~~~~NEED HELP ASAP~~~~

Block A slides into block B along a frictionless surface. They are moving in the direction from left o the right.

Block A= 3kg

Block B= 4kg

Block A velocity before collision =30m/s.

Block B velocity before collision = 15 m/s

The velocity of block B after the collision is 20m/s.


a.) What is the velocity of block A after collision?

b.) Is the collision elastic? Show work to explain answer why or why not.

Answers

Answer 1

Answer:

Block A velocity is 23.33 m/s and the collission is not elastic.

Explanation:

a) m1v1 + m2v2 = m1v1' + m2v2'

Plug in givens

90+60=3v1'+80

solve for v1'= 23.33m/s

b) Find the initial and final kinetic energy of Block B

Ki= 1/2(4)(15)^2 + 1/2(3)(30)^2 = 1800 J

Kf= 1/2(4)(20)^2 + 1/2(3)*(23.33)^2= 1616.433J

Since Ki does not equal Kf the collision is not elastic


Related Questions

Question 9 of 10
According to the law of conservation of momentum, the total initial
momentum equals the total final momentum in a(n)
A. Interacting system
B. System interacting with one other system
C. Isolated system
D. System of balanced forces

Answers

Answer:

The answer is C. Isolated System

Answer:

C. Isolated system

Explanation :

∵According to law of  conservation of momentum ,In an isolated system ,the total momentum remains conserved.

a vessel with mass 10kg intially moving withthe velocicity 12m s along the x axis explodes into three exactly identical pieces Just after the explosion one piece moves with speed 10 m s along the x axis and asecond piece moves with speed 10 m s along the y axis What iis the magnitude of the component of velocity of the third piece along the y axiss

Answers

Answer:

Explanation:

Apply law of conservation of momentum along y-axis.

Initially there was no momentum along y-axis. So there will be nil momentum along y-axis again finally.

Let the mass of each piece after breaking be m .

Momentum of piece moving along positive y-axis

= m x 10 = 10m .

Let the component of velocity of third piece along y-axis be v .

Its momentum along the same direction = m v .

Total momentum along y -axis = 10 m + m v

According to law of conservation of momentum

10 m + mv = 0

v = - 10 m/s .

Component of velocity of the third piece along y-axis will be - 10 m/s .

In other words it will be along negative y-axis with speed of 10 m/s.

Topic: Chapter 10: Projectory or trajectile?
Projectile range analysis:
A projectile is launched from the ground at 10 m/s, at
an angle of 15° above the horizontal and lands 5.1 m away.
What other angle could the projectile be launched at, with the same velocity,
and land 5.1 m away?

90°
75°
45
50°
30°

Answers

Answer:

The other angle is 75⁰

Explanation:

Given;

velocity of the projectile, v = 10 m/s

range of the projectile, R = 5.1 m

angle of projection, 15⁰

The range of a projectile is given as;

[tex]R = \frac{u^2sin(2\theta)}{g}[/tex]

To find another angle of projection to give the same range;

[tex]5.1 = \frac{10^2 sin(2\theta)}{9.81} \\\\100sin(2\theta) = 50\\\\sin(2\theta) = 0.5\\\\2\theta = sin^{-1}(0.5)\\\\2\theta = 30^0\\\\\theta = 15^0\\\\since \ the \ angle \ occurs \ in \ \ the \ first \ quadrant,\ the \ equivalent \ angle \\ is \ calculated \ as;\\\\90- \theta = 15^0\\\\\theta = 90 - 15^0\\\\\theta = 75^0[/tex]

Check:

sin(2θ) = sin(2 x 75) = sin(150) = 0.5

sin(2θ) = sin(2 x 15) = sin(30) = 0.5

In an exciting game, a baseball player manages to safely slide into second base. The mass of the baseball player is 88.9 kg and the coefficient of kinetic friction between the ground and the player is 0.53. (a) Find the magnitude of the frictional force in newtons. N (b) It takes the player 1.7 s to come to rest. What was his initial velocity (in m/s)

Answers

Answer:

Look at explanation

Explanation:

a) Kinetic Friction= μmg

μmg=0.53*88.9*9.8=461.75N

b)  -461.75N=ma

a= -5.19m/s^2

v=v0+at

5.19*1.7=v0

v0=8.81m/s^2

(a) The magnitude of the frictional force will be 461.75N

(b)The initial velocity will be 8.81 m/s.

What is kinetic friction?

