When Marcel finds the distance L from the previous part, it turns out to be greater than Lend, the distance from the pivot to the end of the seesaw. Hence, even with Jacques at the very end of the seesaw, the twins Gilles and Jean exert more torque than Jacques does. Marcel now elects to balance the seesaw by pushing sideways on an ornament (shown in red) that is at height h above the pivot. (Figure 3)With what force in the rightward direction, Fx, should Marcel push? If your expression would give a negative result (using actual values) that just means the force should be toward the left.Express your answer in terms of W, Lend, w, L2, L3, and h.

Answer:

1.48 g

Explanation:

A = A₀ (½)^(t / T)

where A is the final amount,

A₀ is the initial amount,

t is time,

and T is the half life.

A = (5.00 g) (½)^(5.00 a / 2.85 a)

A = 1.48 g

Answer:

Explanation:

If there is only 1 force, the body can never be in equilibrium, providing that the force is not zero (and that would hardly be a force. Zero is possible in math and it means something. It is debatable in physics).

You cannot think of a condition where something is stationary on planet earth and there are not 2 forces or more forces involved.

Think of something like a block of wood sitting on a table. It is not moving, we'll say. Gravity is holding it down, but what is pushing up on it?

The table is. There are 2 forces and they are equal in magnitude, but opposite in direction. That matters.

' A ' looks like the best choice.

Answer:

B.  inverse plot, 0.51 kilograms/meter3

Explanation:

Answer:

A. Ein = 8.05*10^-4 V/m

B. Clockwise sense

Explanation:

A. the magnitude of the electric field induced in the ring is obtaind by using the following formula:

[tex]\int E_{in} \cdot ds=-\frac{d\Phi_B}{dt}[/tex]            (1)

Ein: induced electric field

ds: differential of a path of the ring

ФB: magnetic flux in the ring

The Ein vector is parallel to ds in the complete ring. Furthermore, the area of the ring is constant, hence, you have in the equation (1):

[tex]\int E_{in}ds=E_{in}(2\pi r)=-A\frac{dB}{dt}\\\\E_{in}=-\frac{A}{2\pi r}\frac{dB}{dt}[/tex]   (2)

dB/dt = -0.280T/s     (it is decreasing)

A: area of the ring = π(r/2)^2= (π/4) r^2

r: radius of the ring = 4.60/2 = 2.30 cm

Then, you replace the values of all variables in the equation (2):

[tex]E_{in}=-\frac{(\pi/4)r^2}{2\pi r}\frac{dB}{dt}=\frac{r}{8}\frac{dB}{dt}\\\\E_{in}=-\frac{0.0230m}{8}(-0.280T)=8.05*10^{-4}\frac{V}{m}[/tex]

hence, the induced electric field is 8.05*10^-4 V/m

B. The induced current in the ring produced a magnetic field that is opposite to the magnetic field of the magnet. The, in this case you have that the induced current is in a clockwise sense.

Answer:

D

Explanation:

Now the net force is the applied force minus the frictional force; this is expressed mathematically as:

Fnet= Fappplied - Ffrictional

Now the frictional force is given as ;

Coefficient of friction × normal reaction

Normal reaction is the weight of the human acting in opposite direction.

Normal reaction of the human is ;

35 × 9.8 = 343N { note that weight = m× g and g= 9.8m/S2, a known standard }

Hence the Frictional force =343×0.4 =137.20N

Hence Fnet = 200-137.20 = 62.8N

Fnet = 63N to the nearest whole

The net force on the couch as it slides is  63N.

What is frictional force?

When an object is moving on a rough surface, it experiences opposition. This opposing force is called the friction force.

The friction force is given by

f = coefficient of friction x Normal force

Given, the coefficient of kinetic friction between a couch and the floor is 0.4. If the couch has a mass of 35 kg and you push it with a force of 200 N.

Normal reaction is the weight of the human acting in opposite direction.

Normal reaction N =35 × 9.8 = 343N

Frictional force f =0.4 x 343

                          f =137.20N

The net force will be

Fnet= Fappplied - Ffrictional

Fnet = 200-137.20 = 62.8N

Fnet = 63N

Thus,  the net force on the couch as it slides is  63N.

