The spring in the figure has a spring constant of 1400 N/m . It is compressed 17.0 cm , then launches a 200 g block. The horizontal surface is frictionless, but the block's coefficient of kinetic friction on the incline is 0.210.
What distance d does the block sail through the air?

The Spring In The Figure Has A Spring Constant Of 1400 N/m . It Is Compressed 17.0 Cm , Then Launches

Answers

Answer 1

Use the work-energy theorem to find the velocity of the block when it's released by the spring. The work done by the spring on the block as it's restored to equilibrium is

W = 1/2 kx ²

where k is the spring constant and x is the compression of the spring. So

W = 1/2 (1400 N/m) (0.170 m)² = 20.23 J

This is equal to the block's change in kinetic energy ∆K,

W = ∆K

and since it starts from rest, the initial K is zero, leaving us with

W = 1/2 mv ²

where m is the mass of the block and v is its speed, so that

20.23 J = 1/2 (0.200 kg) v ²

==>   v ≈ 14.2 m/s

The block slides at this speed across the frictionless surface until it hits the incline which introduces friction.

First, you need to find the length of the incline. It forms a 45° angle, and the underlying 45°-45°-90° triangle has a hypotenuse of length √2 (2.0 m) ≈ 2.83 m.

Next, you need to find the total work done on the block as it slides up the incline. Use Newton's second law to examine the forces acting on the block during this phase:

• the net force acting on the block in the direction perpendicular to the incline is

F = n - mg cos(45°) = 0

where n = mg cos(45°) ≈ 1.39 N is the magnitude of the normal force and mg cos(45°) ≈ 1.39 N is the perpendicular component of the block's weight;

• the net force acting on the block parallel to the surface is

F = -f - mg sin(45°) = ma

where f = µn = 0.210n ≈ 0.291 N is the magnitude of kinetic friction, mg sin(45°) ≈ 1.39 N is the parallel component of the weight, and a is the acceleration of the block.

Only the parallel forces do work on the block, and this work is negative because friction and weight oppose the block's sliding up the incline. The total work done on the block is then

W = (-0.291 N - 1.39 N) (2.83 m) ≈ -4.74 J

Use the work-energy theorem again to find the block's new speed v at the top of the incline:

W = ∆K

==>   -4.74 J = 1/2 (0.200 kg) v ² - 1/2 (0.200 kg) (14.2 m/s)²

==>   v ≈ 12.4 m/s

And now this becomes a projectile problem. The block travels a horizontal distance x after being launched at an angle of 45° with initial speed 12.4 m/s after time t according to

x = (12.4 m/s) cos(45°) t

Its height y from the 2.0 m-high surface at time t is given by

y = (12.4 m/s) sin(45°) t - 1/2 gt ²

The block lands on the surface when y = 0, which occurs after t ≈ 1.79 s, at which point the block has covered a distance d15.7 m.

Answer 2

The block sail through the air at the distance of "15.8 m"

Given:

Spring constant,

1400 N/m

Mass,

200 g

Block's coefficient,

0.210

By using Work energy theorem, we get

→ [tex]W_{spring}+W_g+W_f = KE_f-KE_i[/tex]

By substituting the values, we get

→ [tex]\frac{1}{2}\times 1400\times (0.17)^2- (0.2\times 9.8\times 2)-(0.21\times 0.2\times \frac{9.8}{\sqrt{2} }\times \sqrt{2}\times 2 )= \frac{1}{2}\times 0.2\times V_f^2[/tex]

here,

[tex]V_f = 12.44 \ m/s[/tex]

→ [tex]d = \frac{V_f^2 Sin 2 \Theta}{g}[/tex]

     [tex]= \frac{(12.44)^2 Sin 90^{\circ}}{9.8}[/tex]

     [tex]= 15.8 \ m[/tex]

Thus the answer above is right.  

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Related Questions

What is the efficiency of a ramp that is 5.5 m long when used to move a 66 kg object to a height of 110 cm when the object is pushed by a 150 N force .






Answer and I will give you brainiliest

Answers

Explanation:

Energy input = F×d = (150 N)(5.5 m) = 825 J

Energy output = mgh = (66 kg)(9.8 m/s^2)(1.10 m) = 711 J

efficiency = [tex]\dfrac{\text{output}}{\text{input}}[/tex]×100% = 86.2%

Newton's Second Law of Motion
14. Why was it necessary to transfer mass from the cart to the hanger rather than just
simply adding mass to the hanger?

