A 0.454-kg block is attached to a horizontal spring that is at its equilibrium length, and whose force constant is 25.0 N/m. The block rests on a frictionless surface. A 5.70×10−2-kg wad of putty is thrown horizontally at the block, hitting it with a speed of 8.99 m/s and bounced with the same speed of 8.99 m/s in opposite direction. How far does the block compresses the spring?

Answers

Answer 1

The total momentum of the block and putty prior to their collision is

(0.454 kg) (0 m/s) + (5.70 × 10⁻² kg) (8.99 m/s) ≈ 0.512 kg•m/s

and the total momentum after the collision is

(0.454 kg) v + (5.70 × 10⁻² kg) (-8.99 m/s)

where v is the velocity of the block. Momentum is conserved, so

(0.454 kg) v + (5.70 × 10⁻² kg) (-8.99 m/s) = 0.512 kg•m/s

==>   v ≈ 2.26 m/s

The total work done on the block by the spring as it gets compressed by a distance x is equal to the change in the block's kinetic energy:

1/2 (25.0 N/m) x ² = 1/2 (0.454 kg) (2.26 m/s) - 0

==>   x ≈ 0.202 m ≈ 20.2 cm


Related Questions

P Flag question
Matt, shown below, is able to move
the large truck because

Answers

Can we please get the picture

You are asked to design a spring that will give a 1070 kg satellite a speed of 3.75 m/s relative to an orbiting space shuttle. Your spring is to give the satellite a maximum acceleration of 5.00g. The spring's mass, the recoil kinetic energy of the shuttle, and changes in gravitational potential energy will all be negligible.
(a) What must the force constant of the spring be?
(b) What distance must the spring be compressed?

Answers

Answer:

[tex]380697.33\ \text{N/m}[/tex]

[tex]0.138\ \text{m}[/tex]

Explanation:

m = Mass rocket = 1070 kg

v = Velocity of rocket = 3.75 m/s

a = Acceleration of rocket = 5g

g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]

The energy balance of the system is given by

[tex]\dfrac{1}{2}kx^2=\dfrac{1}{2}mv^2\\\Rightarrow kx=\dfrac{mv^2}{x}\\\Rightarrow kx=\dfrac{1070\times 3.75^2}{x}\\\Rightarrow kx=\dfrac{7250}{x}[/tex]

The force balance of the system is given by

[tex]ma=kx\\\Rightarrow m5g=\dfrac{7250}{x}\\\Rightarrow x=\dfrac{7250}{1070\times 5\times 9.81}\\\Rightarrow x=0.138\ \text{m}[/tex]

The distance the spring must be compressed is [tex]0.138\ \text{m}[/tex]

[tex]k=\dfrac{7250}{x^2}\\\Rightarrow k=\dfrac{7250}{0.138^2}\\\Rightarrow k=380697.33\ \text{N/m}[/tex]

The force constant of the spring is [tex]380697.33\ \text{N/m}[/tex].

A soccer ball with mass 0.450 kg is initially moving with speed 2.20 m/s. A soccer player kicks the ball, exerting a constant force of magnitude 38.0 N in the same direction as the ball's motion. Over what distance must her foot be in contact with the ball to increase the ball's speed to 6.00m/s?

Answers

Answer:

0.187 m

Explanation:

We'll begin by calculating the acceleration of the ball. This can be obtained as follow:

Mass (m) = 0.450 Kg

Force (F) = 38 N

Acceleration (a) =?

F = m × a

38 = 0.450 × a

Divide both side by 0.450

a = 38 / 0.450

a = 84.44 m/s²

Finally, we shall determine the distance. This can be obtained as follow:

Initial velocity (u) = 2.20 m/s.

Final velocity (v) = 6 m/s

Acceleration (a) = 84.44 m/s²

Distance (s) =?

v² = u² + 2as

6² = 2.2² + (2 × 84.44 × s)

36 = 4.4 + 168.88s

Collect like terms

36 – 4.84 = 168.88s

31.52 = 168.88s

Divide both side by 168.88

s = 31.52 / 168.88

s = 0.187 m

Thus, the distance is 0.187 m

180 J of work is done when lifting a box up to a shelf that is 3 m high. What is the mass of the box?

Answers

Explanation:

Work done= MGH

180=m×10×3

180/30=m

m=6kg

What is the acceleration of a bicycle that goes from 3 m/s to 1 m/s in 2 seconds?
0.5 m/s2
1.0 m/s2
1.5 m/s2
-1.0 m/s2

Answers

Answer:

a=vf - vi/t

vf=final velocity

vi= initial velocity

t=time period

Now The bicycle went from 3ms to 1ms...

