$

$\boxed {t = 1 ~ second}$

<h3>Second Ball</h3>

$h = H + ut - \frac{1}{2}gt^2$

$0 = 19.6 + 14.7t - \frac{1}{2}(9.8)t^2$

$0 = 19.6 + 14.7t - 4.9t^2$

$4.9t^2 - 14.7t - 19.6 = 0$

$t^2 - 3t - 4 = 0$

$(t - 4)(t + 1) = 0$

$(t - 4) = 0$

$\boxed {t = 4 ~ seconds}$

The difference in the two ball's time in the air is:

$\Delta t = 4 ~ seconds - 1 ~ second$

$\large {\boxed {\Delta t = 3 ~ seconds} }$

<h2>Question B:</h2><h3>First Ball</h3>

$v^2 = u^2 - 2gH$

$v^2 = (-14.7)^2 + 2(-9.8)(-19.6)$

$v^2 = 600.25$

$v = \sqrt {600.25}$

$\boxed {v = 24.5 ~ m/s}$

<h3>Second Ball</h3>

$v^2 = u^2 - 2gH$

$v^2 = (14.7)^2 + 2(-9.8)(-19.6)$

$v^2 = 600.25$

$v = \sqrt {600.25}$

$\boxed {v = 24.5 ~ m/s}$

The velocity of each ball as it strikes the ground is 24.5 m/s

<h2>Question C:</h2><h3>First Ball</h3>

$h = H - ut - \frac{1}{2}gt^2$

$h = 19.6 - 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2$

$\boxed {h = 11.025 ~ m}$

<h3>Second Ball</h3>

$h = H + ut - \frac{1}{2}gt^2$

$h = 19.6 + 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2$

$\boxed {h = 25.725 ~ m}$

The difference in the two ball's height after 0.500 s is:

$\Delta h = 25.725 ~ m - 11.025 ~ m$

$\large {\boxed {\Delta h = 14.7 ~ m} }$

<h3>Learn more</h3>

Velocity of Runner : brainly.com/question/3813437
Kinetic Energy : brainly.com/question/692781
Acceleration : brainly.com/question/2283922
The Speed of Car : brainly.com/question/568302

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle

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2 years ago

in which sealed container would the organisms be able to continuously cycle o2 and co2 gases? a. a jar with a lizard eating a sn

DerKrebs [107]

Correct answer choice is :

C) A jar with snails crawling on living plants

Explanation:

By using the power of sunlight, plants can change carbon dioxide and water into glucose and oxygen in a method described photosynthesis. As photosynthesis needs sunlight, this method only occurs during the day. We usually like to think of this as plants `exhaling in carbon dioxide and `breathing out oxygen. Gas transfer between Alveolar Spaces and Capillaries. The role of the respiratory system is to transfer two gases, oxygen and carbon dioxide. The transfer takes place in the millions of alveoli in the lungs and the capillaries that surround them.

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2 years ago

Read 2 more answers

Submit Quiz

kkurt [141]

Explanation:

the question is unanswerable

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2 years ago

A helicopter (m = 3250 kg) is cruising at a speed of 56.9 m/s atan altitude of 185 m. What is the total mechanical energy of the

taurus [48]

Answer:

The mechanical energy of the helicopter is $1.12\times 10^7\ J$ .

Explanation:

It is given that,

Mass of the helicopter, m = 3250 kg

Speed of the helicopter, v = 56.9 m/s

Position of the helicopter, h = 185 m

The energy possessed by an object due to its motion is called its kinetic energy. It is given by :

$E=\dfrac{1}{2}mv^2$

$E=\dfrac{1}{2}\times 3250\times (56.9)^2$

$E=5.26\times 10^6\ J$

The energy possessed by an object due to its position is called its potential energy. It is given by :

$E=mgh$

$E=3250\times 9.8\times 185$

$E=5.89\times 10^6\ J$

The sum of kinetic and potential energy is called mechanical energy of the system. It is given by :

$M=5.26\times 10^6+5.89\times 10^6$

$M=11.15\times 10^6\ J$

$M=1.12\times 10^7\ J$

So, the mechanical energy of the helicopter is $1.12\times 10^7\ J$ . Hence, this is the required solution.

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2 years ago

To practice Problem-Solving Strategy 23.2 for continuous charge distribution problems. A straight wire of length L has a positiv

Lesechka [4]

Answer:

E = k Q / [d(d+L)]

Explanation:

As the charge distribution is continuous we must use integrals to solve the problem, using the equation of the elective field

E = k ∫ dq/ r² r^

"k" is the Coulomb constant 8.9875 10 9 N / m2 C2, "r" is the distance from the load to the calculation point, "dq" is the charge element and "r^" is a unit ventor from the load element to the point.

Suppose the rod is along the x-axis, let's look for the charge density per unit length, which is constant

λ = Q / L

If we derive from the length we have

λ = dq/dx ⇒ dq = L dx

We have the variation of the cgarge per unit length, now let's calculate the magnitude of the electric field produced by this small segment of charge

dE = k dq / x²2

dE = k λ dx / x²

Let us write the integral limits, the lower is the distance from the point to the nearest end of the rod "d" and the upper is this value plus the length of the rod "del" since with these limits we have all the chosen charge consider

E = k $\int\limits^{d+L}_d {\lambda/x^{2}} \, dx$

We take out the constant magnitudes and perform the integral

E = k λ (-1/x) ${(-1/x)}^{d+L} _{d}$

Evaluating

E = k λ [ 1/d - 1/ (d+L)]

Using λ = Q/L

E = k Q/L [ 1/d - 1/ (d+L)]

let's use a bit of arithmetic to simplify the expression

[ 1/d - 1/ (d+L)] = L /[d(d+L)]

The final result is

E = k Q / [d(d+L)]

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