Two charged particles exert a force of 0.042 N on each other. Suppose that the particles are moved such that they are now one-eighth as far apart. What is the new force between the particles?

Answer:

θ₁ = 0.5 revolution

Explanation:

We will use the conservation of angular momentum as follows:

\(L_1=L_2\\I_1\omega_1=I_2\omega_2\)

where,

I₁ = initial moment of inertia = 18 kg.m²

I₂ = Final moment of inertia = 3.6 kg.m²

ω₁ = initial angular velocity = ?

ω₂ = Final Angular velocity = \(\frac{\theta_2}{t_2} = \frac{2\ rev}{1.2\ s}\) = 1.67 rev/s

Therefore,

\((18\ kg.m^2)\omega_1 = (3.6\ kg.m^2)(1.67\ rev/s)\\\\\omega_1 = \frac{(3.6\ kg.m^2)(1.67\ rev/s)}{(18\ kg.m^2)}\\\\\omega_1 = \frac{\theta_1}{t_1} = 0.333\ rev/s\\\\\theta_1 = (0.333\ rev/s)t_1\)

where,

θ₁ = revolutions if she had not tucked at all = ?

t₁ = time = 1.5 s

Therefore,

\(\theta_1 = (0.333\ rev/s)(1.5\ s)\\\)

θ₁ = 0.5 revolution

Because the mass and displacement are already given in Kg and m, respectively, in the first part of your question, there is no need to convert them. However, in the second part of your question, you must use the given equation to calculate the spring constant.

if the table data is given in grams and  cm you have to convert it using the following conversion,

1. To convert grams to kilograms, we divide the mass values by 1000.

2. To convert centimeters to meters, we divide the displacement values by 100.

But here in the given table it's already given the mass in kg and the displacement in meters (m). so no need to convert it.

Now comes the second part of your question,

To calculate the spring constants for the given data, we can use the equation:

k = -mg/Δx

where:

k is the spring constant (in N/m),

m is the mass (in kg), and

Δx is the displacement of the spring (in m).

Let's calculate the spring constants using the provided data:

Mass (kg): 0.05  0.1  0.2  0.3  0.4  0.5  0.6

Displacement of Spring (m): 0.012  0.027  0.065  0.1  0.135  0.17  0.199

Using the equation

k = -mg/Δx,

we can calculate the spring constant for each data point:

For the first data point (m = 0.05 kg, Δx = 0.012 m):

k = -0.05 kg * 9.8 m/s² / 0.012 m

k ≈ -40.833 N/m

Similarly, we can calculate the spring constants for the other data points:

For the mass of 0.05 kg, the spring constant is approximately -40.833 N/m.

For the mass of 0.1 kg, the spring constant is approximately -18.519 N/m.

For the mass of 0.2 kg, the spring constant is approximately -6.154 N/m.

For the mass of 0.3 kg, the spring constant is approximately -3.267 N/m.

For the mass of 0.4 kg, the spring constant is approximately -2.222 N/m.

For the mass of 0.5 kg, the spring constant is approximately -1.716 N/m.

For the mass of 0.6 kg, the spring constant is approximately -1.449 N/m.

Therefore, In the first part of the question, there is no need to convert the mass into kg and the displacement cm into m because it is already given in kg and m respectively, and in the second part question you have to calculate the spring constant using the given equation.

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Answer:

please mark me as brainliest I'm begging u

Explanation:

The behavior of an object that has been charged is dependent upon whether the object is made of a conductive or a nonconductive material. Conductors are materials that permit electrons to flow freely from particle to particle. An object made of a conducting material will permit charge to be transferred across the entire surface of the object. If charge is transferred to the object at a given location, that charge is quickly distributed across the entire surface of the object. The distribution of charge is the result of electron movement. Since conductors allow for electrons to be transported from particle to particle, a charged object will always distribute its charge until the overall repulsive forces between excess electrons is minimized. If a charged conductor is touched to another object, the conductor can even transfer its charge to that object. The transfer of charge between objects occurs more readily if the second object is made of a conducting material. Conductors allow for charge transfer through the free movement of electrons

The angle that the net force makes with respect to the x- axis, is determined as 55.3⁰.

The angle that the net force makes with respect to the +x axis, is determined as follows;

F = qvBsinθ

Fy =  qv(By)sinθ

Fx =  qv(Bx)sinθ

The angle that the net force makes with respect to the +x axis;

tanθ = Fy/Fx

tanθ = qv(By)sinθ / qv(Bx)sinθ

tanθ = By/Bx

tanθ = 0.065/0.045

tanθ = 1.444

θ = tan⁻¹(1.444)

θ = 55.3⁰

Thus, the angle that the net force makes with respect to the x- axis, is determined as 55.3⁰.

