Calculate the maximum deceleration (in m/s2) of a car that is heading down a 14° slope (one that makes an angle of 14° with the horizontal) under the following road conditions. You may assume that the weight of the car is evenly distributed on all four tires and that the static coefficient of friction is involved—that is, the tires are not allowed to slip during the deceleration.

Answer: [tex]x=\frac{2}{3}t^3+3t+1[/tex]

Explanation:

Given

velocity of object is given by

[tex]v(t)=3+2t^2[/tex]

and we know change of position w.r.t time is velocity

[tex]\Rightarrow \dfrac{dx}{dt}=v[/tex]

[tex]\Rightarrow \dfrac{dx}{dt}=3+2t^2[/tex]

[tex]\Rightarrow dx=(3+2t^2)dt[/tex]

Integrating both sides we get

[tex]\Rightarrow \int_{1}^{x}dx=\int_{0}^{t}(3+2t^2)dt[/tex]

[tex]\Rightarrow x\mid _{1}^{x}=(3t+\frac{2}{3}t^3)\mid _{0}^{t}[/tex]

[tex]\Rightarrow x-1=3(t-0)+\frac{2}{3}(t^3-0)[/tex]

[tex]\Rightarrow x=\frac{2}{3}t^3+3t+1[/tex]

Answer:

The time taken is  [tex]t_{out} = \frac{D}{v__{R}} + v__{S}}}[/tex]

Explanation:

From the question we are told that

     The speed of the current is  [tex]v__{R}}[/tex]

     The speed of the swimmer in direction of current is [tex]v__{S}}[/tex]

      The distance traveled by the swimmer is  [tex]D[/tex]

       The time taken to travel this distance is  [tex]t_{out}[/tex]

      The speed of the swimmer against  direction of current is  [tex]v__{s}}[/tex]

The resultant speed for downstream current is

       [tex]V_{r} = v__{S}} +v__{R}}[/tex]

The time taken can be mathematically represented as

      [tex]t_{out} = \frac{D}{V_{r}}[/tex]

      [tex]t_{out} = \frac{D}{v__{R}} + v__{S}}}[/tex]

Answer:

The time when the ball strikes the ground is closest to  [tex]t_t = 9.4 \ s[/tex]

Explanation:

From the question we are told that

  The time of projection is t = 0.0 s

   The  distance of the point  from the ground  is  [tex]d = 90 \ m[/tex]

    The  initial velocity of the ball is  [tex]v _i = 36 .2 \ m/s[/tex]

generally the time required to reach maximum height is  

      [tex]t_r = \frac{g}{v}[/tex]

Where is the acceleration due to gravity  with value  [tex]g = 9.8 \ m/s^2[/tex]

Substituting values

        [tex]t_r = \frac{36.2}{9.8}[/tex]

        [tex]t_r = 3.69 s[/tex]

when returning the time and velocity at the roof level is  t =  3.69 s and  u = 36.2 m/s this due to the fact that  air resistance is negligible

   The final velocity at which it  hit the ground is

      [tex]v_f^2 = u^2 + 2ag[/tex]

So  

    [tex]v_f = \sqrt{ u^2 + 2gs}[/tex]

substituting values

    [tex]v_f = \sqrt{ 3.69^2 + 2* 9.8 * 90}[/tex]

     [tex]v_f = 55.45 \ m/s[/tex]

The time taken for the ball to move from the roof level to the ground is  

     [tex]t_g = \frac{v-u}{a}[/tex]

substituting values

    [tex]t_g = \frac{55.45 -36.2}{9.8}[/tex]

     [tex]t_g = 1.96 \ s[/tex]

The total time for this travel is  

    [tex]t_t = t_g + 2 t_r[/tex]

     [tex]t_t = 1.96 + 2(3.69)[/tex]

      [tex]t_t = 9.4 \ s[/tex]

Answer:

The amount of water that will power a battery with that rating = 7.35 m³

Explanation:

The power rating for the battery is missing from the question.

