A thin circular wooden hoop of mass m and radius Rrests on a horizontal frictionless plane. A bullet, also of mass m, moving with horizontal velocity v, strikes the hoop and becomes embedded in it as shown in the figure. (a) (2) Calculate the center of mass velocity (b) (2) Calculate the angular momentum of the system about the CM. (c) (32) Calculate the angular velocity w of the hoop. (d) (4) Calculate the kinetic energy of the system, before and after collision (e) (48) Find a point of the hoop which remains at rest after collision

Answers

Answer 1

The velocity of the center of mass of the system is the same as the velocity of the bullet, which is v. The total angular momentum of the system about the center of mass is  Lhoop + Lbullet. he kinetic energy of the system, before and after collision Khoop + Kbullet and Khoop_after + Kbullet_after respectively. The axis of rotation, which is the center of the hoop, will remain at rest after the collision.

The center of mass of the system is at point Q, which is a distance d from point O, where:d = (mP + mO)/(2*m).The velocity of the center of mass is the derivative of the position with respect to time. Since the bullet is moving with constant velocity, the position of its center of mass is also moving with constant velocity. The position of the center of mass of the hoop is not changing, so its velocity is zero. Therefore, the velocity of the center of mass of the system is the same as the velocity of the bullet, which is v.

The angular momentum of the bullet about the center of mass is:Lbullet = Ibulletw + md*v where Ibullet is the moment of inertia of the bullet about its center of mass, and d is the distance from the center of mass to the point of contact between the bullet and the hoop. The total angular momentum of the system about the center of mass is the sum of the angular momenta of the hoop and the bullet: Ltotal = Lhoop + Lbullet

Since angular momentum is conserved, this change in angular momentum must be due to the change in the angular momentum of the hoop. Therefore, we can set the change in angular momentum of the hoop equal to Ltotal and solve for the angular velocity w:

Ltotal = Lhoop + Lbullet

Ltotal = Ihoopw + mRv + Ibulletw + mdv

Ltotal = (Ihoop + Ibullet)w + (mR + md)v

w = Ltotal/(Ihoop + Ibullet) - (mR + md)*v/(Ihoop + Ibullet).

kinetic energy is conserved, this means that the change in kinetic energy of the hoop must be equal to the change in kinetic energy of the bullet. Therefore, we can set the change in kinetic energy of the hoop equal to the change in kinetic energy of the bullet and solve for the kinetic energy of the hoop after the collision:

Khoop_after - Khoop = Kbullet - Kbullet_after

Khoop_after = Khoop + Kbullet - Kbullet_after

The kinetic energy of the system after the collision is the sum of the kinetic energies of the hoop and the bullet after the collision:

Ktotal_after = Khoop_after + Kbullet_after

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Related Questions

An interference pattern is produced by light with a wavelength 550 nm from a distant source incident on two identical parallel slits separated by a distance (between centers) of 0.500 mm .
a. If the slits are very narrow, what would be the angular position of the second- order, two-slit interference maxima?
b. Let the slits have a width 0.300 mm. In terms of the intensity lo at the center of the central maximum, what is the intensity at the angular position in part "a"?

Answers

Answer:

a

 \(\theta = 0.0022 rad\)

b

 \(I = 0.000304 I_o\)

Explanation:

From the question we are told that  

   The  wavelength of the light is \(\lambda = 550 \ nm = 550 *10^{-9} \ m\)

    The  distance of the slit separation is  \(d = 0.500 \ mm = 5.0 *10^{-4} \ m\)

 

Generally the condition for two slit interference  is  

     \(dsin \theta = m \lambda\)

Where m is the order which is given from the question as  m = 2

=>    \(\theta = sin ^{-1} [\frac{m \lambda}{d} ]\)

 substituting values  

      \(\theta = 0.0022 rad\)

Now on the second question  

   The distance of separation of the slit is  

       \(d = 0.300 \ mm = 3.0 *10^{-4} \ m\)

The  intensity at the  the angular position in part "a" is mathematically evaluated as