Answers

Answer:

The weight of the hanging mass provides tension in the string, which helps to accelerate the cart along the track. A small frictional force will resist this motion. 

. If force (F), work (W) and velocity (v) are taken as fundamental quantities.
What is the dimensional formula of time (T)?

Answers

Answer:

∴ [T]=[WF−1V−1]

Hope this answer is right!!

Answer :

[T] = [W(F)^-1(V)^-1]

Kinematics equations tells us the position of an object under constant acceleration increases linearly with time.
A. True
B. False

Answers

Answer:

False.

Explanation:

Suppose that we have an object that moves with constant acceleration A.

Then the acceleration of the object is defined by the equation:

a(t) = A

The acceleration is the rate of change of the velocity, then the velocity equation is given by the integration of the acceleration equation, we will get:

v(t) = A*t + V₀

Where V₀ is the velocity of the object at the time t = 0s.

Now, if we integrate it again, we will get the position equation:

p(t) = (1/2)*A*t^2 + V₀*t + P₀

Where P₀ is the initial position equation.

Here, we can see that the position equation is a quadratic equation (not a linear equation), then the statement is false.

4. Water stands 12.0 m deep in a storage tank whose top is open to the atmosphere at
1.00 atm. The density of water is given as 1000 kg/m² and some pressure conversion
are 1 Pa = 1 N/m² while 1 atm = 101 325 Pa.
a) What is the absolute pressure at the bottom of the tank?
b) What is the gauge pressure at the bottom of the tank?
[4]
[4]​

Answers

Answer:

[tex]P=217600Pa[/tex]

Explanation:

From the question we are told that:

Density [tex]\rho=1000kg/m^3[/tex]

Depth of Water [tex]d=12.0m[/tex]

Generally the equation for Pressure is mathematically given by

 [tex]P=\rho gh[/tex]

 [tex]P=1000*9.8*12[/tex]

 [tex]P=117600N/m^2[/tex]

Therefore

Absolute Pressure=P+P'

Where

P=Pressure under water

P'=Atmospheric Pressure

Therefore

 [tex]P_A=P+P'[/tex]

 [tex]P_A=117,600+10^5[/tex]

 [tex]P=217600Pa[/tex]

Atoms of which two elements could combine with atoms of chromium (Cr) to
form ionic bonds?
O A. F.
B. Au
C. Se
D. Ti
E. Mg

Answers

Answer:

D and E

Ionic bonds are formed between metals and non-metals and both of those are metals

Find the X and Y components of the following:
A. 35 m/s at 57q from the x-axis.

Answers

Explanation:

Given that,

35 m/s at 57° from the x-axis.

Speed, v = 35 m/s

Angle, θ = 57°

Horizontal component,

[tex]v_x=v\cos\theta\\\\=35\times \cos(57)\\\\=19.06 m/s[/tex]

Vertical component,

[tex]v_y=v\sin\theta\\\\v_y=35\times \sin(57)\\\\=29.35\ m/s[/tex]

Hence, this is the required solution.

Which method of powering a vehiclewill help to reduce air pollution
using oil
using biofuels
using gasoline
using diesel fuel

Answers

Answer:

Using biofuels

Explanation:

The emissions of NOx and total VOCs lead to the formation of ozone in the troposphere, the main component of smog. ... Biofuels has a number of health and environmental benefits including improvement in air quality by reducing pollutant gas emissions relative to fossil fuels

How does the density of water change when: (a) it is heated from 0o
C to
4o
C; (b) it is heated from 4o
C to 10o
C ?

Answers

Answer:

[b] it id heated from 4o

Explanation:

A peach pie in a 9.00 in diameter plate is placed upon a rotating tray. Then, the tray is rotated such that the rim of the pie plate moves through a distance of 208 in. Express the angular distance that the pie plate has moved through in revolutions, radians, and degrees.