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Answer:

D. Velocity of the car

Explanation:

The centripetal force acting on the car is given by the following formula:

[tex]F_c=ma_c=m\frac{v^2}{r}[/tex]    (1)

m: mass of the car  = 888 kg

v: tangential speed of the car = 7 m/s

r: radius of the flat circular track = 59 m

By the form of the equation (1) you can notice that the greatest change in the centripetal force is obtained when the velocity v is twice. In fact, you have:

[tex]F_c=m\frac{(2v)^2}{r}=4m\frac{v^2}{r}=4F_c[/tex]

Then, the greatest values of the centripetal force is:

[tex]F_c=4(888kg)\frac{(7m/s)^2}{59m}=2949.96N[/tex]

The greatest change in Fc is obtained by changing the value of the speed

answer

D. Velocity of the car

Answer:

Explanation:

moment of inertia of the rod = 1/3 mL² , m is mass and L is length of rod.

1/3 x .154 x .35²

= .00629

moment of inertia  of putty about the axis of rotation

= m d² , m is mass of putty and d is distance fro axis

= .011 x( .35 / 3 )²

= .00015

Total moment of inertia I = .00644 kgm²

angular momentum of putty about the axis of rotation

= mvRsinθ

m is mass , v is velocity , R is distance where it strikes the rod and θ is angle  with the rod at which putty strikes

= .011 x 9 x .35 / 3 x sin 29

= .0056

Applying conservation of angular momentum

angular momentum of putty = angular momentum of system after of collision

.0056 =  .00644 ω where ω is angular velocity of the rod after collision

ω = .87 rad /s .

Rotational energy

= 1/2 I ω²

I is total moment of inertia

=  .5 x .00644 x .87²

= 2.44 x 10⁻³ J .

Answer:

0

Explanation:

if the volume remains constans, the works is 0 because the equation

W = P . ∆V

P = pressure

∆V = change in volume

Answer:

Increased blood circulation to the body.

Explanation:

plato/edmentum

Answer:

a.[tex]5.66ms^{-2}[/tex]

b.55 m

Explanation:

We are given that

Mass ,m=0.9 kg

Length of string,l=50 cm=[tex]\frac{50}{100}=0.50 m[/tex]

1 m=100 cm

[tex]\theta=30^{\circ}[/tex]

R=55 m

a.Centripetal acceleration

[tex]a_c=gtan\theta[/tex]

[tex]a_c=9.8tan30^{\circ}[/tex]

[tex]a_c=5.66 m/s^2[/tex]

Hence, the radial acceleration of the bus=[tex]5.66m/s^2[/tex]

b. Radius of curve,R=55 m

Answer:

Everytime you do an experiment you need something that is regular. For example if you try and measure how much germs spread in bread. you need 1 bread thats clean and 3 different breads for different molds. So thats called a CONTROL

AAAAAAAAAAAA is the answer

Answer:

They had the same speed.

Explanation:

It won't be velocity, because velocity is a vector quantity. Speed is scalar.

Velocity is the rate of change of displacement. During their periods of travel, the cars definitely had the same velocity.

Velocity is the directional speed of a moving object as an indicator of its rate of change in location as perceived from a certain frame of reference and measured by a specific time standard.

Given that the first car travels 40 meters due east in 5 seconds. Therefore, we can write,

Distance = 40 meters

Time = 5 seconds

Velocity = Distance / Time = 40 meter/ 5 sec = 40 m/sec

Also, given that the second car travels 64 meters due west in 8 seconds. Therefore, we can write,

Distance = 64 meters

Time = 8 seconds

Velocity = Distance / Time = 64 meter/ 8 sec = 8 m/sec

Hence, During their periods of travel, the cars definitely had the same velocity.

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Answer:

Explanation:

  P = VI

  I = P/V = P/120

  R = V/I = V/(P/V) = V^2/P = 14400/P

Toaster: I = 1800/120 = 15 . . . amps

  R = 14400/1800 = 8 . . . ohms

Fry pan: I = 1300/120 = 10.833 . . . amps

  R = 14400/1300 = 11.08 . . . ohms

Lamp: I = 100/120 = 0.833 . . . amps

  R = 14400/100 = 144 . . . ohms

The total current exceeds 20 A, so will blow the fuse.

Answer:

10 Sig Figs

Explanation:

Just start counting at the first non zero after the decimal so in this case the nine, and count all of the numbers including zeros after that.

Answer:

Explanation:

Givens

vi = 20 m/s

vf = 28 m/s

t = 2 seconds

Formula

a = (vf - vi) / t

Solution

a = (28 - 20)/2

a = 8/2

a = 4 m/s^2

Answer:

a. the initial velocity of the cat is 5.95 m/s at 60.2° from the horizontal

b. 0.847 m

Explanation:

a. Using v² = u² + 2as, we find the initial vertical velocity of the cat. Now at the peak height, v = final velocity = 0, u = initial velocity and a = -g = 9.8 m/s², s vertical distance travelled by the cat from its position on the cabinet = Δy = 3.130 m - 1.765 m = 1.365 m.

