Two blocks connected by a string are pulled across a horizontal surface by a force applied to one of the blocks, as shown to the right. The mass of the left block m1 = 1.4 kg and the mass of the right block m2 = 4.9 kg. The angle between the applied force and the horizontal is θ = 54°. The coefficient of kinetic friction between the blocks and the surface is μ = 0.38. Each block has an acceleration of a = 3.6 m/s2 to the right.

Answer:

object would move but it could be difficult to slow down or stop.

Answer:

(A) By reducing friction

Answer:

A

Explanation:

Answer:

the change in kinetic energy of the system is 0.577 J

Explanation:

Given;

mass of the bullet, m₁ = 6.25 g = 0.00625 kg

initial velocity of the bullet, u₁ = 365 m/s

mass of the crate, m₂ = 4.5 kg

initial velocity of the crate, u₂ = 0

distance moved by the system after collision, d = 0.15 m

Determine the final velocity of the system after collision;

m₁u₁ + m₂u₂ = v (m₁ + m₂)

0.00625 x 365 + 4.5 x 0 = v(0.00625 + 4.5)

2.2813 + 0 = v(4.5063)

2.2813 =  v(4.5063)

v = 2.2813 / 4.5063

v = 0.506 m/s

The change in kinetic energy of the system after collision is calculated as;

ΔK.E = ¹/₂ (m₁ + m₂)v²

ΔK.E =  ¹/₂ (4.506) x 0.506²

ΔK.E = 0.577 J

Therefore, the change in kinetic energy of the system is 0.577 J

Answer:

Permittivity, also called electric permittivity, is a constant of proportionality that exists between electric displacement and electric field intensity.

Answer:

You need to know the accuracy to which you can read the ruler:

Suppose that you can read the read the ruler to the nearest milimeter

A = L * W     your calculated area of the rectangle

A + ΔA = (L + ΔL) * (W + ΔW) = L W + L ΔW + W * ΔL + ΔL ΔA

Or ΔA =  L ΔW + W ΔL

Where we have subtracted A = L * W and the term ΔL * ΔA is very small

So (5 + .1) * (2 + .1) - 5 * 2 = .1 * 2 + .1 * 5 = .7 cm^2

Then you report A = 10 cm^2 +- .7 cm^2    including the - sign for completeness

Answer:62

Explanation:

The weight of the sled is 77 N. The force by the hill on the sled is equal to its weight that is 77 N. Then the tension force exerted by the rope on the sled is being 77N sin θ, where θ be the angle of inclination.

Force is an external agent acting on an object to change its motion or to deform it. There are various kinds of force like magnetic force, tension force, frictional force, gravitational force etc.

The weight that an object experience on earth is due to the gravitational force. The force that is exerted by a rope on an object is tension force since it is pulling from a side.

The normal force by the hill on the sled is equal to its weight that is 77 N. The tension force on the sled by the rope is dependent on the angle of inclination θ. If know the angle we can find T by the equation:

T = 77 sin θ.

Find more on tension force:

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Answer:

The sugars produced by photosynthesis can be stored, transported throughout the tree, and converted into energy which is used to power all cellular processes. Respiration occurs when glucose (sugar produced during photosynthesis) combines with oxygen to produce useable cellular energy.

Explanation:

I think this is correct lol.

Answer:

D The chemicals on the right side of a chamical equation

Answer:

The direction of the force on the vertical wire is towards the East or right.

Explanation:

Using Fleming's right hand rule, the current is the middle finger pointing straight up, the magnetic field is the fore-finger pointing Northwards and then the thumb is the direction of the force on the vertical wire.

Following these conventions, the thumb points towards the East. So, the direction of the force on the vertical wire is towards the East or right.

Answer:

0.339 kgm²

Explanation:

We know the period of this pendulum, T = 2π√(I/mgh) where I = moment of inertia of the object about the pivot axis, m = mass of object = 2.15 kg, g = acceleration due to gravity = 9.8 m/s² and h = distance of center of mass of object from pivot point = 0.163 m.