Complete Question

Assuming 100% efficient energy conversion how much water stored behind a 50 centimeter high hydroelectric dam would be required to charged the battery with power rating, 12 V, 50 Ampere-minutes

Solution

Potential energy possessed by water at that height = mgH

m = mass of the water = ρV

ρ = density of water = 1000 kg/m³

V = volume of water = ?

g = acceleration due to gravity = 9.8 m/s²

H = height of water = 50 cm = 0.5 m

Potential energy = ρVgH = 1000 × V × 9.8 × 0.5 = (4900V) J

Energy of the battery = qV

q = 50 A.h = 50 × 60 = 3,000 C

V = 12 V

qV = 3,000 × 12 = 36,000 J

Energy = 36,000 J

At a 100% conversion rate, the energy of the water totally powers the battery

(4900V) = (36,000)

4900V = 36,000

V = (36,000/4900)

V = 7.35 m³

Hope this Helps!!!

Answer:

For an object to be an equilibrium it must be experiencing no acceleration.

Explanation:

Hope it helps.

Answer:

F = 0.314 N

Explanation:

In order to calculate the applied force to the ball by the ground, you first calculate the speed of the ball just before it hits the ground. You use the following formula:

[tex]v^2=v_o^2+2gy[/tex]        (1)

y: height from the ball starts its motion = 1.8 m

vo: initial velocity = 0 m/s

g: gravitational acceleration =  9.8 m/s^2

v: final velocity of the ball = ?

You replace the values of the parameters in the equation (1):

[tex]v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(1.8m)}=5.93\frac{m}{s}[/tex]

Next, you take into account that the force exerted by the ground on the ball is given by the change, on time, of the linear momentum of the ball, that is:

[tex]F=\frac{\Delta p}{\Delta t}=m\frac{\Delta v}{\Delta t}=m\frac{v_2-v_1}{\Delta t}[/tex]      (2)

m: mass of the ball = 20g = 20*10^-3 kg

v1: velocity of the ball just before it hits the ground = 5.93m/s

v2: velocity of the ball after it impacts the ground (80% of v1):

0.8(5.93m/s) = 4.75 m/s

Δt: time interval o which the ground applies the force on the ball = 75*10^-3 s

You replace the values of the parameters in the equation (2):

[tex]F=(20*10^{-3}kg)\frac{4.75m/s-5.93m/s}{75*10^{-3}s}=-0.314N[/tex]

The minus sign means that the force is applied against the initial direction of the motion of the ball.

The applied force by the ground on the bouncy ball is 0.314 N

Answer:

Coefficient of kinetic friction = 0.235

Explanation:

Given:

Mass of crate = 330 kg

1st force = 430 N

2nd force = 330 N

Find:

Coefficient of kinetic friction.

Computation:

We know that, velocity is constant.

So, acceleration (a) = 0

So, net force (f) = 430 N + 330 N

Net force (f) = 760 N

F = μmg

μ = f / mg                                   [∵ g = 9.8]

μ = 760 / [330 × 9.8]

μ = 760 / [3,234]

μ = 0.235

Coefficient of kinetic friction = 0.235

Answer:

The answer is option 2.

Explanation:

Both sides are pulling the rope with equal force where the rope doesn't move. So they have a balanced forces.

Answer:

Vf = 78.64 m/s

Explanation:

The rocket is travelling upward at a constant acceleration of 3.99 m/s² until it runs out of fuel. So, in order to calculate its velocity at the point, where it runs out of fuel, we can simply use 3rd equation of motion:

2as = Vf² - Vi²

where,

a = acceleration = 3.99 m/s²

s = distance or height covered by rocket till fuel runs out = 775 m

Vf = Final Velocity = ?

Vi = Initial velocity = 0 m/s   (Since, rocket starts from rest)

Therefore,

2(3.99 m/s²)(775 m) = Vf² - (0 m/s)²

Vf = √(6184.5 m²/s²)

Vf = 78.64 m/s

Answer:

a) t = 27.00 h

b) B = 6.84 MeV/nucleon

Explanation:

a) The time can be calculated using the following equation:

[tex] R = R_{0}e^{-\lambda*t} [/tex]

Where:

R: is the radiation measured at time t

R₀: is the initial radiation

λ: is the decay constant

t: is the time

The decay constant can be calculated as follows:

[tex] t_{1/2} = \frac{ln(2)}{\lambda} [/tex]

Where:

t(1/2): is the half life = 4.5 h

[tex] \lambda = \frac{ln(2)}{t_{1/2}} = \frac{ln(2)}{4.5 h} = 0.154 h^{-1} [/tex]

We have that the radiation measured is 64 times the maximum permissible level, thus R₀ = 64R:  