Substituting these variables into the equation, we have

0² = u² + 2(-9.8m/s²) × 1.365 m

-u² = -26.754 m²/s²

u = √26.754 m²/s²

u = 5.17 m/s

To find its initial horizontal velocity, u₁ we first find the time t it takes to reach the peak height from

v = u + at. where the variables mean the same as above.

substituting the values, we have

0 = 5.17 m/s +(-9.8m/s²)t

-5.17 m/s = -9.8m/s²t

t = -5.17 m/s ÷ (-9.8m/s²)

= 0.53 s

Now, the horizontal distance d = u₁t = 1.560 m

u₁ = d/t = 1.560 m/0.53 s = 2.96 m/s

So, the initial velocity of the cat is V = √(u² + u₁²)

= √((5.17 m/s)² + (2.96 m/s)²)

= √(26.729(m/s)² + 8.762(m/s)²)

= √(35.491 (m/s)²)

= 5.95 m/s

its direction θ = tan⁻¹(5.17 m/s ÷ 2.96 m/s) = 60.2°

So, the initial velocity of the cat is 5.95 m/s at 60.2° from the horizontal

(b)

First, we find the time t' it takes the cat to land on the bed from d' = u₁t'

where d' = horizontal distance of cabinet from bed = 3.582 m

u₁ = horizontal velocity = 2.96 m/s

t' = d'/u₁

= 3.582 m/2.96 m/s

= 1.21 s

The vertical between the bed and cabinet which is the vertical distance moved by the cat is gotten from Δy = ut' +1/2at'²

substituting u = initial vertical velocity = 5.17 m/s, t' = 1.21 s and a = -g = -9.8 m/s² into Δy, we have  

Δy = ut' +1/2at'² = 5.17 m/s × 1.21 s +1/2(- 9.8 m/s²) × (1.21 s)² = 6.256 - 7.174 = -0.918 m

Δy = y₂ - y₁

Since our initial position is the position of the cabinet above the ground = y₁ = 1.765 m

y₂ = position of bed above ground.

Δy = y₂ - y₁ = -0.918 m

y₂ - 1.765 m = -0.918 m

y₂ = 1.765 m - 0.918 m

= 0.847 m

Answer:

The centripetal acceleration of the traveler is [tex]40.98\ m/s^2[/tex]

Explanation:

It is given that, A carousel has a diameter of 6.0-m and completes one rotation every 1.7 s.

We need to find the centripetal acceleration of the traveler. It is given by the formula as follows :

[tex]a=\dfrac{v^2}{r}[/tex]

r is radius of carousel

[tex]v=\dfrac{2\pi r}{T}[/tex]

So,

[tex]a=\dfrac{4\pi ^2r}{T^2}[/tex]

Plugging all the values we get :

[tex]a=\dfrac{4\pi ^2\times 3}{(1.7)^2}\\\\a=40.98\ m/s^2[/tex]

So, the centripetal acceleration of the traveler is [tex]40.98\ m/s^2[/tex].

Answer:

a) [tex]F = 210.803\,N[/tex], b) [tex]W_{F} = 779.971\,J[/tex], c) [tex]W_{f} = 235.683\,J[/tex], d) [tex]W_{N} = 0\,J[/tex]; [tex]W_{g} = 544.289\,J[/tex], e) [tex]W_{net} = 0\,J[/tex]

Explanation:

a) The net force exerted on the crate is:

[tex]\Sigma F = F - m\cdot g \cdot \sin \theta - \mu_{k}\cdot m\cdot g \cdot \cos \theta = 0[/tex]

The magnitud of the force that the work must apply to the crate is:

[tex]F = m\cdot g \cdot \sin \theta + \mu_{k}\cdot m\cdot g \cdot \cos \theta[/tex]

[tex]F = (30\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot \sin 30^{\circ} + 0.25 \cdot (30\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot \cos 30^{\circ}[/tex]

[tex]F = 210.803\,N[/tex]

b) The work done on the crate due to the external force is:

[tex]W_{F} = (210.803\,N)\cdot (3.7\,m)[/tex]

[tex]W_{F} = 779.971\,J[/tex]

c) The work done on the crate due to the external force is:

[tex]W_{f} = (63.698\,N)\cdot (3.7\,m)[/tex]

[tex]W_{f} = 235.683\,J[/tex]

d) The work done on the crate due the normal force is zero, since such force is perpendicular to the motion direction.