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Answer:

Both.

Explanation:

Given that a block and a ball have the same mass and move with the same initial velocity across a floor and then encounter identical ramps. The block slides without friction and the ball rolls without slipping. 1)Which one makes it furthest up the ramp ?

Since both of them have the same mass and the same initial velocity, then, they will both have the same kinetic energy.

That is,

K.E = 1/2mv^2

Friction is a force that opposes motion. And since the frictional force is zero,

Both of them will accelerate from Newton's law.

F = ma

We can therefore conclude that both of them will make it further up the ramp.

the force the woman exerts to do a push-up at constant speed is 333 N.

the work the woman does is 152 J.

her useful power output is 76 W.

(a) To calculate the force the woman exerts to do a push-up, we need to use torque methods. The woman is doing a push-up at constant speed, which means that the net torque on her body is zero. The only torque acting on her body is due to her weight W, which acts at the center of mass of her body. The distance between her center of mass and her hands is 0.76 m, and the angle between her body and the horizontal is 45 degrees.

The torque due to her weight about her hands is given by:

τ = r x W = (0.76 m) x (cos 45°)(W)

where r is the distance between her hands and her center of mass and cos 45° is the component of the distance perpendicular to the weight vector. Since the woman is at constant speed, the torque she exerts about her hands must be equal and opposite to the torque due to her weight. Therefore:

τ = (0.76 m)(cos 45°)(W) = (1/2)(W)(0.76 m)

Solving for W, we get:

W = 2(τ/0.76 m) = 2[(0.5)(mg)(0.76 m)/(0.76 m cos 45°)] = 333 N

Therefore, the force the woman exerts to do a push-up at constant speed is 333 N.

(b) The work the woman does is equal to the change in her potential energy as her center of mass rises. The woman's mass is not given, so we will assume a value of 60 kg. The gravitational potential energy of the woman is given by:

U = mgh

where m is the mass of the woman, g is the acceleration due to gravity (9.81 m/s^2), and h is the height her center of mass rises (0.26 m). Therefore:

U = (60 kg)(9.81 m/s^2)(0.26 m) = 152 J

Therefore, the work the woman does is 152 J.

(c) The useful power output of the woman is the work she does per unit time, taking into account the work done in lowering her body. Each push-up involves two phases: lifting her body and lowering her body. When she lowers her body, the work done is negative, as the force she exerts is in the opposite direction to the displacement. The work done in lowering her body is equal to the work done in lifting her body, so the total work done in one push-up is zero.

The woman does 30 push-ups in 1 minute, which means she does one push-up every 2 seconds. Therefore, the useful power output of the woman is:

P = (152 J)/(2 s) = 76 W

Therefore, her useful power output is 76 W.

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Reduce the distance by a factor of 4 change the distance between two charged particles to increase the electric force between them by a factor of 16.

Thus, When charged things interact with other objects, there is an electric force present in the system.

The electric force between them is appealing because positive charges are attracted to negative charges. For two positive charges or two negative charges, the electric force is repellent.

A typical illustration of this is what happens when two balloons are rubbed on a blanket. When you rub the balloons against the blanket, electrons from the blanket transfer to the balloons, leaving the blanket positively charged and the balloons negatively charged.

Thus, Reduce the distance by a factor of 4 change the distance between two charged particles to increase the electric force between them by a factor of 16.

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Answer:

We believe that everyone deserves access to clear, factual information that helps them educate themselves on the issues of the day and the things that pique their curiosity. This month we’re aiming to raise 3,000 new gifts in order to keep our 2023 coverage, including everything on Vox.com, free — because we believe everyone deserves the chance to access information that helps them learn about the world. This year, alone, our generous supporters helped keep over 2,500 articles, 100 videos, and 650 Vox podcasts that have informed and educated millions of people around the world free. We’re so close to our goal of raising 3,000 new gifts by December 31. If you value our work, please help us close the gap by making a one-time gift to Vox today.

Answer: positive

Explanation:

Gravity can be defined as the force with which the body is attracted towards the center of the earth, or towards any other body. If the force acting on the body is in the direction of displacement then the word done by the applicable force is positive. This causes the free fall of the ball under the influence of gravity is also positive.

Hi there!