A force that acts among sliding parts is referred to as kinetic friction. A body moving on the surface is subjected to a force that opposes its progressive motion

The size of the force will be determined by the kinetic friction coefficient between the two materials.

The given data in the problem is;

μ is the coefficient of kinetic friction= 0.53.

m is the mass = 88.9 kg

g is the acceleration due to gravity= 9.81 m/s²

v is the speed =?

The formula for friction force is;

[tex]\rm F= \mu R \\\\ R=mg \\\\ F= \mu mg \\\\\ F=0.53 \times 88.9 \times 9.81 \\\\ F= 461.75 \ N[/tex]

Mechanical force is found as;

F=ma

-461.75=(88.9)a

(-ve shows the -ve work done)

a=-5.19 m/s

From the Newton's first equation of motion;

v=u+at

0=u+at

u=-at

u=(- (-5.19)(1.7)

u=8.81 m/s²

To learn more about the kinetic friction refer to;

https://brainly.com/question/13754413

#SPJ2

3
Select the correct answer.
What is a substance?

Answers

Answer:

physical material from which something is made or which has discrete existence

Explanation:

A closely wound, circular coil with radius 2.70 cm has 800 turns. What must the current in the coil be if the magnetic field at the center of the coil is 0.0750 T

Answers

Answer:

Approximately 4.029 A

Explanation:

We can use the formula that the B field of a few loops all with current in same direction is permeability of free space (mu)* current * Number or loops divided by 2*radius. You are given B field, radius(convert into meters), number of loops and mu is 4pi * 10^-7. Solve for current and you get 4.029 Amperes.

3. Calculate the force it would take to accelerate a 50 ka bike at a rate of 3 m/s2 (6 points)

Answers

Answer:

150 N

Explanation:

Given that,

Acceleration (a) = 3 m/s²Mass of the bike (m) = 50 kg

We are asked to calculate force required.

[tex]\longrightarrow[/tex] F = ma

[tex]\longrightarrow[/tex] F = (50 × 3) N

[tex]\longrightarrow[/tex] F = 150 N

the Period T of oscillation of a Single Pendulum depends on the length l, and acceleration g. Determine the exact form of the dependence.​

Answers

Answer:

[tex]{ \tt{check \: in \: the \: pic}}[/tex]

In 2009 Usain Bolt set the world record time by running 100 meters in 9.58 s. Assume that during this race he ran in a straight line with constant acceleration a. What would be the required constant acceleration a

Answers

Usain Bolt I think, sorry if it’s not right

In 2009 Usain Bolt set the world record time by running 100 meters in 9.58 seconds, assuming that he ran this race with constant acceleration, then the required constant acceleration would have been

What are the three equations of motion?

There are three equations of motion given by  Newton

v = u + at

S = ut + 1/2×a×t²

v² - u² = 2×a×s

By using the second equation of motion given by Newton,

S = ut + 1/2at²

100= 0 + 0.5*a*9.58²

a = 2.17 meters / second²

Thus,the required constant acceleration of Usain Bolt would have been 2.17 meters / second².

To learn more about equations of motion here, refer to the link;

brainly.com/question/5955789

#SPJ2

A Catapult throws a payload in a circle with an arm that is 65.0 cm long. At a certain instant, the arm is rotating at 8.0 rad/s and the angular speed is increasing at 40.0 rad/s2. For this instant, find the magnitude of the acceleration of the payload.

Answers

Answer:

The acceleration of the payload is 26 m/s2.

Explanation:

length, L = 65 cm =  0.65 m

angular acceleration = 40 rad/s^2

The acceleration is given by

a = angular acceleration x length

a = 40 x 0.65

a = 26 m/s^2

g A computer is reading data from a rotating CD-ROM. At a point that is 0.0189 m from the center of the disk, the centripetal acceleration is 241 m/s2. What is the centripetal acceleration at a point that is 0.0897 m from the center of the disc?

Answers

Answer:

the centripetal acceleration at a point that is 0.0897 m from the center of the disc is 1143.8 m/s²

Explanation:

Given the data in the question;

centripetal acceleration a[tex]_c[/tex]₁ = 241 m/s²

radius r₁ = 0.0189 m

radius r₂ = 0.0897 m

centripetal acceleration a[tex]_c[/tex]₂ = ? m/s²

since the rotational period will be the same for the two disk,

we use the centripetal acceleration formula a[tex]_c[/tex] = (4π²r/T²) to find the rotational period for the first disk.