1.. It decelerated

2... Since it went from 3 to 1... 1 is the final velocity while 3 is the initial Velocity

Applying the Formula

a= 1-3/2

= -2/2

a= -1ms-²

Option D

The chart lists the masses of four planets.
Planetary Masses
Planet
Mass
Neptune
1.02 x 1026
Uranus
8.68 x 1025
Mars
6.42 x 1023
Venus
4.87 X 1024
According to evidence that supports Einstein's general
theory of relativity, which list shows the planets that
would cause curvature in space-time from the least
amount of curvature to the greatest?
O Mars, Venus, Uranus, Neptune
O Neptune, Uranus, Venus, Mars
O Neptune, Uranus, Mars, Venus
O Venus, Mars, Uranus, Neptune

Answers

Answer:

I think it's A

Explanation:

It's definitely not B on edge

The correct option for the given question about Einstein's general

theory of relativity is Option A) Mars, Venus, Uranus, Neptune.

What is the Einstein's general theory of relativity?Albert Einstein established that the rules of physics apply to all non-accelerating observers in his theory of special relativity.He also demonstrated that the speed of light in a vacuum remains constant regardless of the velocity of an observer.The theory may be used to anticipate everything and describes how objects behave in space and time. For instance: that if there are black holes or not, if Gravity can causes light to bend, The way Mercury behaves when it is in orbit and many more interesting things.

As a Conclusion, we can state that the planets who would cause curvature in space time from the least amount of curvature to the greatest will be in order Mars, Venus, Uranus, Neptune.

Learn more about Einstein's general theory of relativity here:

https://brainly.com/question/13311063

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wegut.
c) A body weighs 1.2N on the moon and 120N on
the earth. Calculate the density of the moon,
taking acceleration of free fall as 10ms on the
earth surface and gravitational constant as
6.67 x 10Nm’kg. The radius of the moon is
2740 km.​

Answers

We/Wm = ge/gm = 120N/1.2N

or

gm = ge/100 = 0.1 m/s^2

density = mass/volume = 3M/(4pir^3)

Re-arranging this equation, we get

M/r^2 = (4/3)×pi×(density)×r

From Newton's universal law of gravitation, the acceleration due to gravity on the moon gm is

gm = G(M/r^2) = G×(4/3)×pi×(density)×r

Solving for density, we get the expression

density = 3gm/(4×pi×G×r)

= 3(0.1)/(4×3.14×6.67×10^-11×2.74×10^6)

= 130.6 kg/m^3

two 0.5 kg carts, one red and one green, sit about half a meter apart on a low friction track, you push on the red one with the constant force of 4N for 0.17m and then remove your hand. the cart moves 0.33 m on the track and then strikes the green cart. what is the work done by you on the two cart system?​

Answers

Answer:

The work done by you on the two cart system is 2 N-m

Explanation:

Work done is the product of force and displacement.

W = F * D

Substituting the given values we get -

W =

[tex]4 * (0.17+0.33)\\= 2[/tex]

The work done by you on the two cart system is 2 N-m

A bicycle possesses 1000 units of momentum. what would be the bicycle's momentum if,
A.its velocity is doubled
B. its mass is tripled

Answers

the answer is A. it’s velocity’s doubled

A playground merry-go-round has a mass of 120 kg and a radius of 1.80 m and it is rotating with an angular velocity of 0.400 rev/s. What is its angular velocity (in rev/s) after a 22.0 kg child gets onto it by grabbing its outer edge

Answers

Answer:

The final angular velocity is rev/s is 0.293 rev/s.