We can begin by finding the total time taken for the ball to reach the net using the equation:

dₓ = vₓt

12.9 = 44.7t

12.9/44.7 = t = 0.289 s

Now, we can use the following equation to solve for displacement in the Y direction:

d = y₀ + vit + 1/2at²

There is no initial vertical velocity, so:

d = y₀ + 1/2at²

Plug in known values:

d = 1.28 + 1/2(-9.8)(0.289²)

d = 0.87m

Thus, since 0.87 m < 0.914 m, the tennis ball does NOT make it over the net.

1. The maximum height above its launch level is 1.45 m

2. The time of flight of the ball is 1.1 s

1. How do I determine the maximum height?

From the question given above, the following data were obtained:

The maximum height can be obatianed as follow:

H = u²Sine²θ / 2g

H = [5.5² × (Sine 76)²] / (2 × 9.8)

Maximum height = 1.45 m

How do I determine the time of flight?

The time of flight of the ball can be obtained as follow:

T = 2uSineθ / g

T = [2 × 5.5 × Sine 76] / 9.8

Time of flight = 1.1 s

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Answer:

At the equator؛ the sun is always at 90 degerees in the equator at that day.

Answer:

Sound Waves

Light Waves

Explanation:

THIS IS THE COMPLETE QUESTION BELOW

With what minimum speed must you toss a 130 g ball straight up to just touch the 14-m-high roof of the gymnasium if you release the ball 1.1 m above the ground

And what speed does the ball hit the ground? Solve this problem using energy.

Answer

a)minimum speed must you toss a 130 g is 15.9090m/s

b)speed the ball hit the ground is 16.57m/s

Explanation:

a)We know that For any closed/isolated system, the total energy is CONSERVED.

K.E. lost by the ball=The change in P.E of the ball at 1.1m above the ground as well as the P.E. of the ball at 14 m-high roof

This statement can be expressed as the expression below from K.E and P.E energy formula

P.E. = mgh

K.E. = (1/2)mv^2

Therefore,

(mgh1 - mgh2)=(1/2)mv^2

Where h2=the ball height above the ground=1.1m

h1=ball height at roof of the gymnasium= 14m

Then if we substitute we have

[(10) x (0.14) x (9.81)] - [(1.1) x (0.14) x (9.81)] = (1/2)(0.14)(v^2)

16.45137=0.065V^2

V=15.9090m/s

minimum speed must you toss a 130 g is 15.9090m/s

b)To calculate the speed the ball hit the ground?

This is the highest point (14m-high roof),and the type of the energy the ball possesses is Po.tential energy only.

At the lowest point (ground), the energy the ball possesses is K.E. only.

P.E at 10m-high roof = K.E. at ground.

(14) x (0.13) x (9.81) = (1/2) x (0.13) x v^2

17.8542= 0.065V^2

V= 16.57

Therefore,And speed the ball hit the ground is 16.57m/s

Answer:29.1

Explanation:

acellus!!!!

Answer:

Explanation:

200.34

The power to stop the car with kinetic energy of a car is  \(8*10^{6} J\) as it travels along a horizontal road is  \(8*10^{5} watt\), option B

Kinetic energy can be seen as one that is been recorded when an object is able to move from a place , in a broad term we can say this is the energy that can be attributed to that of someone leaving a place and go to another place hence we can see it as the one in the motion.

The definition of energy as the "power to accomplish work" refers to the capacity to apply a force that moves an object. Even if the word is vague, it is clear what energy actually means: it is the force that causes objects to move. The two types  can be attributed to the one we know which are kinetic and potential energy.

\(Power \frac{Energy}{time}\)

\(Energy = 8*10^{6} J\)

\(time = 10 s\)

\(Power = \frac{8*10^{6} J}{10}\)

\(power = 8*10^{5} watt\)

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proper question;

The kinetic energy of a car is 8 × 106 J as it travels along a horizontal road. How much power is required to stop the car in 10 s? (A) zero joules (B) 8  105 J (C) 8  107 J (D)8  104 J (E) 8  106 J​

Answer:

false

Explanation:

A vector quantity is the magnitude of a given quantity? True Or False

Answer:

The fields are at right angles to each other and to the direction of the wave.

Explanation:

Answer:

Her average speed is: 100 m/min

Explanation:

Recall that the formula for average speed is given by:

Speed = Distance / time

Then in our case, this is

Speed = 300 m / 3 min = 100 m/min

Answer:

e

Explanation:

Answer:

On the way to the Moon the Apollo astronauts reached a point where the. Moon's gravitational pull became stronger, than the Earth's. (a) Determine the distance of ...