[tex]W_{N} = 0\,J[/tex]

And, the work done by gravity is:

[tex]W_{g} = (147.105\,N)\cdot (3.7\,m)[/tex]

[tex]W_{g} = 544.289\,J[/tex]

e) Lastly, the total work done is:

[tex]W_{net} = W_{F} - W_{f} - W_{g} - W_{N}[/tex]

[tex]W_{net} = 779.971\,J - 235.683\,J - 0\,N - 544.289\,J[/tex]

[tex]W_{net} = 0\,J[/tex]

Answer:

false

Explanation:

Answer:

A , D , E

Explanation:

Solution:-

- Consider the two identical objects with mass ( m ).

- The stiffness of the springs are ( k1 and k2 ).

- Both the spring store 55.0 J of potential energy.

- We will apply the principle of energy conservation on both the systems. In both cases the spring stores 55.0 Joules of energy. Once released, the objects gain kinetic energy with a consequent loss of potential energy in either spring.

- The maximum speed ( v ) is attained when all the potential energy is converted to kinetic energy.

- Apply Energy conservation for spring with stiffness ( k1 ).

                         ΔU = ΔEk

                         55.0 = 0.5*m*v^2

                         v = √ ( 110 / m )

- Apply Energy conservation for spring with stiffness ( k2 ).

                         ΔU = ΔEk

                         55.0 = 0.5*m*v^2

                         v = √ ( 110 / m )

Answer: Both objects will have the same maximum speed ( A )

- We are told that one spring is more stiff as compared to the other one. The measure of stiffness is proportionally quantified by the spring constant. To mathematically express we can write it as:

                         k1 > k2

Where,

                 k1: The stiff spring

                 k2: The flexible spring

Answer: The stiff spring has a larger spring constant than the flexible spring. ( D )

- We will assume that the spring with constant ( k1 ) undergoes a displacement ( x1 ) and the spring with constant ( k2 ) undergoes a displacement ( x2 ). The potential energy stored in both spring is the same. Hence,

                      U1 = U2

                      0.5*( k1 ) * ( x1 )^2 = 0.5*( k2 ) * ( x2 )^2

                      [ k1 / k2 ] = [ x2 / x1 ]^2

Since,

                     k1 > k2 , then [ k1 / k2 ] > 1    

Then,

                     [ x2 / x1 ]^2 > 1

                     [ x2 / x1 ] > 1

                     x2 > x1                  

Answer: The flexible spring ( x2 ) was compressed more than the stiff spring ( x1 ). ( E )

Answer:

The maximum speed v that the car can go without flying off the track = 15.29 m/s

Explanation:

let us first lay out the information clearly:

mass of car (m) = 410 kg

radius of race track (r) = 83.4 m

coefficient of friction (μ) = 0.286

acceleration due to gravity (g) = 9.8 m/s²

maximum speed = v m/s

For a body in a constant circular motion, the centripetal for (F) acting on the body is given by:

F = mass × ω

where:

F = maximum centripetal force = mass × μ × g

ω = angular acceleration = [tex]\frac{(velocity)^2}{radius}[/tex]

∴ F = mass × ω

m × μ × g = m × [tex]\frac{v^{2} }{r}[/tex]

410 × 0.286 × 9.8 = 410 × [tex]\frac{v^{2} }{83.4}[/tex]

since 410 is on both sides, they will cancel out:

0.286 × 9.8 = [tex]\frac{v^{2} }{83.4}[/tex]

2.8028 = [tex]\frac{v^{2} }{83.4}[/tex]

now, we cross-multiply the equation

2.8028 × 83.4 = [tex]v^{2}[/tex]

[tex]v^{2}[/tex] = 233.754

∴ v = √(233.754)

v = 15.29 m/s

Therefore, the maximum speed v that the car can go without flying off the track = 15.29 m/s

Answer:

the correct answer is B

Explanation:

We analyze this exercise a little, the block goes into the air and is under the acceleration of gravity. The ball is fired by the hand and is describing a parabolic movement, subjected to the acceleration of gravity.

For the ball to hit the block we must have the distance the ball goes up equal to the distance the block moves, therefore we must shoot the ball at the block at its highest point.