      \(I = I_o [\frac{sin \beta}{\beta} ]^2\)

Where  \(\beta\) is mathematically evaluated as

       \(\beta = \frac{\pi * d * sin(\theta )}{\lambda }\)

  substituting values

     \(\beta = \frac{3.142 * 3*10^{-4} * sin(0.0022 )}{550 *10^{-9} }\)

    \(\beta = 0.06581\)

So the intensity is  

    \(I = I_o [\frac{sin (0.06581)}{0.06581} ]^2\)

   \(I = 0.000304 I_o\)

A small block sits at one end of a flat board that is 2.50 m long. The coefficients of friction between the block and the board are μs = 0.450 and μk = 0.400. The end of the board where the block sits is slowly raised until the angle the board makes with the horizontal is α0, and then the block starts to slide down the board.

Answers

Hi there!

In order for a block to begin sliding, the force due to STATIC friction must be overcome.

In this instance, the following forces are acting on the block ALONG the axis of the incline.

Force due to gravity (Fg)Force due to STATIC friction (Fs)

Force due to gravity:

On an incline, the component of the force due to gravity contributing to the object's downward movement is equivalent to the horizontal (sine) component.

\(F_g = Mgsin\theta\)

Force due to static friction:
The force due to friction is equivalent to the normal force multiplied by the coefficient of friction.

The normal force is the cosine component (perpendicular to the incline), so:
\(N = Mgcos\theta\\\\F_s = \mu_sMgcos\theta\)

To find the minimum angle for the block to begin sliding, we can set the two forces equal to 0. They work in opposite directions (let down the incline be negative and up the incline be positive).

\(\Sigma F = F_s - F_g\\\\0 = F_s - F_g\\\\0 = \mu_sMgcos\theta - Mgsin\theta\\\\Mgsin\theta = \mu_sMgcos\theta\)

Cancel out 'Mg' and rearrange to solve for theta.

\(sin\theta = \mu_scos\theta\\\\tan\theta = \mu_s\\\\\theta = tan^{-1}(\mu_s) = tan^{-1}(.45) = \boxed{24.228^o}\)

A 1.40-kg block is on a frictionless, 30 ∘ inclined plane. The block is attached to a spring (k = 40.0 N/m ) that is fixed to a wall at the bottom of the incline. A light string attached to the block runs over a frictionless pulley to a 60.0-g suspended mass. The suspended mass is given an initial downward speed of 1.60 m/s .
How far does it drop before coming to rest? (Assume the spring is unlimited in how far it can stretch.)
Express your answer using two significant figures.

Answers

Answer:

0.5

Explanation:

because the block is attached to the pulley of the string

Four cylindrical wires of different sizes are made of the same material. Which of the following combinations of length and cross-sectional area of one of the wires will result in the smallest resistance?
a. Length Area
3L 3a
b. Length Area
3L 6a
c. Length Area
6L 3a
d. Length Area
6L 6a

Answers

Answer:

Explanation:

For resistance of a wire , the formula is as follows .

R = ρ L/S

where ρ is specific resistance , L is length and S is cross sectional area of wire .

for first wire resistance

R₁ =  ρ 3L/3a = ρ L/a

for second wire , resistance

R₂ = ρ 3L/6a

= .5 ρ L/a

For 3 rd wire resistance

R₃ = ρ 6L/3a

= 2ρ L/a

For fourth wire , resistance

R₄ = ρ 6L/6a

=  ρ L/a

So the smallest resistance is of second wire .

Its resistance is .5 ρ L/a

I NEED HELP WITH THIS PLEASE

I NEED HELP WITH THIS PLEASE

Answers

Answer:

D

Explanation:

One of the factors of increasing the rate of a reaction is increasing concentration. Therefore adding more people increases the number of people on the dance floor , therefore increasing the concentration increases the rate of reaction.

if an 80kg object has 632 J of PE, how high is it

Answers

The object is on 81 cm high.

What is potential energy?

The energy that an item has or acquires when its location changes as a result of being in a gravitational field is known as gravitational potential energy. Gravitational potential energy can be defined as an energy that has a connection to gravitational force or gravity.