Answers

Answer:

a)  [tex]X_1=7.36rev[/tex]

b)  [tex]X_2=46.22radians[/tex]

c) [tex]X_3=2649.6^o[/tex]

Explanation:

From the question we are told that:

Diameter [tex]d=9.00[/tex]

Distance [tex]x=208[/tex]

Generally the equation for circumference of a circle is mathematically given by

[tex]C=2 \pi r\\\\C=2*\pi*4.5[/tex]

[tex]C=28.3[/tex]

Therefore

Angular distance that the pie plate has moved through in revolutions is

[tex]X_1=\frac{x}{C}[/tex]

[tex]X_1=\frac{208}{28.3}[/tex]

[tex]X_1=7.36rev[/tex]

Generally Angular distance that the pie plate has moved through in radians is

[tex]X_2= 7.36rev* 2 \pi[/tex]

[tex]X_2=46.22radians[/tex]

Generally Angular distance that the pie plate has moved through in degrees is

[tex]X_3=7.36rev* 360[/tex]

[tex]X_3=2649.6^o[/tex]

Which of the following statements is false?

Weight is a vector quantity

Weight is measured in newtons. N

The weight of an object is the same on the Earth and the moon​

Answers

Answer:

the weight of an object is the same on earth and moon

Explanation:

bcoz weight depends on both mass and gravity

since the gravity of earth and moon is different then the weight is also different

mass doesn't change not weight

A child is playing in a park on a rotating cylinder of radius, r , is set in rotation at an angular speed of w. The Base of the cylinder is slowly moved away, leasing the child suspended against the wall in a vertical position.

What Is the minimum coefficient of friction between the child's clothing and wall is needed to prevent it from falling .

Answers

Answer:

[tex]\mathbf{\mu_s = \dfrac{g}{\omega^2r}}[/tex]

Explanation:

From the given information:

The force applied to the child should be at equilibrium in order to maintain him vertically hung on the wall.

Also, the frictional force acting on the child against gravitational pull is:

[tex]F_f = \mu _sN[/tex]

where,

the centripetal force [tex]F_c[/tex] acting outward on the child is equal to the normal force.

[tex]F_c= N[/tex]

SO,

[tex]F_f = \mu_s F_c[/tex]

Since the centripetal force [tex]F_c = \dfrac{mv^2}{r}[/tex]

Then:

[tex]F_f = \dfrac{ \mu_s \times mv^2}{r}[/tex]

Using Newton's law, the frictional force must be equal to the weight

[tex]F_f = W[/tex]

[tex]\dfrac{ \mu_s \times mv^2}{r} = mg[/tex]

[tex]\dfrac{ \mu_s v^2}{r} = g[/tex]

Recall that:

The angular speed [tex]\omega = \dfrac{v}{r}[/tex]

Therefore;

[tex]g = \mu_s \omega^2 r[/tex]

Making the coefficient of friction [tex]\mu_s[/tex] the subject of the formula:

[tex]\mathbf{\mu_s = \dfrac{g}{\omega^2r}}[/tex]

state what is meant by graviration potential at a point in an orbit 6.5×10^7

Answers

Explanation:

The gravitational potential at a point in a gravitational field is the work done per unit mass that would have to be done by some externally applied force to bring a massive object to that point from some defined position of zero potential,

what is the main limitation of debye huckel theory​

Answers

Answer:

Explanation:

For very low values of the ionic strength the value of the denominator in the expression above becomes nearly equal to one. In this situation the mean activity coefficient is proportional to the square root of the ionic strength. This is known as the Debye–Hückel limiting law.

(a) From an atomic point of view, why do you have to heat a solid to melt it? (b) If you have a solid and a liquid at room temperature, what conclusion can you draw about the relative strengths of their inter-atomic forces?

Answers

Answer:

A. & B. Heat energy is needed to convert solid into a liquid because heat energy increases the kinetic energy of the particles. The heat energy that it used to change 1 kg of solid into liquid at atmospheric pressure and at its melting point is called the latent heat of fusion.

The unit of distance used in astronomy is the light year, defined as the distance travelled by light in one calender year. How far away from earth (in km) is a star if its distance is quoted as 10 light years?