Since T = 2π√(I/mgh), making I subject of the formula, we have

I = mghT²/4π²

Now since it takes 241 s to complete 113 cycles, then it takes 241 s/113 cycles to complete one cycle.

So, T = 241 s/113 = 2.133 s

So, Substituting the values of the variables into I, we have

I = mghT²/4π²

I = 2.15 kg × 9.8 m/s² × 0.163 m × (2.133 s)²/4π²

I = 15.63/4π² kgm²

I = 0.396 kgm²

Now from the parallel axis theorem, I = I' + mh² where I' = moment of inertia of object with respect to its center of mass about an axis parallel to the pivot axis

I' = I - mh²

I' = 0.396 kgm² - 2.15 kg × (0.163 m)²

I' = 0.396 kgm² - 0.057 kgm²

I' = 0.339 kgm²

Answer:

gas, liquid, solid

sound cannot travel in space

Answer:

1. gas , liquid , gas

2.sound cannot travel in space

Answer:

The number of electrons that surround the nucleus will determine whether or not it is electrically charged or electrically neutral

Explanation:

Answer:

The work and energy is inter- related because if we are energetic then only we are able to work.

People who are weak and not energetic cannot work continuously.

Explanation:

hope this helps you....

Answer:

19.5 m/s

87.8 m

Explanation:

The acceleration of block one is:

∑F = ma

-m₁gμ = m₁a

a = -gμ

a = -(9.8 m/s²) (0.22)

a = -2.16 m/s²

The velocity of block one just before the collision is:

v² = v₀² + 2aΔx

v² = (8.25 m/s)² + 2 (-2.16 m/s²) (2.3 m)

v = 7.63 m/s

Momentum is conserved, so the velocity of block two just after the collision is:

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

m₁u₁ = m₂v₂

(18.5 kg) (7.63 m/s) = (7.25 kg) v

v = 19.5 m/s

The acceleration of block two is also -2.16 m/s², so the distance is:

v² = v₀² + 2aΔx

(0 m/s)² = (19.5 m/s)² + 2 (-2.16 m/s²) Δx

Δx = 87.8 m

Explanation:

By using conservation of linear momentum and also by equating work done to kinetic energy,  [tex]V_{2}[/tex]  = 15.36 m/s and [tex]d_{2}[/tex] = 4.32 meters

Parameters given are :

[tex]m_{1}[/tex] = 19.5 kg

friction, μk = 0.35

distance d = 2.6 m

mass [tex]m_{2}[/tex] = 8.25 kg.

initial velocity of [tex]U_{1}[/tex] = 6.5 m/s.

a.) Since we assumed that the block one stops after it collides with block two, the final velocity for block one will be zero. That is, [tex]V_{1}[/tex] = 0 so its final momentum = 0

Let us also assume that block two was initially at rest. Therefore, it initial velocity and its momentum will be equal to zero.

The formula to use will be :

[tex]m_{1}U_{1} = m_{2}V_{2}[/tex]

Substitute all the parameters into the formula above

19.5 x 6.5 = 8.25[tex]V_{2}[/tex]

Make [tex]V_{2}[/tex] the subject of formula

[tex]V_{2}[/tex] = 126.75/8.25

[tex]V_{2}[/tex] = 15.36 m/s

b.) Let us first calculate the work done in by block one.

The K.E = [tex]1/2mU^{2}[/tex]

substitute its mass and velocity into the formula

K.E = 1/2 x 19.5 x [tex]6.5^{2}[/tex]

K.E = 411.94 Joule

The work done = Kinetic energy

But the resultant Force F = force f - friction

where Frictional force = 0.35 x 19.5 x 9.8

Frictional force = 66.89N

Work done will be the product of resultant Force F and the distance travelled

(F - 66.89) x 2.6 = 411.94

F - 66.89 = 411.94/2.6

F - 66.89 = 158.44

F = 225.3 N

The second block will experience the same force which is equal to 225.3N

Find the kinetic energy of the second block.