Given that: an 80kg object has 632 J of Potential energy.

Let the height is h: then,

mgh = PE

80 kg × 9.8× m/s^2 × h = 632 Joule

h = 0.81 m = 81 cm.

The object is on 81 cm high.

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A substance has a mass of 15,000 kg and a volume of 30 m 3 . What is the density of thesubstance?

Answers

ANSWER

500 kg/m³

EXPLANATION

Given:

• The mass of the substance, m = 15,000 kg

,

• The volume of the substance, V = 30 m³

Find:

• The density of the substance, ρ

The density of a substance of mass m and volume V is,

\(\rho=\frac{m}{V}\)

Substitute the known values and solve,

\(\rho=\frac{15,000kg}{30m^3}=500kg/m^3\)

Hence, the density of the substance is 500 kg/m³.

What is curved space?

Answers

Curved space often refers to a spatial geometry which is not "flat", where a flat space is described by Euclidean geometry. Curved spaces can generally be described by Riemannian geometry though some simple cases can be described in other ways.

(8.6) A space station in deep space is spun like a giant wheel to produce centrifugal force so the occupants experience artificial gravity of one g. How does a clock at the rim of the space station compare with one residing at the hub? What does this say about the behavior of a clock sitting on the surface of a planet with a surface gravity of one g?

Answers

The way that the clock at the rim of the space station compare with one residing at the hub is that Rim clock will always runs slower.

The behavior of a clock sitting on the surface of a planet with a surface gravity of one g is the same.

Is there a timepiece on the ISS?

For instance, timekeeping on the International Space Station (ISS) is slightly slower than that of reference clocks on Earth. This explains why astronauts on the International Space Station (ISS) mature more slowly, lagging by 0.007 seconds every six months.

Ground-based atomic clocks have served as the industry standard for timekeeping since the 1950s. These clocks use the extremely steady and accurate frequencies of light emitted by particular atoms to keep time.

Therefore, Satellite clocks run slower due to more velocity but faster due to reduced gravity. Clocks move more slowly at the ISS orbital height because velocity dominates. At GPS altitude, clocks move more quickly due to the influence of gravity.

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A 6.30-N force is applied to a 4.5-kg object to accelerate it rightwards. Neglecting any frictional forces, what is the acceleration of the object?

Answers

Answer:

1.4 m/s2

Explanation:

i took the test

Brainliest if correct Question 6 of 10
According to the law of conservation of energy, which statement must be
true?
A. There is only one form of energy.
B. Energy that is transformed is neither destroyed nor created.
C. The total energy in a system can only increase over time.
D. Energy can change only from nuclear to chemical

Answers

Answer:

B. Energy that is transformed is neither destroyed nor created.

Explanation:

Hope this helped! :)

What kind of energy do electric and magnetic fields contain?
A
thermal energy
B
electrical energy
C
kinetic energy
D
pot

Answers

Answer:

Electromagnetic waves

Electromagnetic waves bring energy into a system by virtue of their electric and magnetic fields. These fields can exert forces and move charges in the system and, thus, do work on them. However, there is energy in an electromagnetic wave itself, whether it is absorbed or not.

So the answer is B electrical energy

Thank you and please rate me as brainliest as it will help me to level up

if the mass of an object were doubled, its weight would be

Answers

Weight of a body is defined as the force with which it is attached to the earth.

mass is the total quantity of matter present.

the weight of something is given as mass * acceleration

weight of an object is directly proportional to the mass of the object.  so, if mass doubles, weight will also double.

hope this helps :)

- jeron

Increasing the surfactant concentration above the critical micellar concentration
will result in: Select one:
1.An increase in surface tension
2. A decrease in surface tension
3. No change in surface tension
4.None of the above​

Answers

Answer:

Explanation:no change in surface tension

An increase in the surfactant concentration above the critical micellar concentration will result in no change in surface tension.

In water-gas interface, surfactant reduces the surface tension of water by adsorbing at the liquid–gas interface.