Answers

Answer:

9.7 trillion kilometers

Explanation:

The answer is 9.7 Trillion km

In an experiment to measure the value of pi, the following results are obtained for pi:
3.14 3.11 3.20 3.06 3.08
a) Calculate the Mean Value.
b) Calculate the Average Deviation from the Mean
c) Calculate the: True Standard Deviation
d) Calculate the: Standard Deviation of the Mean.
e) If the correct value is 3.14159, calculate the: Percent Error

Answers

Answer:

a)[tex]MV=3.118[/tex]      

b)[tex]AD=0.0416[/tex]  

c)[tex]TSD=0.00302[/tex]  

d)[tex]SD=0.00242[/tex]  

e)[tex]%E=0.75%[/tex][tex]PE=0.75\%[/tex]

Explanation:

a)

The mean value equation is:

[tex]MV=\frac{\Sum x}{n}[/tex]

Where:

x represents the values of pin is the number of pi values

[tex]MV=\frac{3.14+3.11+3.20+3.06+3.08}{5}[/tex]

[tex]MV=3.118[/tex]      

b)

The average deviation equations is:

[tex]AD=\frac{\Sum |x-MV|}{n}[/tex]    

[tex]AD=\frac{0.208}{5}[/tex]  

[tex]AD=0.0416[/tex]  

c)  

The true Standard Deviation equation is:

[tex]TSD=\sqrt{\frac{\Sum (x-MV)^{2}}{n-1}}[/tex]    

[tex]TSD=\sqrt{\frac{0.01208}{5-1}}[/tex]      

[tex]TSD=0.00302[/tex]    

d)

The  Standard Deviation equation is:

[tex]TSD=\sqrt{\frac{\Sum (x-MV)^{2}}{n}}[/tex]    

[tex]TSD=\sqrt{\frac{0.01208}{5}}[/tex]      

[tex]SD=0.00242[/tex]    

e)

The percent error equation is:

[tex]%E=\frac{|Measured value - True value|}{|True value|}[/tex][tex]PE=\frac{|Measured value - True value|}{True value}[/tex]

[tex]%E=\frac{|Measured value - True value|}{|True value|}[/tex][tex]PE=\frac{|3.118-3.14159|}{3.14159}*100[/tex]

[tex]%E=\frac{|Measured value - True value|}{|True value|}[/tex][tex]PE=0.75\%[/tex]

[tex]%E=0.75%[/tex]

I hope it helps you!

         

If distance between two charges increased by 2 times then force

Answers

Explanation:

The size of the force varies inversely as the square of the distance between the two charges. Therefore, if the distance between the two charges is doubled, the attraction or repulsion becomes weaker, decreasing to one-fourth of the original value.

Give a quantitative definition of being in contact.

Answers

Two things are said to be in contact if the smallest distance between a point in one of them and a point in the other one is zero.

An airplane flies between two points on the ground that are 500 km apart. The destination is directly north of the point of origin of the flight. The plane flies with an airspeed of 120 m/s. If a constant wind blows at 10 m/s toward the west during the flight, what direction must the plane fly relative to the air to arrive at the destination

Answers

Answer:

The right solution is "4.8° east of north".

Explanation:

Given:

Distance,

= 500 km

Speed,

[tex]\vec{v}=120 \ m/s[/tex]

Wind (towards west),

[tex]v_0=10 \ m/s[/tex]

According to the question, we get

The angle will be:

⇒ [tex]\Theta=Cos^{-1}(\frac{v_0}{v_1} )[/tex]

       [tex]=Cos^{-1}(\frac{10}{120} )[/tex]

       [tex]=85.21[/tex] (north of east)

hence,

The direction must be:

⇒ [tex]\Theta'=90-85.21[/tex]

        [tex]=4.79^{\circ}[/tex]

or,

        [tex]=4.8^{\circ}[/tex] (east of north)

If the car falls down the side of the cliff, what is happening to the gravitational potential energy of the falling car? (Assume the bottom of the cliff is zero)

Answers

Answer:

Sentences with many clauses and phrases are difficult to understand because the clauses and phrases typically _____.

modify other clauses and phrases in the sentence

refer to other sentences in the passage

make it hard to determine where the sentence ends

change the intended meaning of the sentence

Explanation:

What is the "best" explanation for why the universe is the way it is?

A) god created the universe
B) there is a multiverse and this one happens to be perfect for life.
C) this is the only universe and it happens to be perfect for life.
D) It is all in illusion and none of it exists.
E) none of the above, they are all just guesses.

I know the answer I just wanna see what you guys think.
i will give brainly if you get it right.

Answers

Easy- A) God created the universe.

A meter stick has a mass of 0.30 kg and balances at its center. When a small chain is suspended from one end, the balance point moves 28.0 cm toward the end with the chain. Determine the mass of the chain.