Explanation:

Given;

mass of the merry-go-round, m₁ = 120 kg

radius of the merry-go-round, r = 1.8 m

initial angular velocity, ω = 0.4 rev/s

mass of the child, m₂ = 22 kg

Apply the principle of conservation angular momentum to determine the final angular velocity;

[tex]I_i= I_f\\\\\frac{1}{2} m_1r^2 \omega _i = \frac{1}{2} m_1r^2 \omega _f + m_2r^2 \omega _f\\\\ \frac{1}{2} m_1r^2 \omega _i =( \frac{1}{2} m_1r^2 + m_2r^2 )\omega _f\\\\\omega _f = \frac{ \frac{1}{2} m_1r^2 \omega _i}{\frac{1}{2} m_1r^2 + m_2r^2} \\\\\omega _f = \frac{ \frac{1}{2} m_1 \omega _i}{\frac{1}{2} m_1 + m_2}\\\\\omega _f = \frac{0.5 \ \times \ 120\ kg \ \times \ 0.4\ rev/s}{0.5 \ \times 120\ kg \ \ + \ \ 22 \ kg} \\\\\omega _f = 0.293 \ rev/s\\[/tex]

Therefore, the final angular velocity is rev/s is 0.293 rev/s.

A wall in a house contains a single window. The window consists of a single pane of glass whose area is 0.10 m2 and whose thickness is 8 mm. Treat the wall as a slab of the insulating material Styrofoam whose area and thickness are 11 m2 and 0.15 m, respectively. Heat is lost via conduction through the wall and the window. The temperature difference between the inside and outside is the same for the wall and the window. Of the total heat lost by the wall and the window, what is the percentage lost by the window

Answers

Answer:

the percentage of heat lost by the window is 93.18%

Explanation:

Given the data in the question;

Area of glass [tex]A_{glass[/tex] = 0.10 m²

Thickness of glass [tex]t_{glass[/tex] = 8 mm = 0.008 m

Area of Styrofoam [tex]A_{styrofoam[/tex] = 11 m²

Thickness of Styrofoam [tex]t_{styrofoam[/tex] = 0.15 m

we know that;

Thermal conductivity of glass [tex]k_{glass[/tex] =  0.80 J/smC°

Thermal conductivity of Styrofoam  [tex]k_{styrofoam[/tex] =  0.010 J/smC°

Now, temperature difference between outside and inside the walls and window is ΔT

So, In time t, heat lost due to conduction in the window will be;

[tex]Q_{glass[/tex] = [[tex]k_{glass[/tex] × [tex]A_{glass[/tex]  × ΔTt] / [tex]t_{glass[/tex]

we substitute

[tex]Q_{glass[/tex] = [ 0.80 × 0.10 × (ΔT)t] / 0.008

[tex]Q_{glass[/tex] = [ 0.80 × 0.10 × (ΔT)t] / 0.008  

[tex]Q_{glass[/tex] = 10(ΔT)t J

Also, the heat lost due to conduction in the wall be;

[tex]Q_{styrofoam[/tex] =  [[tex]k_{styrofoam[/tex] × [tex]A_{styrofoam[/tex] × ΔTt] / [tex]t_{styrofoam[/tex]

we substitute

[tex]Q_{styrofoam[/tex] =  [ 0.010 × 11 × ΔTt] / 0.15

[tex]Q_{styrofoam[/tex] =  0.7333(ΔT)t J

Now, Net heat lost in the wall and window is;

Q = [tex]Q_{glass[/tex] + [tex]Q_{styrofoam[/tex]

Q = 10(ΔT)t J + 0.7333(ΔT)t J

Q = 10.7333(ΔT)t J

So, the percentage of heat lost by the windows will be;

% of heat lost = [tex]Q_{glass[/tex] / Q

= 10(ΔT)t J / 10.7333(ΔT)t J

= 0.93167

= ( 0.93167 × 100 )%

= 93.18%

Therefore, the percentage of heat lost by the window is 93.18%

What is the light speed formula?

Answers

If ' c ' is the speed of light, then the formula for it is . . .

c = 299,792,458 meters per second

NO LINKS; The graph shows the motion of a train first moving, then stopping, then traveling again at a slower speed. Calculate the average speed for the entire trip.

20 m/s

8.3 m/s

10 m/s

0 m/s

Answers

Would it be like 38.3m/s?