Also, in oil-water interface, surfactant reduces the interfacial tension between oil and water by adsorbing at the oil-water interface.

The concentration of the surfactant can increase to a level called critical micellar concentration, which is an important characteristic of a surfactant.

As the concentration of the surfactant increases before critical micellar concentration, the surface tension changes strongly with an increase in the concentration of the surfactant. After reaching the critical micellar concentration, any further increase in the concentration will result in no change of the surface tension, that is the surface tension will be constant.

Thus, increasing the surfactant concentration above the critical micellar concentration will result in no change in surface tension.

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Tamara is playing a game of billiards. The white cue ball has a mass of 0.17 kg and all the other balls have a mass of 0.16 kg. The cue ball is moving at a velocity of 6 m/s when it collides with the number seven ball. If the cue ball comes to a complete stop after the collision, then what will the resulting velocity be on the number seven ball?​

Answers

Answer: (0.17 * 6) + ( 0.16 * 0 ) = (0.17 * 0) + ( 0.16 *  v2)

v2 = 6.375 m/s

Explanation:

The answer is v2 = 6.375 m/s

(I did the equation and everything but just in case, check with someone else so you don't get the wrong answer)

An average hole drift velocity of 103 cm/sec results when 2 V is applied across a 1 cm long semiconductor bar. What is the hole mobility inside the bar

Answers

Answer:

ε = 2 V/cm

Explanation:

To calculate the mobility inside this bar, we just need to apply the expression that let us determine the mobility. This expression is the following:

ε = ΔV / L

Where:

ε: Hole mobility inside the bar

ΔV: voltage applied in the bar

L: Length of the bar

We already have the voltage and the length so replacing in the above expression we have:

ε = 2 V / 1 cm

ε = 2 V/cm

The data of the speed can be used for further calculations, but in this part its not necessary.

Hope this helps

A hollow, thick-walled, conducting cylinder carries a current of 11.2 A and has an inner radius ri = r and outer radius ro = 3r/2, where r = 4.90 mm. Determine the magnitude of the magnetic field at the following distances from the center of the cylinder.

(a) ra = r/2 T

(b) rb = 5r/4 T

(c) rc = 2r T

Answers

Answer:

a. 0 T b. 1.03 × 10⁻⁴ T c. 2.29 × 10⁻⁴ T

Explanation:

a. Using ampere's law ∫B.ds = μ₀i

for ra = r/2, i = 0 (since no current is enclosed) and

∫B.ds = B∫ds = B(2πr/2) = Bπr

So, Bπr = 0

B = 0 T

b. rb = 5r/4

WE find the current enclosed between r = r and r = 5r/4. The total current density in the holow thick-walled conducting cylinder is J = i/[π(3r/2)² - πr²] = i/[9πr²/4 - πr²] = i/8πr²/4 = i/2πr².

Since the current density is constant, we find the current, i' enclosed between  r = r and r = 5r/4. J = i'//[π(5r/4)² - πr²] = i'/[25πr²/16 - πr²] = i'/9πr²/16 = 16i'/9πr²

So, i/2πr² = 16i'/9πr²

i' = 9i/32

Using ampere's law ∫B.ds = μ₀i'

∫B.ds = = B∫ds = B × 2π(5r/4) = 5Bπr/2

5Bπr/2 = μ₀(9i/32)

B = 9μ₀i/32πr × 2/5

B = 9μ₀i/80πr

Substituting r = 4.90 mm = 4.90 × 10⁻³ m and i = 11.2 A, we have

B = 9μ₀i/80πr

= 9 × 4π × 10⁻⁷ H/m × 11.2 A/80π(4.90 × 10⁻³ m)

= 403.2/392 × 10⁻⁴ T

= 1.029 × 10⁻⁴ T

≅ 1.03 × 10⁻⁴ T

c. rc = 2r

Using ampere's law ∫B.ds = μ₀i

∫B.ds = = B∫ds = B × 2π(2r) = 4Bπr

4Bπr = μ₀i  (since i = current enclosed = 11.2 A)