Answers

Answer:

M L1 = m L2       torques must be zero around the fulcrum

M = m L2 / L1 = .3 kg * 28 cm / 22 cm = .382 kg

Rank the six combinations of electric charges on the basis of the electric force acting on q1.

a.
q1 = -1nC
q2= +1nC
q3= +1nC

b.
q1 = -1nC
q2= -1nC
q3= -1nC

c.
q1 = +1nC
q2= +1nC
q3= -1nC

d.
q1 = +1nC
q2= -1nC
q3= +1nC

Answers

The anwser for this question would be D. Q1 = +1nC , Rank the six combinations of the electric charges on the basis of the electric force acting on q1.

Part C
When only one color of light reflects from a piece of paper, what happens to the other colors of light?
Remember that light is energy, and energy cannot be created or destroyed.
В І ох
х
Font Sizes
A
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Characters used: 0 / 15000
Plz help me yall I’m struggling

Answers

Answer:

the other colours get absorbed by the paper

Answer:

The other colors are absorbed by the paper and not reflected.

THE ANSWER!!! Please

Answers

Answer:

I think -7 N. Netforce is 3N-10N= -7N

Explanation:

Một học sinh làm thí nghiệm sóng dừng trên dây cao su dài L với hai đầu A và B cố định . Xét điểm M trên dây sao cho khi sợi dây duỗi thẳng thì M cách B một khoảng a < L/2 . Khi tần số sóng là f = f1 = 60 Hz thì trên dây có sóng dừng và lúc này M là một điểm bụng . Tiếp tục tăng dần tần số thì lần tiếp theo có sóng dừng ứng với f = f2=72 Hz và lúc này M không phải là điểm bụng cũng không phải điểm nút . Thay đổi tần số trong phạm vi từ 73 Hz đến 180 Hz , người ta nhận thấy với f = fo thì trên dây có sóng dừng và lúc này M là điểm nút . Lúc đó , tính từ B ( không tính nút tại B ) thì M có thể là nút thứ ?

Answers

Have suxhebeuxhsbendixbebendue bride. Did e did e end Rudd. R

the minimum charge on any object cannot be less than​

Answers

Answer:

1.6 x 10^{-19} Coulombs

Explanation:

In Physics, the standard unit of measurement of a charge is Coulombs and it's denoted by C. Also, the symbol for denoting a charge is Q.

In Chemistry, electrons can be defined as subatomic particles that are negatively charged and as such has a magnitude of -1.

The minimum charge on any object such as an electron cannot be less than​ 1.6 x 10^{-19} Coulombs and it's usually referred to as the fundamental unit of charge.

I was reading an old thermodynamics textbook and came across this equation describing change in internal energy. What does the (dU/dT)V and (dU/dV)T mean? I think it’s partial derivatives, but I’ve never seen a derivative sub (something). Please help

Answers

Explanation:

I remember that notation! The expression

[tex]dQ = dU = (\dfrac{\partial U}{\partial T})_{V} dT+ (\dfrac{\partial U}{\partial V})_{T}dV[/tex]

is the 1st law of thermodynamics and it refers to the heat supplied to the system dQ which is also a change in its internal energy dU. The first term is the partial derivative of the internal energy U with respect to temperature T while the volume V is kept constant, as denoted by the subscript V. The 2nd term is similar but this time, temperature is kept constant while its volume partial derivative is being taken.

Ah, memories!

The only variable we found that affects the speed of a wave on a string was the tension of the string. How does this relate to how a musician tunes a stringed instrument?

Answers

Answer:

We know that in a string with two fixed points (like the one you will find in a guitar, where the string is "fixed" at the bridge and at the nut) the only thing that defines the speed (the frequency) at which the string vibrates is the tension (the length is also important, but for the 6 strings in the guitar all of them have the same length, but, as you know, when you press in a given fret the note changes, this happens because you are changing the length of the string, finally, the mass of the string is also important, but all the strings have almost the same mass, so we can ignore this).

Then for the open notes (the notes that you play when you don't fret any note) the only thing that defines the note that will sound is the tension of the string.

And a musician can tune the stringed instrument by changing the tension of each string using a tuner, which is the mechanism in the headstock of the instrument). As more tense is the string, higher will be the open note.

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