Let's write the kinematic equation for the two bodies

The block. At the highest point of the path

      y = - ½ g t2

The ball, in its vertical movement

     y = vo t - ½ g t2

therefore the correct answer is B

Answer:

a. 900 J

b. 0.383

Explanation:

According to the question, the given data is as follows

Horizontal force = 150 N

Packing crate = 40.0 kg

Distance = 6.00 m

Based on the above information

a. The work done by the 150-N force is

[tex]W = F x = \mu N x = \mu\ m\ g\ x[/tex]

[tex]W = 150 \times 6[/tex]

= 900 J

b. Now the coefficient of kinetic friction between the crate and surface is

[tex]\mu = \frac {F}{m\timesg}[/tex]

[tex]= \frac{150}{40\times 9.8}[/tex]

= .383

We simply applied the above formulas so that each one part could calculate

We want to find the work and kinetic friction for the given situation. The solutions are:

Here we have a horizontal force of 150N pushing a 40.0 kg packing crate a distance of 6.00m at a constant speed.

a) First we want to find the work, it is given by the force applied times the distance moved, so the work is just:

W = 150N*6.00m = 900 N*m

b) Now we want to find the coefficient of kinetic friction, it must be such that the kinetic friction force is equal to the pushing force, in this way there is no net force, and then there is no acceleration.

Remember that the friction force is:

F = m*g*μ

Where:

Then we must solve:

150N = 40kg*(9.8 m/s^2)*μ = 392N*μ

150N/392N = 0.38

So the coefficient of kinetic friction between the crate and the surface is 0.38

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Soundproofing material is required for blocking sound during some works like recording voice in the studio. Image D represents the interaction of a sound wave with soundproofing material in a recording studio.

Soundproofing is done by absorbing the sound. A very much used material for this is a dense foam.

Foam and like materials absorbs sound and it travels directly into the soft surface resulting in soundproofing.

Thus, the correct option is C, as the D image is showing the absorption.

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Answer: C.D

Explanation:...

Complete question

The complete question is shown on the first uploaded image  

Answer:

The velocity is  [tex]v = c* \sqrt{1 - \frac{1}{n^2} }[/tex]

Explanation:

From the question we are told that

           a = nb

The length of the minor axis  of  the symbol of the Federation, a circle, seen by the observer at velocity v must be equal to the minor axis(b) of the  Empire's symbol, (an ellipse)

Now this length seen by the observer can be mathematically represented as

        [tex]h = t \sqrt{1 - \frac{v^2}{c^2} }[/tex]

Here t  is the actual length of the major axis of of the  Empire's symbol, (an ellipse)

So t = a = nb

and  b is the length of the minor axis of the symbol of the Federation, (a circle) when seen by an observer at velocity v which from the question must be the length of the minor axis of the of the  Empire's symbol, (an ellipse)

 i.e    h = b

So

    [tex]b = nb [\sqrt{1 - \frac{v^2}{c^2} } ][/tex]  

     [tex][\frac{1}{n} ]^2 = 1 - \frac{v^2}{c^2}[/tex]

      [tex]v^2 =c^2 [1- \frac{1}{n^2} ][/tex]

       [tex]v^2 =c^2 [\frac{n^2 -1}{n^2} ][/tex]

        [tex]v = c* \sqrt{1 - \frac{1}{n^2} }[/tex]

Answer:

The minimum angular spread of the laser beam is 3.32 × [tex]10^{-6}[/tex] radians (or 3.32 μrad).

Explanation:

The minimum angular spread of a wave is the ratio of its narrowest diameter to its wavelength.

From Rayleigh's formula,

Angular spread = 1.22 (wavelength ÷ diameter)

                          = 1.22 (λ ÷ D)

Given that:

diameter, D = 0.18 m and wavelength, λ = 490 nm, then;

Angular spread of the laser beam = 1.22 (λ ÷ D)

                         = 1.22[tex](\frac{490*10^{-9} }{0.18})[/tex]

                         = 1.22× 2.7222 × [tex]10^{-6}[/tex]

                        = 3.3211 × [tex]10^{-6}[/tex] rad

The minimum angular spread of the laser beam is 3.32 × [tex]10^{-6}[/tex] radians.