B =  μ₀i/4πr

= 4π × 10⁻⁷ H/m × 11.2 A/4π(4.90 × 10⁻³ m)

= 2.286 × 10⁻⁴ T

≅ 2.29 × 10⁻⁴ T

Problem Try to answer the following questions:(a) What is the maximum height above ground reached by the ball?(b) What are the magnitude and the direction of the velocity of the ball just before it hits the ground? Show Your Problem Solving Steps: Show these below:1) Draw a Sketch2) Choose origin, coordinate direction3) Inventory List – What is known?4) Write the kinematics equation(s) and solution of Part (a):5) Write the kinematics equation(s) and solution of Part (b):Problem 3 A small ball is launched at an angle of 30.0 degrees above the horizontal. It reaches a maximum height of 2.5 m with respect to the launch position. Find (a) the initial velocity of the ball when it’s launched and (b) its range, defined as the horizontal distance traveled until it returns to his original height. As always you can ignore air resistance.(a) Initial velocity [Hints: How is v0 related to vx0 and vy0. How can you use the information given to calculate either or both of the components of the initial velocity?](b) Range [Hints: This problem is very similar to today’s Lab Challenge except that for the challenge the ball will land at a different height.]

Problem Try to answer the following questions:(a) What is the maximum height above ground reached by

Answers

3.

\(\begin{gathered} \theta=30^{\circ} \\ y_{\max }=2.5m \end{gathered}\)

a)

\(\begin{gathered} y_{\max }=\frac{v^2\sin ^2(\theta)}{g} \\ \end{gathered}\)

Solve for v:

\(\begin{gathered} v=\sqrt[]{\frac{y_{\max }\cdot g}{\sin ^2(\theta)}} \\ v=98\cdot\frac{m}{s} \end{gathered}\)

b)

\(\begin{gathered} r=\frac{v^2}{9}\sin (2\theta) \\ r=\frac{98^2}{9.8}\cdot\sin (2\cdot30) \\ r=\frac{98^2}{9.8}\sin (60) \\ r=848.7m \end{gathered}\)

Suppose a 44-turn coil lies in the plane of the page in a uniform magnetic field that is directed out of the page. The coil originally has an area of 0.350 m2. It is stretched to have no area in 0.100 s. What is the magnitude (in V) and direction (as seen from above) of the average induced emf if the uniform magnetic field has a strength of 1.00 T

Answers

Answer:

Explanation:

Dude no one can answer this on this app

List the 6 questions you may apply to formulating a logical, reasonable perspective to any situation.

Answers

Steps to formulate a logical, reasonable perspective to any situation are: gather information, identify problem, analyze the situation, consider assumptions, generate solutions, evaluate options, consider your values, make decision and monitor and adjust

What are the 6 questions that may be applied to formulate logical, reasonable perspective to any situation?

Here are the six questions that you can apply to formulating a logical, reasonable perspective to any situation:

What are the issues that should be addressed?

What are the relevant facts and data related to this problem or issue?

What assumptions am I making about the problem or issue?

What are the possible solutions or outcomes, and what are the pros and cons of each?

What are my values and priorities related to this problem or issue?

What additional information do I need to make an informed decision or come to a reasonable conclusion?

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A trolley of mass 5.0 kg is moving at 1.0 ms to the right. A constant force of 25 N acts to the left for 0.75 seconds.

Calculate the change of kinetic energy of the trolley.
(Show Work)

Answers

Answer:

change in kinetic energy of the trolley is 53.91 J.

Explanation:

mass of the trolley, m = 5.0 kg

initial velocity of the trolley, u = 1.0 m/s

external force on the trolley, F = 25 N

time of force action, t = 0.75 s

The final velocity of the trolley at the end of 0.75 s is calculated as follows;

\(F = \frac{m(v-u)}{t} \\\\25 = \frac{5(v-1)}{0.75}\\\\5(v-1) = 18.75\\\\v-1 = \frac{18.75}{5} \\\\v-1 = 3.75\\\\v = 4.75 \ m/s \ in \ the \ direction \ of \ the \ applied \ force\)

The change in kinetic energy of the trolley is calculated as;

Δ K.E = ¹/₂m(v² - u²)

Δ K.E = ¹/₂ x 5(4.75² - 1²)

Δ K.E = 53.91 J.