a[tex]_c[/tex]₁ = (4π²r₁/T²)

make T² subject of formula

T² = 4π²r₁ / a[tex]_c[/tex]₁

we substitute

T² = ( 4 × π² × 0.0189 )  / 241  

T² = 0.00309602528 s²

Now we use the same formula to find a[tex]_c[/tex]₂

a[tex]_c[/tex]₂ = ( 4π²r₂ / T² )

we substitute

a[tex]_c[/tex]₂ = ( 4 × π² × 0.0897 )  / 0.00309602528

a[tex]_c[/tex]₂ = 1143.8 m/s²

Therefore, the centripetal acceleration at a point that is 0.0897 m from the center of the disc is 1143.8 m/s²

A 10-cm-long spring is attached to the ceiling. When a 2.0 kg mass is hung from it, the spring stretches to a length of 15 cm.What is the spring constant k?How long is the spring when a 4.0 kg mass is suspended from it?

Answers

As the spring is stretched, it exerts an upward restoring force f. At maximum extension, Newton's second law gives

F = f - mg = 0   ==>   f = (2.0 kg) (9.8 m/s²) = 19.6 N

By Hooke's law, if k is the spring constant, then

f = kx   ==>   k = f/x = (19.6 N) / (0.15 m) ≈ 130 N/m

A 4.0 kg mass would cause the spring to exert a force of

f = (4.0 kg) (9.8 m/s²) = 39.2 N

which would result in the spring stretching a distance x such that

39.2 N = (130 N/m) x   ==>   x = (39.2 N) / (130 N/m) ≈ 0.30 m ≈ 30 cm

1. A message signal m(t) has a bandwidth of 5kHz and a peak magnitude of 2V. Estimate the bandwidth of the signal u(t) obtained when m(t) frequency modulates a carrier with a) kf = 10 Hz/V, b) kf = 100 Hz/V, and c) kf = 1000 Hz/V.

Answers

Answer:

3v at 5.3 herts

Explanation:

PLEASE HELP ME WITH THIS ONE QUESTION
What is the rest energy of a proton? (c = 2.9979 x 10^9 m/s, mp = 1.6726 x 10^-27)

Answers

Answer:

multiply mp and c^2

Explanation:

e=mc^2

A 165-km-long high-voltage transmission line 2.00 cm in diameter carries a steady current of 1,015 A. If the conductor is copper with a free charge density of 8.50 1028 electrons per cubic meter, how many years does it take one electron to travel the full length of the cable?

Answers

Answer:

22.1 years

Explanation:

Since the current in the wire is I = nevA where n = electron density = 8.50 × 10²⁸ electrons/cm³ × 10⁶ cm³/m³= 8.50 × 10³⁴ electrons/m³, e = electron charge = 1.602 × 10⁻¹⁹ C, v = drift velocity of electrons and A = cross-sectional area of wire = πd²/4 where d = diameter of wire = 2.00 cm = 2 × 10⁻² m

Making v subject of the formula, we have

v = I/neA

So, v = I/neπd²/4

v = 4I/neπd²

Since I = 1,015 A, substituting the values of the other variables into the equation, we have

v = 4I/neπd²

v = 4(1,015 A)/[8.50 × 10³⁴ electrons/m³ × 1.602 × 10⁻¹⁹ C × π ×(2 × 10⁻² m)²]

v = 4(1,015 A)/[8.50 × 10³⁴ electrons/m³ × 1.602 × 10⁻¹⁹ C × π × 4 × 10⁻⁴ m²]

v = (1,015 A)/[42.779 × 10¹¹ electronsC/m]

v = 23.73 × 10⁻¹¹ m/s

v = 2.373 × 10⁻¹⁰ m/s

Since distance d = speed, v × time, t

d = vt

So, the time it takes one electron to travel the full length of the cable is t = d/v

Since d = distance moved by free charge = length of transmission line = 165 km = 165 × 10³ m and v = drift velocity of charge = 2.373 × 10⁻¹⁰ m/s

t = 165 × 10³ m/2.373 × 10⁻¹⁰ m/s

t = 69.54 × 10⁷ s

t = 6.954 × 10⁸ s

Since we have 365 × 24 hr/day × 60 min/hr × 60 s/min = 31536000 s in a year = 3.1536 × 10⁷ s

So,  6.954 × 10⁸ s =  6.954 × 10⁸ s × 1yr/3.1536 × 10⁷ s = 2.21 × 10 yrs = 22.1 years

It will take one electron 22.1 years to travel the full length of the cable

A proton is released from rest at the positive plate of a parallel-plate capacitor. It crosses the capacitor and reaches the negative plate with a speed of 54000 m/s. The experiment is repeated with a He+ ion (charge e, mass 4 u).What is the ion's speed at the negative plate?