Explanation:

Clock a remains in place and clock b is carried around the earth (40,000 km). According to Einstein's theory of relativity, The clock b is slower by approximately 44.6 seconds.

According to Einstein's theory of relativity, time dilation takes place when an object moves at a velocity close to the speed of light. The closer the velocity is to the speed of light, the more time slows down. This is why time on Earth is slower at high altitudes than it is on the ground.

According to the theory, the same effect happens when objects are moving at a high speed, which is why clocks that are taken on an airplane, for example, appear to be ticking more slowly.

1. The following equation is used to determine the time dilation:

t = t0 / √(1 – v²/c²),

where t is the time elapsed, t0 is the time at rest, v is the velocity, and c is the speed of light. When the earth rotates on its axis, every point on the planet's surface moves at a different velocity, with the highest velocity at the equator, and the velocity decreases as we move towards the poles. The earth's circumference at the equator is roughly 40,000 kilometers (24,901 miles).
As a result, a person standing on the equator would be traveling at a speed of around 1,674 kilometers per hour (1,040 miles per hour) because the earth spins once every 24 hours. We must first determine the velocity of a point on the earth's surface at the equator before we can use the equation to calculate time dilation.

2. We use the formula

v = 2πr / T,

where v is velocity, r is the radius of the earth, and T is the time it takes the earth to complete one rotation. The formula is as follows:

v = 2πr / Tv

= 2 x 3.14 x 6,378 km / 24 hv

= 1,674 km/h

3. Substituting these values into the equation, we get:

t = t0 / √(1 – v²/c²)t = t0 / √(1 – (1,674 m/s)² / (299,792,458 m/s)²)t = t0 / √(1 – 2.8 x 10^-8)t = t0 / 0.9999999714

This means that the clock on the equator will tick slightly slower than it would at rest. The difference in time can be calculated by subtracting the two values:

t – t0 = t0 / 0.9999999714 – t0t – t0 = t0 (1 – 0.9999999714)t – t0 = 0.0000000286 t0

4. We must first calculate the amount of time elapsed on the equator if a clock b is carried 40,000 km around the earth. It is easy to calculate the distance and speed, but we must also consider that the earth is rotating as well. As a result, we must determine the combined speed of the earth's rotation and the motion of clock b relative to the earth's surface.

5. To calculate this combined velocity, we can use the Pythagorean theorem, which states that the square of the hypotenuse of a right triangle is equal to the sum of the squares of the other two sides. If we imagine the velocity of the earth's rotation as the base of the triangle and the velocity of clock b as the height of the triangle, we can use this theorem to calculate the combined velocity as follows:

combined velocity = √(1,674² + vclock²)

where v clock is the velocity of clock b. Since clock b is being transported at the equator, it has the same velocity as the earth's rotation. As a result, we can substitute 1,674 km/h for v clock:

combined velocity = √(1,674² + 1,674²)

combined velocity = √(2 x 1,674²)

combined velocity = 2,367 km/h

6. Substituting the combined velocity into the equation for time dilation, we obtain:

t – t0 = t0 (1 – √(1 – v²/c²))t – t0 = t0 (1 – √(1 – (2,367 km/h)² / (299,792,458 m/s)²))t – t0

= t0 (1 – √(1 – 1.579 x 10^-11))t – t0

= t0 (1 – 0.999999999920215)t – t0

= 0.000000000079785 t0

Converting this value to seconds, we get:

0.000000000079785 t0 = 79.785 ns

Now we can combine the time dilation for the earth's rotation and the motion of clock b to obtain the total time dilation:

t – t0 = 0.0000000286 t0 + 0.000000000079785 t0t – t0 = 0.000000028679785 t0

Substituting the value of t0 (one second) into the equation, we get:

t – 1 = 0.000000028679785 seconds

Therefore, clock b will be approximately 44.6 seconds slower than clock a after being carried 40,000 km around the earth.

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The number of distinct ways of placing four indistinguishable balls into five distinguishable boxes is 20.

To solve this problem, we can use the stars and bars method, which is a combinatorial technique used to count the number of ways to distribute indistinguishable objects into distinguishable containers. In this case, we have four indistinguishable balls and five distinguishable boxes.

The stars and bars method works by representing each ball as a star and using bars to separate the balls into different boxes. For example, if we wanted to distribute two balls into three boxes, we could use the following diagram:

* | * * | *

In this diagram, the first and last bars represent the boundaries of the containers, while the stars represent the balls.