Therefore, change in kinetic energy of the trolley is 53.91 J.

HI PLEASE HELP ON QUESTION ASAP USING AVERAGE (MEAN) TO ANSWER QUESTION! IF UR ANSWER AND EXPLAINATION IS CORRECT ILL RATE YOU FIVE STARS, A THANKS AND MAYBE EVEN BRAINLIEST. PLEASE MAKE SURE YOU ANSWER MY QUESTION USING AVERAGES.
1) a meal for 6 cost £12 per person. as it is one of the diners birthday , the other 5 decided to pay for his meal. how much do each of the five friends need to pay?

Answers

Each of the five friends needs to pay £12 to cover the cost of their own meals and contribute towards the birthday person's meal. Using mean allows us to distribute the cost equally among the friends, ensuring a fair division of expenses for the meal.

To determine how much each of the five friends needs to pay, we can use the concept of averages (mean) and divide the total cost by the number of people paying.

In this scenario, the total cost of the meal for 6 people is £12 per person. Since the other 5 friends have decided to pay for the birthday person's meal, they will collectively cover the cost of their own meals plus the birthday person's meal.

To calculate the total cost covered by the five friends, we can subtract the cost of one person's meal (since the birthday person's meal is being paid by the group) from the total cost. The cost of one person's meal is £12.

Total cost covered by the five friends = Total cost - Cost of one person's meal

= (£12 x 6) - £12

= £72 - £12

= £60

Now, to find out how much each of the five friends needs to pay, we divide the total cost covered by the five friends (£60) by the number of friends (5).

Amount each friend needs to pay = Total cost covered by the five friends / Number of friends

= £60 / 5

= £12

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find the image of (4 "-1)" obtained by translating 4 units up followed by a reflection over the y-axis

find the image of (4 "-1)" obtained by translating 4 units up followed by a reflection over the y-axis

Answers

Answer:(-4,3)

Explanation: They didn’t show the whole graph so it looks confusing but it’s not.

Find the emitted power per square meter of peak intensity for a 3000 K object that emits thermal radiation. Express your answer to three significant figures and include the appropriate units. H MÅ 0 = ? P= Value Units Submit Request Answer oblem 5.58 SUDITI NcqucstATIS WEL

Answers

The peak wavelength for peak intensity of 3000 K is 967 nm and peak wavelength for peak intensity of 50,000K will be 58nm.

What is Power?

Power is the amount of energy which is transferred or converted per unit of time taken. In the International System of Units, the unit of power is the watt (W), which is equal to one joule per second.

Stefan-Boltzmann constant = 5.67 × 10⁻⁸ Wm²

Peak intensity = 2000 K

P = (5.67 × 10⁻⁸) × (3000)⁴

P = 4.59 × 10⁶

P = 459 2700 watts of power per square meter.

Peak intensity = 3000 K

Using Wein law,

λ max = 0.29/ 7

λ max = 0.29/ 3000 = 9.67 × 10⁻⁵ = 967 nm

Same method as above will be used for 50,000 K

P = 3.56 × 10¹¹ W/m²

P = 356.00.000000000 W/m²

λ (Peak) = 58nm

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1.
is the rate that velocity changes
O Acceleration
O
Time
O
Distance
O
Mass

Answers

Answer:

a) acceleration

Explanation:

Acceleration is, by definition, the change of an object's velocity.

What is the index of refraction for a material if the speed of light in that material is 1.862x108 m/s?

Answers

The index of refraction for the material is 1.612.

The index of refraction (n) of a material can be calculated using the formula:

n = c / v

where c is the speed of light in vacuum and v is the speed of light in the material.