Answers

4058 m longest serving

vector A has a magnitude of 8 unit make an angle of 45° with posetive x axis vector B also has the same magnitude of 8 unit along negative x axis find the magnitude of A+B?​

Answers

Answer:

45 × 8 units = A + B as formular

collisions may take place between​

Answers

Answer:

Collisions may take place between the reactants.

Explanation:

The collision frequency must be greater than the frequency factor for the reaction. A collision between the reactants must occur.

how can scientific method solve real world problems examples

Answers

The scientific method is nothing more than a process for discovering answers. While the name refers to “science,” this method of problem solving can be used for any type of problem

A 0.495-kg hockey puck, moving east with a speed of 4.50 m/s , has a head-on collision with a 0.720-kg puck initially at rest. Assuming a perfectly elastic collision, what will be the speed (magnitude of the velocity) of each object after the collision?

Answers

Answer:

a)  [tex]v_1=-0.833m/s[/tex]

b)  [tex]V_2=12.5m/s[/tex]

Explanation:

From the question we are told that:

Hockey puck Mass [tex]m_1=0.495kg[/tex]

Hockey puck Speed [tex]u_1=4.50m/s[/tex]

Puck Mass [tex]m_2=0.720kg[/tex]

Assuming

Initial speed of Puck [tex]u_2=0[/tex]

Generally the equation for Speed of First Puck is mathematically given by

 [tex]v_1=(\frac{m_1-m_2}{m_1+m_2})*v_1+(\frac{2m_2}{m_1+m_2})u_2[/tex]

 [tex]v_1=(\frac{0.495-0.720}{0.495+0.720})*4.50+0[/tex]

 [tex]v_1=-0.833m/s[/tex]

Generally the equation for Speed of Second Puck is mathematically given by

 [tex]V_2=(\frac{2m_1}{m_1+m_2})u_2-(\frac{m_1-m_2}{m_1+m_2})v[/tex]

 [tex]V_2=(\frac{2*0.495}{0.495*0.720})*4.50-0[/tex]

 [tex]V_2=12.5m/s[/tex]

What are the differences among elements, compounds, and mixtures?

Answers

Answer:

Elements have a characteristic number of electrons and protons.Both Hydrogen(H) and oxygen(O) are two different elements.

••••••••••••••••

Compounds are chemical substances where the atoms atoms of two different elements are combined together. It is made of .Hydrogen(H) and Oxygen(O) both qr4 naturally gases,but they react to form water(H2O),which is liquid compound.

•••••••••••••••

A mixture is made of atleast two parts》 solid,liquid or gas.The difference is that it's not a chemical substance that's bonded by other elements. 

------------------------------

Hope it helps...

Have a great day!!!

Answer: Elements have a characteristic number of electrons and protons. Both Hydrogen(H) and oxygen(O) are two different elements. Compounds are chemical substances where the atoms atoms of two different elements are combined together. It is made of.Hydrogen(H) and Oxygen(O) both qr4 naturally gases, but they react to form water(H2O), which is liquid compound. A mixture is made of at least two parts solid, liquid, or gas. The difference is that it's not a chemical substance that's bonded by other elements.

uppose that 3 J of work is needed to stretch a spring from its natural length of 32 cm to a length of 49 cm. (a) How much work (in J) is needed to stretch the spring from 37 cm to 45 cm

Answers

Answer:

0.113 J

Explanation:

Applying,

w = ke²/2................. Equation 1

Where w = workdone in stretching the spring, k = spring constant, e = extension

make k the subject of the equation

k = 2w/e²................ Equation 2

From the question,

Given: w = 3 J, e = 49-32 = 17 cm = 0.17 m

Substitute these values into equation 2

k = (2×3)/0.17²

k = 6/0.17

k = 35.29 N/m

(a) if the spring from 37 cm to 45 cm,

Then,

w = ke²/2

Given: e = 45-37 = 8 cm = 0.08

w = 35.29(0.08²)/2

w = 0.113 J

Why are objects measured?​

Answers

In order to find out how long/wide/heavy/high/dense/deep/ massive/voluminous/reflective/opaque/ tansparent/warm/cold/hard/soft/ malleable/flexible/rigid/radioactive/old/ valuable/symmetrical/flat/regular/ irregular they are.