[tex] \frac{R}{64R} = e^{-\lambda*t} [/tex]                      

[tex] t = -\frac{ln(1/64)}{\lambda} = -\frac{ln(1/64)}{0.154 h^{-1}} = 27.00 h [/tex]            

b) The binding energy (B) can be calculated using the following equation:

[tex]B = \frac{(Z*m_{p} + N*m_{n} - M_{A})}{A}*931.49 MeV/u[/tex]

Where:

Z: is the number of protons = 2 (for [tex]^{4}_{2}He[/tex])

[tex]m_{p}[/tex]: is the proton mass = 1.00730 u

N: is the number of neutrons = 2 (for [tex]^{4}_{2}He[/tex])

[tex]m_{n}[/tex]: is the neutron mass = 1.00869 u  

[tex]M_{A}[/tex]: is the mass of the He atom = 4.002602 u

A =  N + Z = 2 + 2 = 4    

The binding energy of [tex]^{4}_{2}He[/tex] is:

[tex]B = \frac{(2*1.00730 + 2*1.00869 - 4.002602)}{4}*931.49 MeV/u = 7.35\cdot 10^{-3} u*931.49 MeV/u = 6.84 MeV/nucleon[/tex]

Hence, the binding energy per nucleon is 6.84 MeV.

I hope it helps you!

Answer:

The distance covered by puck A before collision is  [tex]z = 8.56 \ m[/tex]

Explanation:

From the question we are told that

   The label on the two hockey pucks is  A and  B

    The distance between the  two hockey pucks is D   18.0 m

     The speed of puck A is  [tex]v_A = 3.90 \ m/s[/tex]

        The speed of puck B is  [tex]v_B = 4.30 \ m/s[/tex]

The distance covered by puck A is mathematically represented as

     [tex]z = v_A * t[/tex]

  =>  [tex]t = \frac{z}{v_A}[/tex]

 The distance covered by puck B  is  mathematically represented as

      [tex]18 - z = v_B * t[/tex]

=>   [tex]t = \frac{18 - z}{v_B}[/tex]

Since the time take before collision is the same

        [tex]\frac{18 - z}{V_B} = \frac{z}{v_A}[/tex]

substituting values

          [tex]\frac{18 -z }{4.3} = \frac{z}{3.90}[/tex]

=>      [tex]70.2 - 3.90 z = 4.3 z[/tex]

=>       [tex]z = 8.56 \ m[/tex]

Answer:

1. 25 Km

2. zero

3. 27.7 m/s

Explanation:

Data provided in the question:

Circumference of the track = 12.5 km

Speed of the car = 100 Km/h

Time for which car travels = 15 minutes = [tex]\frac {15}{60}[/tex] hr

Now,

1. Distance traveled = Speed × Time

= 100 × [tex]\frac{15}{60}[/tex]

= 25 Km

2. The distance traveled is 2 times the circumference of the track (i.e 2 × 12.5 = 25 Km)

Which means that the car is again at the initial position

Therefore, The displacement is zero.

3. Speed of car in Km/hr = 100 Km/h

now,

1 Km = 1000 m

1 hr = 3600 seconds

therefore,

100 Km/h = [tex]100\times\frac{1000}{3600}[/tex] m/s

= 27.7 m/s

Hence, the speed of car in m/s = 27.7

Answer:

The  normal force N1 exerted by the floor is  [tex]N_1 = 951 \ N[/tex]

The  normal force N2 exerted by the wall is  [tex]N_2= 616.43 \ N[/tex]

The frictional force exerted by the wall is  [tex]f = N_2 = 616.43 \ N[/tex]  

Explanation:

From the question we are told that

    The length of the ladder is  [tex]L = 4.6 \ m[/tex]

    The weight of the ladder  is

    The distance of the ladder position on the wall from the floor is  [tex]D = 3.75 \ m[/tex]

     The mass of the person is  [tex]m = 90 kg[/tex]

Applying Pythagoras theorem

The length of the position the ladder on the ground from the base of the wall is

    [tex]A = \sqrt{L^ 2 - D^2}[/tex]

substituting values

    [tex]A = \sqrt{(4.6^2)-(3.75^2)}[/tex]

    [tex]A = 2.66 \ m[/tex]

  In order the for the ladder not to shift from the ground the sum of the moment about the position of the ladder on the ground must be equal to zero this is mathematically represented as