The second bar separates the first two balls from the last ball, indicating that the first two balls are in the first container and the last ball is in the third container.

To distribute four balls into five boxes, we need to use three bars to separate the balls into four groups. We have a total of six spaces to place the bars (including the boundaries), and we need to choose three of them to place the bars.

Therefore, the number of distinct ways to place four indistinguishable balls into five distinguishable boxes is the same as the number of ways to choose three spaces out of six, which is:

6 choose 3 = (6!)/(3!3!) = 20

Therefore, there are 20 distinct ways to place four indistinguishable balls into five distinguishable boxes using the stars and bars method.

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Answer:

B. The maximum angle decreases

Explanation:

If θ be the maximum angle of a slope that allows a crate placed on it to remain at rest , following condition exists .

tanθ = μ , θ is called angle of repose . μ is coefficient of static friction .

So the tan of angle of repose θ is proportional to coefficient of static friction.

If coefficient of static friction is less than .7 , naturally angle of repose will also become less ,ie,  it at lower angle of inclination , the object will start slipping .

Answer:

B. combustion

Explanation:

Combustion process will produce the least amount of energy to power an aircraft carrier.

Combustion is a chemical change in which a fuel combines with oxygen to liberate energy from chemical energy.

Nuclear fusion, radioactive decay and nuclear fission are nuclear changes. They take place in the nucleus of atoms. They usually liberate energy in several orders compared to combustion.

it’s radioactive decay

Answer:

a) the total electric potential is 2282000 V

b) the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

Explanation:

Given the data in the question and as illustrated in the image below;

a) Determine the total electric potential (in V) at the origin.

We know that; electric potential due to multiple charges is equal to sum of electric potentials due to individual charges

so

Electric potential at p in the diagram 1 below is;

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

we know that; Coulomb constant, k = 9 × 10⁹ C

q1 = 4.60 uC = 4.60 × 10⁻⁶ C

r1 = 1.25 cm = 0.0125 m

q2 = -2.06 uC = -2.06 × 10⁻⁶ C

location x2 = −1.80 cm; so r2 = 1.80 cm = 0.018 m

so we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0125 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.018 )

Vp = (3312000) + ( -1030000 )

Vp = 3312000 -1030000

Vp = 2282000 V

Therefore, the total electric potential is 2282000 V

b)

the total electric potential (in V) at the point with coordinates (0, 1.50 cm).

As illustrated in the second image;

r1² = 0.015² + 0.0125²

r1 = √[ 0.015² + 0.0125² ]

r1 = √0.00038125

r1 = 0.0195

Also

r2² = 0.015² + 0.018²

r2 = √[ 0.015² + 0.018² ]

r2 = √0.000549

r2 = 0.0234

Now, Electric Potential at P in the second image below will be;

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0195 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.0234 )

Vp = 2123076.923 + ( -762962.962 )

Vp = 2123076.923 -792307.692

Vp =  1330769.23 V

Therefore, the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

a) The total electric potential is 2282000 V

b) The total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

The electric potential is defined as the amount of work energy needed to move a unit of electric charge from a reference point to a specific point in an electric field.

Given the data in the question and as illustrated in the image below;

a) Determine the total electric potential (in V) at the origin.

We know that; electric potential due to multiple charges is equal to sum of electric potentials due to individual charges

Electric potential at p in diagram 1 below is;

\(V_P=V_1+V_2\)

\(Vp = \dfrac{kq_1}{r_1} + \dfrac{kq_2}{r_2}\)

we know that; the Coulomb constant, k = 9 × 10⁹ C

q1 = 4.60 uC = 4.60 × 10⁻⁶ C

r1 = 1.25 cm = 0.0125 m

q2 = -2.06 uC = -2.06 × 10⁻⁶ C

location x2 = −1.80 cm; so r2 = 1.80 cm = 0.018 m

so we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0125 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.018 )

Vp = (3312000) + ( -1030000 )

Vp = 3312000 -1030000

Vp = 2282000 V

Therefore, the total electric potential is 2282000 V

b)The total electric potential (in V) at the point with coordinates (0, 1.50 cm).