In a way that you can easily and conveniently describe to other people.

why is the water drawn from the bottom of the dam rather than the top?​

Answers

Answer:

because minerals can be gotten from the bottom

Explanation:

it's self explanatory

Think about a thermos bottle. It consists of an inner bottle with a shiny silver surface separated from an outer container by a space with no air. In what ways does it block conduction, convection, and radiation?

Answers

Answer:

Radiation

Explanation:

Conduction, convection and radiation are the three modes of heat transfer.

1. Conduction: When the one end is heated of any rod, the heat transfer to the other end by the vibrational motion of the molecules, it is called conduction.

The heat transfer in a solid is due to the conduction.

2. Convection: When the liquid or gas is heated, the molecules which is in contact to the heat, heated first and due to the decrease in density they moves up and the molecules on the upper side are higher in density so they moves down. These are called convection currents. The process continues till the entire liquid becomes heated. It generally takes place in liquids and gases.  

3. Radiation: The process of heat transfer in which no molecules takes place is called radiation. The heat coming from sun is due to the radiation. It does not require any medium.  

In the thermos bottle, as there is no air between the two layers, so the heat transfer is due to the radiation.  

A flat, 75-turn, coil is oriented with its plane perpendicular to a uniform magnetic field that varies steadily from 0.00 To 1.20 T in 20.0 ms. The diameter of each coil is 10 cm. Calculate the emf induced in the coil during this time, in volts.

Answers

40 m long what is 6 m on

If an object, initially at rest, accelerates at the rate of 25m/s2, what will the magnitude of the displacement be after 50s

Answers

Answer:

31250 meters

Explanation:

Given data

Intitially at rest, the velocity will be

u= 0m/s

acceleration a= 25m/s^2

Time= 50s

We know that the expression for the displacement is given as

S=U+ 1/2at^2

S= 0+ 1/2*25*50^2

S= 12.5*2500

S=31250 meters

Hence the displacement is 31250 meters

Two positive charges, 91 = 5 x 10-'[C] and q2 =1 x 10-9 [C], are
separated by a distance of d=0.05 m. At location 'P' between the
two charges, the net electric field is found to be zero.

b. [10 points] The distance between charge qı and location 'P' is
considered to be 'x'. Find the value of 'x' in [cm]

Answers

Answer:

wareffctgggyyggghhhh

g Three masses are located in the x- y plane as follows: a mass of 6 kg is located at (0 m, 0 m), a mass of 4 kg is located at (3 m, 0 m), and a mass of 2 kg is located at (0 m, 3 m). Where is the center of mass of the system

Answers

Answer:

Xcm = (6 * 0 + 4 & 3 + 2 * 0) / 12 = 1

Ycm = (6 * 0 + 4 * 0 + 2 * 3) / 12 = 1/2

(Xcm , Ycm) = (1 , 1/2)

Using definition of center of mass

Consider a tall building of height 200.0 m. A stone A is dropped from the top (from the cornice of the building). One second later another stone B is thrown vertically up from the point on the ground just below the point from where stone A is dropped.Birthstones meet at half the height of the tower. (a) Find the initial velocity of vertical throw of stone B.(b) Find the velocities of A and B, just before they meet.

Answers

Answer:

a)  v₀ = 44.27 m / s, b) stone A  v = 44.276 m / s,  stone B   v = 0.006 m / s

Explanation:

a) This is a kinematics exercise, let's start by finding the time it takes for stone A to reach half the height of the building y = 100 m

          y = y₀ + v₀ t - ½ gt²

as the stone is released its initial velocity is zero

         y- y₀ = 0 - ½ g t²

         t = [tex]\sqrt{ -2(y-y_o)/g}[/tex]

         t = [tex]\sqrt{ -2(100-200)/9.8}[/tex]

         t = 4.518 s

now we can find the initial velocity of stone B to reach this height at the same time

         y = y₀ + v₀ t - ½ g t²

stone B leaves the floor so its initial height is zero

         100 = 0 + v₀ 4.518 - ½ 9.8 4.518²

         100 = 4.518 v₀ - 100.02

         v₀ = [tex]\frac{100-100.02}{4.518}[/tex]

         v₀ = 44.27 m / s

   

b) the speed of the two stones at the meeting point

stone A

          v = v₀ - gt

          v = 0 - 9.8 4.518

          v = 44.276 m / s

stone B

          v = v₀ -g t

          v = 44.27 - 9.8 4.518

          v = 0.006 m / s

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