        [tex]\sum M = 0 = N_2 * D - [\frac{1}{2} * W_L ] * [(mg) *A ][/tex]

         [tex]\sum M = 0 = N_2 * 3.75 - [\frac{1}{2} * 69.0 ] * [(90*9.8) * \frac{4.6}{2.66} ][/tex]

        [tex]N_2 * 3.75 =2311.62[/tex]

        [tex]N_2 * 3.75 =2311.62[/tex]

        [tex]N_2= 616.43 \ N[/tex]

Now the force exerted by the floor on the ladder is mathematically represented as

           [tex]N_1 = W_L + (m * g )[/tex]

substituting values

          [tex]N_1 = 951 \ N[/tex]

Now the horizontal forces acting on the ladder are [tex]N_2 \ and \ f[/tex] and they are in opposite direction so

     [tex]f = N_2 = 616.43 \ N[/tex]  

Answer:

Explanation:

At height h , potential energy of coaster car  having mass m = mgh .

The car will lose potential energy and gain kinetic energy.

height lost by car when it is at the top of loop of radius R

= h - 2R

potential energy lost = mg ( h - 2R )

kinetic energy gained = mg ( h - 2R )

kinetic energy = 0 + mg ( h - 2R )

= mg ( h - 2R )

b )

For the car to remain in contact with the track , if v be the minimum velocity

centripetal force at top = mg

m v² / R = mg

v² = gR

kinetic energy = 1/2 mv²

= 1/2 m x gR

= mgR /

If h be the minimum height that can give this kinetic energy

mg ( h - 2R ) = mgR / 2

h - 2R = R / 2

h = 2.5 R .

Answer:

It will take 7.75 seconds for the car to go 155m

Explanation:

From the question, we can understand that the distance covered by the moving car is got by a product of its speed and the time it travels.

i.e distance = speed X time.

However, in this case, we have the distance travelled and the speed of the car, and we are looking for the time of travel

TO solve this, we will simply make the travel time the subject of the formula in the equation above.

i.e time = distance / speed

time = 155/20= 7.75 seconds.

Hence, it will take 7.75 seconds for the car to go 155m

Answer:

a) R = 2.5 Ω, b) R = 1 Ω, c)     R = 2R / 3 Ω

Explanation:

The resistance configuration can be in series or in parallel, for each one the equivalent resistance can be calculated

series, the equivalent resistance is the sum of the resistances

parallel, the inverse of the equivalent resistance is the inverse of the sum of the resistances

let's apply these principles to each case

case a)

    equivalent series resistance

         R₁ = 1 +4 = 5 ohm

         R₂ = 2 +3 = 5 ohn

these two are in parallel

        1 / R = 1/5 +1/5

        1 / R = 2/5

         R = 2.5 Ω

case B

we solve the parallel

       1 / R₁ = ½ + ½ = 1

        R₁ = 1 Ω

we solve the resistors in series

       R₂ = 1 + 1

       R₂ = 2 Ω

finally we solve the last parallel

       1 / R = ½ +1/2 = 1

        R = 1 Ω

case C

we solve house resistance pair in series

     R₁ = R + 2R = 3R

we go to the next mesh

     R₂ = R + 2R = 3R

     R₃ = R + 2R = 3R

last mesh

     R₄ = R + R = 2R

now we solve the parallel of this equivalent resistance

     1 / R = 1 / R₁ + 1 / R₂ + 1 / R₃ + 1 / R₄

     1 / R = 1 / 3R + 1 / 3R + 1 / 3R + 1 / 2R

      1 / R = 3 / 3R + 1 / 2R = 1 / R + 1 / 2R

     1 / R = 3 / 2R

      R = 2R / 3 Ω

Answer:7.5V

Explanation:

Ohm's law, V=IR

so, V=0.3×25

V=7.5V

Answer:

7.5 V

Explanation:

Answer: B, Companies passed on production and transportation costs to consumers

Explanation:

A higher oil price occurred when companies passed on production and transportation costs to consumers.

The oil producing companies spend so much money in producing crude oil from the reservoirs to the surface. They also spend money in processing and transporting the crude oil to the end users or consumers.

The final price of the oil depends on the total amount spent by these companies in producing the hydrocarbons.

Thus, a higher oil price occurred when companies passed on production and transportation costs to consumers.