Answer: 0 m/s. that is your answer i hope this help sorry if i am wrong

Explanation:

What does the area under the curve on a velocity-versus-time graph represent? ... then slows down to travel the last 40 miles in three hours. ... 20. A bicyclist travels the first 700 m of a trip at an average speed of 8 m/s, travels the ... to complete the trip at an average speed, for the entire trip of 440 km/h. ... 125) v1= 0 m/s.

how does light travel across the universe to earth?

Answers

Light travels at the speed of 186,000 miles per second. If you were to travel around the earth it would be 7.5 times in a second

P Flag question
Matt, shown below, is able to move
the large truck because

Answers

Answer:

he has more pushing force than the truck there for pushing it forword.

Explanation:

Please help I’ll mark you brainliest

Answers

Answer:

Percentage:

Rr = 50% because it's 2/4 (for both or 25% each since you have them separate)

rr = also 50%, because it's also 2/4.

Phenotype:

Rr = heterozygous

rr = "hozygous" recessive

In addition, RR is "hozygous" dominant

Explanation:

They said the hozygous is a swearword LOL.

What is the correct calculation for voltage if the resistance is 3 ohms and the current is 4 amps?
3 x 4 = 12 volts
3 + 4 = 7 volts
4= 3 = 1.33 volts
4-3 = 1 volt

Answers

Answer:

iek

Explanation:

Answer:

the answer to your questions above is 3 x 4 = 12 volts

Explanation:

Voltage =  Current × Resistance

The plank you are going to walk has a length of 8.76 m and a mass of 16.35 kg. The plank is not bolted down but is to be placed so that 3.17 m of the plank are actually in contact with the boat. The rest of the board overhangs the jelly fish infested waters. You have a mass of 60.3 kg. What is the minimum mass you must put on the end of the board that is still on the ship so that you can successfully walk to the end of the plank without have the board rotate you into the water

Answers

Answer:

Explanation:

The center of gravity  will act at the middle point that is at 4.38 m from the end of the plank that is on boat . 16.35 x 9.8 N will act at this point .

The plank will turn around a point at 3.17 m from the end of plank that is on boat .

Suppose the weight required be W . It will be placed at the end of the plank that is on boat . The position of man is on the extreme end of the plank that is over water . So its distance from turning point will be 8.76 - 3.17 = 5.59 m .

Taking torque about turning point of all the forces like , weight of plank , weight of the person walking and force due to W

W   x 3.17 =  16.35 x ( 4.38 - 3.17 ) + 60.3 x ( 8.76 - 3.17 )

W   x 3.17 =  19.78  + 337.07 = 356.85

W = 112.57 kg .

What is the answer to question 4

Answers

In order to make it all the way from the opening to the detector, a wave has to travel through air, glass, water, plastic, and vacuum.

-- The siren and the tuning fork are sounds.

-- Sound cannot travel through vacuum.

-- That knocks out choices B, C, and D.

The answer is A.

Note:  Making a big exception here.  We don't do test questions on Brainly. That would be cheating. Don't let me catch you doing it again.

Your ear is capable of differentiating sounds that arrive at each ear just 0.34 ms apart, which is useful in determining where low frequency sound is originating from. (a) Suppose a low-frequency sound source is placed to the right of a person, whose ears are approximately 18 cm apart, and the speed of sound generated is 340 m/s. How long is the interval between when the sound arrives at the right ear and the sound arrives at the left ear

Answers

Answer:

Δt = 5.29 x 10⁻⁴ s = 0.529 ms

Explanation:

The simple formula of the distance covered in uniform motion can be used to find the interval between when the sound arrives at the right ear and the sound arrives at the left ear.

[tex]\Delta s = v\Delta t\\\\\Delta t = \frac{\Delta s}{v}[/tex]

where,

Δt = required time interval = ?

Δs = distance between ears = 18 cm = 0.18 m

v = speed of sound = 340 m/s

Therefore,

[tex]\Delta t = \frac{0.18\ m}{340\ m/s}[/tex]

Δt = 5.29 x 10⁻⁴ s = 0.529 ms

Please please help me please please help

Answers

Answer:

Materials like air, water, and clear glass are called transparent. When light encounters transparent materials, almost all of it passes directly through them. ... When light strikes translucent materials, only some of the light passes through them. The light does not pass directly through the materials.