As illustrated in the second image;

\(r_1^2=0.015^2+0.0125^2\)

\(r_1 = \sqrt{[ 0.015^2 + 0.0125^2 ]\)

\(r_1 = \sqrt{0.00038125}\)

\(r_1 = 0.0195\)

Also

\(r_2^2 = 0.015^2 + 0.018^2\)

\(r_2 = \sqrt{0.015^2 + 0.018^2}\)

\(r_2 = \sqrt{0.000549\)

\(r_2 = 0.0234\)

Now, Electric Potential at P in the second image below will be;

Vp = V1 + V2

\(Vp = \dfrac{kq_1}{r_1} + \dfrac{kq_2}{r_2}\)

we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0195 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.0234 )

Vp = 2123076.923 + ( -762962.962 )

Vp = 2123076.923 -792307.692

Vp =  1330769.23 V

Therefore, the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

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Hello!

a)
For a car on an incline, we only have the normal force and force of gravity acting on the car.

The car is only experiencing a net force caused by the sine component of the force of gravity vector, which causes it to slide down the incline towards the center of the curve.

Or, as an equation:

\(F_{net} = Mgsin\phi\)

This net force produces a centripetal force. Recall the equation for centripetal force:
\(F_c = \frac{mv^2}{r}\)

In reference to the 15° angle of the incline, the cosine component of the centripetal force is equivalent to the sine component of the force due to gravity (both parallel to the incline). So:
\(F_c cos\phi = Mgsin\phi \\\\\frac{mv^2}{r}cos\phi = mgsin\phi\)

Cancel out 'm' and solve for 'v'.

\(\frac{v^2}{r}cos\phi = gsin\phi\\\\v^2 = gr \frac{sin\phi}{cos\phi}\\\\v = \sqrt{grtan\phi}\)

Plug in the given values and solve.

\(v = \sqrt{(9.8)(85)tan(15)} = \boxed{14.94 \frac{m}{s}}\)

b)
Begin by converting 20.0 km/h to m/s.

\(\frac{20 km}{hr} * \frac{1 hr}{3600 s} * \frac{1000m}{1 km} = 5.556 \frac{m}{s}\)

For this situation, we also have the force of friction present along the axis of the sine component of the force of gravity that contributes to the net force.

Recall the equation of kinetic friction:
\(F_f = \mu_k N\)

In this situation, we have the sine (vertical) component of the centripetal force as well as the cosine component of the force of gravity making up the normal force, so:
\(F_f = \mu_k (\frac{mv^2}{r}sin\phi + mgcos\phi)\)

If a curve is banked at a slower speed than appropriate, the car will tend to slide towards the center. Thus, this force of friction points up the incline, opposite to the force due to gravity. We can do another summation of forces like above.

\(\frac{mv^2}{r} cos\phi= mgsin\phi - \mu_k (\frac{mv^2}{r}sin\phi + mgcos\phi)\)

Cancel out 'm' and simplify the equation further to solve for μ.

\(\frac{v^2}{r} cos\phi= gsin\phi - \mu_k (\frac{v^2}{r}sin\phi + gcos\phi)\\\\\mu_k (\frac{v^2}{r}sin\phi + gcos\phi)= gsin\phi - \frac{v^2}{r} cos\phi\\\\\mu_k = \frac{gsin\phi - \frac{v^2}{r} cos\phi}{(\frac{v^2}{r}sin\phi + gcos\phi)}\)

Plug in values.

\(\mu_k = \frac{9.8sin(15) - \frac{5.556^2}{85} cos(15)}{\frac{5.556^2}{85}sin(15) + 9.8cos(15)} = \boxed{0.2286}\)

Answer:

The last option is the only correct one if you like to multiply

The second last option is good if you like to divide.

Explanation:

Each fraction in the last two options has a value of 1

example

dividing by 1

15 cm /(100 cm/ 1 m) = 0.15 m          0.15 m / (1000 m/ 1km) = 0.00015 km

and

multiplying by 1

15 cm(1 m / 100cm) = 0.15 m         0.15m(1 km/1000m) = 0.00015 km

only one of the two fractions in each of the top two options has a value of 1.

Answer:

Gradient of the hill is 53.33m/km which is lesser than the 500 m/km. So, he will be able to climb the hill conveniently.

Explanation: Given that

Maximum gradient = 500 m/km

Total distance = 1.5 km

Starting elevation = 20 m

Final elevation = 100 m

Gradient = change in elevation/ total distance.

Now, substitute the values into the formula.

Gradient = (100m - 20m)/1.5km

= 80m/1.5km

= 53.33m/km

Gradient of the hill is 53.33m/km which is lesser than the 500 m/km. So, he will be able to climb the hill conveniently.