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Answer:

25m

Explanation:

Let's assume the Jeep attains a velocity of 36km/h ; a constant speed same with that of the car.

While the Jeep is accelerating to that speed, the car with that speed passes it.

Now we can calculate the time taken for the Jeep to attain the velocity of 36km/h on her constant acceleration.

This time is t = v/a; from Newton's Law of Motion:

a = V-U / t ; a-acceleration

V is final velocity = 36km/h

U is initial velocity 0 since the body starts from rest.

Hence t = 36000/3600 ÷ 4 = 2.5s

Note conversting from km/h to m/s we multiply by 1000/3600.

But the distance covered by the car while the Jeep just accelerates is

S = U × t = 10× 2.5 = 25m.

Note From Newton's law of Motion, distance for constant speed is defined as: U × t

Hence the Car would be 25m off the starting point just as the Jeep accelerates. It would overtake the Jeep when it just covers 25m from the Jeep starting point.

Answer:

E) She can perform no measurement to determine this quantity.

Explanation:

A spacecraft is a machine used to fly in outer space.

According to Isaac Newton's third law of motion, every action produces an equal and opposite reaction. When fuel is shoot out of one end of the rocket, the rocket moves forward for which no air is required.

As Leah is moving in a spaceship at a constant velocity away from a group of stars, she cannot measure to determine this quantity.

Answer:

0.002kg and 0.01cm

Explanation:

500 leaves has a thickness is 5cm

Means I leaf has a thickness of 5/500= 0.01cm

Similarly the mass of one leaf would be 1/500 =0.002kg

Answer:

  c.  in the absence of an unbalanced force, an object at rest will stay at rest and an object in motion will stay in motion.

Explanation:

First law: things keep doing what they are doing, unless force is applied.

Answer:

d = 41.91 m

Explanation:

In order to calculate the distance traveled by the train while superman passes it, you write the equations of motion for both superman and train:

For train, you have a motion with constant speed. You write the equation of motion of the position of the front of the train:

[tex]x=x_o+v_1t[/tex]    (1)

xo: initial position of the front of the train = 500m

v1: speed of the train = 50km/h

For superman, you take into account that the motion is an accelerated motion (you assume superman is at the origin of coordinates):

[tex]x'=v_2t+\frac{1}{2}at^2[/tex]   (1)

v2: initial speed of superman = 100km/h

a: acceleration = 10m/s^2

When superman passes the train, both positions x and x' will be equal. Hence, you equal the equations (1) and (2) and you calculate the time t. But before you convert the units of the velocities v1 and v2 to m/s:

[tex]v_1=50\frac{km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s}=13.88\frac{m}{s}\\\\v_2=100\frac{km}{h}=\frac{1000m}{1km}*\frac{1h}{3600s}=27.77\frac{m}{s}[/tex]

Thus, you equal x=x'

[tex]x=x'\\\\x_o+v_1t=v_2t+\frac{1}{2}at^2\\\\500m+(13.88m/s)t=(27.77m/s)t+\frac{1}{2}(10m/s^2)t^2\\\\(50\frac{m}{s^2})t^2+(13.89\frac{m}{s})t-500m=0[/tex]

You solve the last equation for t by using the quadratic formula:

[tex]t_{1,2}=\frac{-13.89\pm \sqrt{(13.89)^2-4(50)(-500)}}{2(50)}\\\\t_{1,2}=\frac{-13.89\pm 316.53}{100}\\\\t_1=3.02s\\\\t_2=-3.30s[/tex]

You only use t1 = 3.02s because negative times do not have physical meaning.

Next, you replace this value of t in the equation (1) to calculate the position of the train (for when superman just passed it):

[tex]x=500m+(13.88m/s)(3.02s)=541.91m[/tex]

x is the position of the front of the train, then, the dstance traveled by the train is:

d = 541.91m - 500m = 41.91 m

Answer:

a. 78 degree

Explanation:

According to Snell's Law, we have:

(ni)(Sin θi) = (nr)(Sin θr)

where,

ni = Refractive index of medium on which light is incident

ni = Refractive index of ethyl alcohol = 1.361

nr = Refractive index of medium from which light is refracted

nr = Refractive index of ethyl alcohol = 1.333

θi = Angle of Incidence

θr = Angle of refraction

So, the Angle of Incidence is know as the Critical Angle (θc), when the refracted angle becomes 90°. This is the case of total internal reflection. That is:

θi = θc

when, θr = 90°

Therefore, Snell's Law becomes:

(1.361)(Sin θc) = (1.333)(Sin 90°)

Sin θc = 1.333/1.361

θc = Sin⁻¹ (0.9794)

θc = 78.35° = 78° (Approximately)

Therefore, correct answer will be:

a. 78 degree

The angle relative to the normal interface of the two liquids at which the light is totally reflected is 78 degrees.