Answer:

(A) object B has the largest mass because it has the least acceleration

(B) object C has the smallest mass because it has the largest acceleration

(C) mass A : mass B = 3 : 5

Explanation:

Given;

acceleration of object A = 5 m/s²

acceleration of object B = 3 m/s²

acceleration of object C = 7 m/s²

A constant force, F

According to Newton's second law of motion;

F = ma

m = F / a

Mass of object A:

m = F / 5

Mass of object B:

m = F / 3

Mass of object  C:

m = F / 7

(A). Which object has the largest mass:

object B has the largest mass because it has the least acceleration

(B). Which object has the smallest mass:

object C has the smallest mass because it has the largest acceleration

(C). What is the ratio of mass A to mass B;

mass A = F / 5

mass B = F / 3

[tex]mass \ A : \ mass \ B = \frac{F}{5} : \frac{F}{3} \\\\\frac{mass \ A}{mass \ B} = \frac{F}{5} * \frac{3}{F}= \frac{3}{5} \\\\mass \ A : \ mass \ B = 3: 5[/tex]

A. The Object B has largest mass.

B. The Object A has smallest mass.

C. The ratio of mass A to mass B is, [tex]\frac{3}{5}[/tex]

The second law states that the acceleration of an object is dependent upon two variables - the net force acting upon the object and the mass of the object.

                   [tex]F=ma\\\\m=\frac{F}{a}[/tex]

                     [tex]\frac{m_{A}}{m_{B}} =\frac{a_{B}}{a_{A}} =\frac{3}{5}[/tex]

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Answer:

Q = 3877 KJ

Explanation:

Since, the system is closed and isolated. Therefore, the law of conservation of energy can be written as:

Heat Absorbed By Water (Q) = Heat required to raise the temperature of water (Q₁) + Heat required to convert water to steam (Q₂)

Q = Q₁ + Q₂   ----- equation (1)

Now, for Q₁:

Q₁ = m C ΔT

where,

m = Mass of Water = 1.5 kg

C = Specific Heat of Water = 4187 J/kg.°C

ΔT = Change in Temperature of Water = T₂ - T₁ = 100°C - 22°C = 78°C

Therefore,

Q₁ = (1.5 kg)(4187 J/kg.°C)(78°C)

Q₁ = 490 x 10³ J =490 KJ

Now, for Q₂:

Q₂ = m H

where,

m = Mass of Water = 1.5 kg

H = Heat of Vaporization of Water = 2258 KJ/kg

Therefore,

Q₂ = (1.5 kg)(2258 KJ/kg)

Q₂ = 3387 KJ

Substituting the values in equation (1), we get:

Q = Q₁ + Q₂

Q = 490 KJ + 3387 KJ

Q = 3877 KJ

The question is incomplete. Here is the complete question.

Calculate the maximum deceleration  of a car that is heading down a 14° slope (one that makes an anlge of 14° with the horizontal) under the following road conditions. You may assum that the weight of the car is evenlydistributed on all four tires and that the sttic coefficient of friction is involved - that is, the tires are not allowed to slip during the deceleration. (Ignore rolling) Calculate for a car: (a) On a dry concrete. (b) On a wet concrete. (c) On ice, assuming that μs = 0.100, the same as for shoes on ice.

Answer: (a) a = - 11.05 m/s²; (b) a = - 10.64 m/s²; (c) a = - 9.84m/s²

Explanation: The image in the attachment describe the forces acting on the car. Observing that, we know that:

[tex]F_{net}[/tex] = - [tex]W_x[/tex] - [tex]f_s[/tex]

The [tex]W_x[/tex] is a x-component of force due to gravity (W) and, in this case, is given by: [tex]W_x[/tex] = W.sin(14)

W is described as: W = m.g

Force due to friction ([tex]f_s[/tex]) is given by: [tex]f_s[/tex] = μs.N

N is the normal force and, in the system, is equivalent of [tex]W_y[/tex], so:

[tex]W_y[/tex] = m.g.cos(14)

Therefore, the formula will be:

[tex]F_{net}[/tex] = - [tex]W_x[/tex] - [tex]f_s[/tex]

m.a = - (m.g.sin14) - (μs.mg.cos14)

a = - g (sin14 + μscos 14)

a) For dry concrete, μs = 1:

a = - g (sin14 + μscos 14)

a = - 9.8 (sin14 + 1.cos14)

a = - 11.05 m/s²

b) For wet concrete, μs = 0.7:

a = - g (sin14 + μscos 14)

a = - 9.8 (sin 14 + 0.7.cos14)

a = - 10.64 m/s²

c) For ice, μs = 0.1:

a = - g (sin14 + μscos 14)

a = - 9.8 (sin14 + 0.1cos14)

a = - 9.84 m/s²