Given that the speed of light in the material is 1.862x\(10^8\) m/s, we can substitute the values into the formula:

n = (3.00x\(10^8 m/s) / (1.862x10^8 m/s\))

Simplifying the expression:

n = 1.612

Therefore, the index of refraction for the material is approximately 1.612.

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PLEASE HELP NO LINKS NEED HELP FAST
Use the scenario to answer the question.

An astronomer discovers a new galaxy using a telescope. The astronomer wants to investigate how the galaxy is moving relative to the Milky Way galaxy.

In one or two sentences, make a hypothesis about the movement of the galaxy and explain at least one way to test the hypothesis.

Answers

Answer:

The galaxies outside of our own are moving away from us, and the ones that are farthest away are moving the fastest. This means that no matter what galaxy you happen to be in, all the other galaxies are moving away from you

Explanation:

The hypothesis about the movement of the galaxy is that galaxies are moving far from each other continuously.

What is the milky way galaxy?

The milky way galaxy is a galaxy that contains over a hundred billion stars and it also includes our solar system.  Its name describes its appearance when viewed from the earth. All the individual stars in the whole sky are a portion of the Milky Way Galaxy, the term "Milky Way" is because of the band of light.

The astronomer has discovered a new galaxy which means our universe is continuously expanding. This is because the universe encloses everything that exists.

Galaxies are moving in space and since the universe space is continuously expanding so the galaxies continuously move from each other. The farther the galaxy is from the milky way which is an observable part, the faster will be moving the galaxy and the closer the galaxy is to the milky way, the slower will be movement of the galaxy.

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How much force will this car experience if it collides with a wall and the collision lasts for
0.05 s?

Answers

The amount of force after 0.05 seconds will be 528000 N.

What is a force?

A force is an influence that can cause an object's motion to change. A force can cause a mass object to change its velocity, or accelerate. Intuitively, a force can be described as a push or a pull.

Given that a car of mass 1500kg collides with a wall. the initial and final speed of the car is -15m/s and 2.6m/s. if the collision lasts for 0.05 seconds.

The force will be calculated as:-

Mc x V₁ + F x t = Mc x V₂

Here, V₁ = -15 m /s and V₂ = 2.6 m /s.

Mc x V₁ + F x t = Mc x V₂

-( 1500 x 15 ) + F x 0.05 = 1500 x 2.6

F x 0.05 = 22500 + 3900

F = 26400 / 0.05

F = 528000 N

Therefore, the force after the collision will be 528000 N.

The complete question is given below.

A car of mass 1500kg collides with a wall. the initial and final speed of the car is -15m/s and 2.6m/s. if the collision lasts for 0.05 seconds. find the impulse caused by the collision and the average force on the car​.

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A 4.9-m wide swimming pool is filled to the top. The bottom of the pool becomes completely shaded in the afternoon when the sun is 25 degrees above the horizon. How deep is the pool?

Answers

Answer:

Explanation:

We can solve this problem using trigonometry. Let's draw a diagram:

          |\

          | \

          |  \   <- Sun rays

          |   \

          |    \

          |     \

          |      \

          ---------

           Pool

The angle between the sun rays and the horizontal line is 90 - 25 = 65 degrees. Let's call the depth of the pool "d". We want to find the value of "d" that makes the bottom of the pool completely shaded.

We can see that the triangle formed by the sun rays, the top edge of the pool, and the bottom edge of the shaded area is a right triangle. The angle between the sun rays and the top edge of the pool is also 65 degrees, because the top edge is parallel to the ground.

Using trigonometry, we can write:

tan(65 degrees) = d / 4.9 m

Solving for "d", we get:

d = 4.9 m * tan(65 degrees)

Using a calculator, we find:

d ≈ 13.7 m

Therefore, the pool is approximately 13.7 meters deep.

EVERYONE What are you thoughts on Nolan Donkor. He says weird things and does mean stuff to everyone in my school

Answers

Answer:

well i think if hes mean to everyone in school, then he must be going through something at home or he just feels insecure about something and likes to bully others for it. Bullies just bully people because they have nothing better to do and cant find happiness. Tell Nolan to get a life

Explanation:

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