From the information given;

According to Snell's Law of refraction;

[tex]\mathbf{n_1 sin \theta _c = n_2 sin \theta_r}[/tex]

[tex]\mathbf{1.361 \times sin \theta _c = 1.333 \times sin 90}[/tex]

[tex]\mathbf{ sin \theta _c =\dfrac{ 1.333 \times sin 90}{1.361}}[/tex]

[tex]\mathbf{ sin \theta _c =\dfrac{ 1.333 \times 1}{1.361}}[/tex]

[tex]\mathbf{ \theta _c = sin^{-1} (0.9794)}[/tex]

[tex]\mathbf{ \theta _c =78.35^0}[/tex]

[tex]\mathbf{ \theta _c \simeq78^0}[/tex]

Therefore, we can conclude that the angle relative to the normal interface of the two liquids at which the light is totally reflected is 78 degrees.

Learn more about Snell Law of refraction here:

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Answer:

107 N, option d

Explanation:

Given that

mass of the ball, m = 0.2 kg

initial velocity of the ball, u = 20 m/s

final velocity of the ball, v = -12 m/s

time taken, Δt = 60 ms

Solving this question makes us remember "Impulse Theorem"

It states that, "that the product between the average force applied and the duration of the collision is equal to the change in momentum of the object"

Mathematically, it is represented as

FΔt = m(v - u), where

F = the average force

Δt = time taken

m = mass of the ball

v = final velocity of the ball

u = initial velocity of the ball

From the question we were given, if we substitute the values in it, we have

F = ?

Δt = 60 ms = 0.06s

m = 0.2 kg

v = -12 m/s

u = 20 m/s

F = 0.2(-12 - 20) / 0.06

F = (0.2 * -32) / 0.06

F = -6.4 / 0.06

F = -106.7 N

Thus, the magnitude is 107 N

Answer:

Explanation:

Time taken to accelerate to 28 m /s

= 28 / 2 = 14 s

a ) Total length of time in motion

= 14 + 41 + 5

= 60 s .

b )

Distance covered while accelerating

s = ut + 1/2 at²

= 0 + .5 x 2 x 14²

= 196 m .

Distance covered while moving in uniform motion

= 28 x 41

= 1148 m

distance covered while decelerating

v = u - at

0 = 28 - a x 5

a = 5.6 m / s²

v² = u² - 2 a s

0 = 28² - 2 x 5.6 x s

s = 28² / 2 x 5.6

= 70 m .

Total distance covered

= 196 + 1148 + 70

= 1414 m

total time taken = 60 s

average velocity

= 1414 / 60

= 23.56 m /s .

Answer:

55.3

Explanation:

The computation of the number of bright-dark-bright fringe shifts observed is shown below:

[tex]\triangle m = \frac{2d}{\lambda} (n - 1)[/tex]

where

d = [tex]3.95 \times 10^{-2}m[/tex]

[tex]\lambda = 400 \times 10^{-9}m[/tex]

n = 1.00028

Now placing these values to the above formula

So, the  number of bright-dark-bright fringe shifts observed is

[tex]= \frac{2 \times3.95 \times 10^{-2}m}{400 \times 10^{-9}m} (1.00028 - 1)[/tex]

= 55.3

We simply applied the above formula so that the number of bright dark bright fringe shifts could come

Answer:

6,650 newtons

Explanation:

The computation of the force exerted on it by static friction is shown below:

Data provided in the question

Mass of car = m = 1,900 kg

speed = v = 14 m/s

radius = r = 56 m

Let us assume friction force be f

And, the Coefficient of friction = [tex]\mu[/tex]= 0.88

As we know that

[tex]f = \frac{mv^2}{r}[/tex]

[tex]= \frac{1,900 \times 14^2}{56}[/tex]

= 6,650 newtons

We simply applied the above formula so that the force exerted could come