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2 years ago

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A block of mass 12.2 kg is sliding at an initial velocity of 3.9 m/s in the positive x-direction. The surface has a coefficient

of kinetic friction of 0.173. (indicate the direction with signs of your answer) (a) What is the force of kinetic friction in N acting on the block? (b) What is the block's acceleration in /s^2? (c) How Far will it slide (in m) before coming to rest? Plz answer as soon as possible

1 answer:

Studentka2010 [4]2 years ago

5 0

Answer:

Explanation:

a) Force of friction = μ R where μ is coefficient of kinetic friction and R is reaction force

R = mg where m is mass of the block

Force of friction F = μ x mg

= .173 x 12.2 x 9.8

= 20.68 N

b ) Only force of friction is acting on the body so

deceleration = force / mass = 20.68 / 12.2 = 1.7 m /s²

acceleration = - 1.7 m /s²

c )

v² = u² - 2 a s

v = 0 , u = 3.9 m /s

a = 1.7 m /s

0 = 3.9² - 2 x 1.7 x s

s = 4.47 m

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A car's velocity changes from 40 m/s to 30 m/s in 40 seconds. What is the

valina [46]

Answer:

- 0.25m/s²

Explanation:

Given parameters:

Initial velocity = 40m/s

Final velocity = 30m/s

Time taken = 40s

Unknown:

Acceleration of the car = ?

Solution:

Acceleration is the rate of change of velocity with time.

So;

Acceleration = $\frac{Final velocity - Initial velocity }{time }$

Acceleration = $\frac{30 - 40}{40}$ = - 0.25m/s²

This implies that the car will decelerate at a rate of 0.25m/s²

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3 years ago

A 4-kg hammer is lifted to a height of 10 m and dropped from rest. What was the velocity (in m/s) of the hammer when it was at a

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Answer:

v = 10.84 m/s

Explanation:

using the equation of motion:

v^2 = (v0)^2 + 2×a(r - r0)

<em>due to the hammer starting from rest, vo = 0 m/s and a = g , g is the gravitational acceleration.</em>

v^2 = 2×g(r - r0)

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= 10.84 m/s

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2 years ago

A wire with radius 23 cm has a current of 7 A which is distributed uniformly through its cross sectional area. If you were to us

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Answer:

The magnetic induction of the magnetic field is 0.0005293 mT

Explanation:

Data given

I = 7 A = the total current in the wire

r = 23 cm = the radius of the wire = 0.23 meter

r' = 2cm = the measurement point, which should be inside the wire = 0.02 meter

Let's consider the current density is constant in the wire, ⇒ the current enclosed is a function of the enclosed area

I(enclosed) = Jπ r ²

we can consider the current density as the total current over the whole area:

I(enclosed) = I / (πr ²) * πr' ²

I(enclosed) = (I* r'²)/ (r ²)

with I = total current in the wire = 7A

With r = the radius of wire = 0.23 meter

with r' = the distance of point from the center of wire 0.02 meter

We plug this into ampere's law:

∮ *B *dl =μ 0 * (I* r'²)/ (r ²)

with B = Magnetic flux density (in Tesla) or magnetic induction

with dl = an infinitesimal element (a differential) of the curve C

with µ0 = the magnectic constant = 4π*10^−7 H/m

We can simplify this, by using an Amperian loop can write this as:

B *( 2 π r') = μ 0 * (I* r'²)/ (r ²)

Because the circumference of a circle is 2 π r , when we integrate over length at a distance r ′ from the center of wire whose crossection is a circle we get 2 π r ′

When we isolate B, we get:

B = µo *(Ir'/2 π r ²)

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3 years ago

"a grindstone of radius 4.0 m is initially spinning with an angular speed of 8.0 rad/s. the angular speed is then increased to 1

kotegsom [21]

The average angular speed of the grindstone is 10 rad/s

$\texttt{ }$

<h3>Further explanation</h3>

<em>Let's recall </em><em>Angular Speed</em><em> formula as follows:</em>

$\boxed{ \omega = \omega_o + \alpha t }$

$\boxed{ \theta = \omega_o t + \frac{1}{2} \alpha t^2 }$

$\boxed{ \omega^2 = \omega_o^2 + 2 \alpha \theta }$

$\boxed{ \theta = \frac{( \omega + \omega_o )}{2} t }$

<em>where :</em>

<em>ω = final angular speed ( rad/s )</em>

<em>ω₀ = initial angular speed ( rad/s )</em>

<em>α = angular acceleration ( rad/s² )</em>

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

<em>θ = angular displacement ( rad )</em>

$\texttt{ }$

<u>Given:</u>

radius of the grindstone = R = 4.0 m

initial angular speed = ω₀ = 8.0 rad/s

final angular speed = ω = 12 rad/s

elapsed time = t = 4.0 seconds

<u>Asked:</u>

average angular speed = ?

<u>Solution:</u>

<em>Firstly , we will calculate </em><em>angular displacement </em><em>as follows:</em>

$\theta = \frac{( \omega + \omega_o )}{2} t$

$\theta = \frac{ ( 12 + 8.0 ) }{2} \times 4.0$

$\theta = 10 \times 4.0$

$\boxed {\theta = 40 \texttt{ rad}}$

$\texttt{ }$

<em>Next , we could calculate the </em><em>average angular speed</em><em> as follows:</em>

$\texttt{average angular speed} = \theta \div t$

$\texttt{average angular speed} = 40 \div 4.0$

$\boxed{\texttt{average angular speed} = 10 \texttt{ rad/s}}$

$\texttt{ }$

<h3>Learn more</h3>

Impacts of Gravity : brainly.com/question/5330244
Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
The Acceleration Due To Gravity : brainly.com/question/4189441
Moment of Inertia : brainly.com/question/13796477
The Ratio of the Moments of Inertia : brainly.com/question/2176655

$\texttt{ }$

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Rotational Dynamics

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3 years ago

Read 2 more answers

The Mars Curiosity rover was required to land on the surface of Mars with a velocity of 1 m/s. Given the mass of the landing veh

Aliun [14]

Answer:

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Explanation:

From the question we are told that

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The drag coefficient is $C__{D}} = 0.5$

The atmospheric density of Earth is $\rho = 1.2 \ kg/m^3$

The acceleration due to gravity in Mars is $g_m = 3.689 \ m/s^2$

Generally the Mars atmosphere density is mathematically represented as

$\rho_m = 0.71 * \rho$

=> $\rho_m = 0.71 * 1.2$

=> $\rho_m = 0.852 \ kg/m^3$

Generally the drag force on the rover and the parachute is mathematically represented as

$F__{D}} = m * g_{m}$

=> $F__{D}} = 2270 * 3.689$

=> $F__{D}} = 8374 \ N$

Gnerally this drag force is mathematically represented as

$F__{D}} = C__{D}} * A * \frac{\rho_m * v^2 }{2}$

Here A is the frontal area

So

$A = \frac{2 * F__D }{ C__D} * \rho_m * v^2 }$

=> $A = \frac{2 * 8374 }{ 0.5 * 0.852 * 1 ^2 }$

=> $A = 39315 \ m^2$

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3 years ago