Explanation:

A playground merry-go-round with a radius of 2.0 m and a rotational inertia of 100 kg m2 is rotating at 3.0 rad/s. A child with a mass of 22 kg jumps onto the edge of the merry-go-round, traveling radially inward. What is the new angular speed of the merry-go-round

Answers

Answer:

The new angular speed of the merry-go-round = [tex]1.6rad/sec[/tex]

Explanation:

From angular momentum conservation

[tex]Iw_1 = (I + mr^2)w_2\\\\3*100 = (100 + 22*2^2)w_2\\\\300 = (100 + 88)w_2\\\\w_2 = \frac{300}{188}\\\\w_2 = 1.6rad/sec[/tex]

For more information on angular momentum, visit

https://brainly.com/subject/physics

What is the answer to this problem

Answers

Answer:

Material that allow the electrons to move freely in order to produce a current

Please mark as brainliest if answer is right

Have a great day, be safe and healthy  

Thank u  

XD  

material that allow the electrons to move, it is B

What happens when sulfur reacts with potassium?
O A. Electrons move from the sulfur atoms to the potassium atoms.
B. Electrons become delocalized among the atoms.
O C. Electrons are shared between the potassium atoms and the sulfur
atoms.
O D. Electrons move from the potassium atoms to the sulfur atoms.

Answers

Answer: Electrons move from the potassium atoms to the sulfur atoms.

Explanation:  2 K + S ⇒ 2 K^+ + S^2-   . Usually metals donor electrons to non-metals

A force is described as?

Answers

Explanation:

A push or pull of an objecr

[tex]\huge{\textbf{\textsf{{\color{navy}{An}}{\purple{sw}}{\pink{er}} {\color{pink}{:}}}}}[/tex]

A force is described as push or pull on an object.

ThanksHope it helps.

A flat circular coil of wire having 400 turns and diameter 6.0 cm carries a current of 7.0 A. It is placed in a magnetic field of with the plane of the coil making an angle of 30° with the magnetic field. What is the magnitude of the magnetic torque on the coil?​

Answers

Answer:

6.8 N.m

Explanation:

The computation of the magnitude of the magnetic torque on the coil is given below:

Given that

n = 400

d =  6.0 cm

Current  is I = 7.0 A

Angle is [tex]\theta[/tex] = 30 degree

Now

We know that

the magnitude of the magnetic torque is

= nIABsin[tex]\theta[/tex]

= (400) (7.0) π ÷ 4 (0.06m)^2 sin(90° - 30°)

As

[tex]\theta[/tex] = (90° - Ф)

=  (400) (7.0) π ÷ 4 (0.06m)^2 sin 60°

= 6.8 N.m

Five lamp, each labbled "6V,3W" are operated at normal brightness. What is the total energy supplied to the lamps in five seconds.​

Answers

Answer:

E = 75 J

Explanation:

First, we will calculate the total power consumed by the five lamps:

[tex]Total\ Power = P = (5)(Power\ of\ one\ lamp)\\P = (5)(3\ W)\\P = 15\ W[/tex]

Now, the energy supply can be calculated as follows:

[tex]E = Pt[/tex]

where,

E = Energy = ?

t = time = 5 s

Therefore,

E = (15 W)(5 s)

E = 75 J

Why does one side of the Moon’s surface always remain much colder than the other side?
A) The colder side of the Moon has a cloud cover over it.
B) The colder side of the Moon is entirely covered with ice.
C) The colder side of the Moon always faces away from the Sun.
D) The colder side of the Moon has many high-altitude mountains.

Answers

Answer:

it's C

Explanation:

The colder side of the Moon always faces away from the Sun.

Please help!!!!!! Motion and Forces​

Answers

157.5J
KE=1/2 mv^2
KE= 1/2(35kg)(3m/s)^2
KE=(17.5kg)(9m^2/s^2)
KE= 157.5J

A wave is a disturbance that carries

A. water from one place to another.
B. sound from one place to another.
C. matter from one place to another.
D. energy from one place to another.

Answers

Answer:

(D) energy from one place to another

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