K.E =  [tex]1/2mV^{2}[/tex]

K.E = 0.5 x 8.25 x 15.36^2

K.E = 973.2

Using The work done = Kinetic energy

225.3[tex]d_{2}[/tex] = 973.2

[tex]d_{2}[/tex] = 973.2/225.3

[tex]d_{2}[/tex] = 4.32 meters

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Answer:the speed of the clay immediately before impact =72.58m/s

Explanation:

Given that  

mass of the stick clay, M₁= 14.0 g = 0.014 kg

mass of the block ,M₂= 90 g = 0.09 kg

Therefore the total mass= (M₁+M₂) = 104g = 0.104 kg

Also, distance, s = 7.50 m

coefficient of friction μ= 0.650

Acceleration due to gravity ,g = 9.8 m/s²

Using the Work- Energy theorem,

change in kinetic energy =  work done

final kinetic energy(K₂) - initial  kinetic energy(K₁) =   force, F x coefficient of friction, μ x distance,s

The final kinetic energy is zero  because after the impact,  the block with the clay comes to a stop after 7.50m

kinetic energy =Work done

0.5 x m x v²=coefficient of friction,  μ x force(F)  x  distance,s(Since force = m g )

0.5 x m x v²= μ x m x g x s

0.5 x 0.104 x v² = 0.650 x 0.104x 9.8 x 7.5

v²= 0.650 x 0.104x 9.8 x 7.5 / 0.5 x 0.104

v²==95.55

V = 9.77 m/s

Using the  conservation of momentum formulae where

M₁ V₁ + M₂ V₂ = (M₁ + M₂ ) V

Since V₂  which is the velocity of block  is zero as the  block is initially at rest, We now have that

M₁ V₁ = (M₁ + M₂ ) V

0.014 kg x V₁ = 0.104 x 9.77

V₁=0.104 x 9.77 / 0.014

V=72.58m/s

Answer:

They have the same acceleration of 0.5m/s2 (please note m/s2 is the unit for acceleration and 2 is the power of s)

Explanation:

acceleration= velocity ÷ time

and the time is said to be 10seconds

velocity of car will be the new velocity- the initial velocity = 27-22= 5

acceleration= 5÷10

acceleration= 0.5

hope this helped

now hit that crown button :)

Answer:

D. Weight

Explanation:

Hope that helps:)

Answer:

a. of liquid at the time bubbles first emerge slowly from the liquid.

Explanation:

Boiling point of liquid happens due to heat energy. This is an exothermic reaction as heat is released in to the environment. The initial boiling vapors slowly move away from the liquid and as the temperature increases the vapors start moving quickly.

Answer:

4va

12va

2jk

1jk

2jk

Answer:

Clara has speed of 80m/min

Explanation:

Clara was jogging at 600 m in 5 minutes. She stopped suddenly which reduced her velocity and then she waited for 10 minutes so that her friends comes near her. She stopped to catch her friend. During this 10 minutes the velocity of Clara is zero. She started to walk again at a slower speed of 80m/min.

Answer:

I think it is transpiration

Answer:

transpiration is the right answer

Answer:

Workdone = -2730 J

Explanation:

Formula for workdone is;

W = Force × Displacement

Now, according to Newton's 3rd law of motion, to every action, there is an equal and opposite reaction.

In the question given, we are told that a force gauge placed halfway down the rope reads out a constant 2100 Newtons while Magnus pulls the bus. This means that the force exerted by the rope on Magnus acts in an opposite direction to that which Magnus does to the rope.

Therefore, the force will be in the negative direction.

So;

Workdone = -2100 N × 1.3 m

Workdone = -2730 J

Answer:

The correct answer is option D.

Explanation:

It is stated in the question that constriction of the vocal tract at a place where the standing wave of a formant exhibits minimum-amplitude pressure oscillations generally causes the formant to drop in frequency so to increase formant frequency, the vocal should expand where the standing wave of a formant exhibits minimum-amplitude pressure oscillations. The answer is D.

I hope this helps.

Answer:

a)  x = 660 m, b)  λ = 0.330 m, c) Δλ = 0.1 cm, d)    Δf = 104

Explanation:

a) the distance in which the train of waves extends can be obtained from the uniform movement

             v = x / t

             x = v t

             x = 330 2

             x = 660 m

b) the speed of sound is related to the wavelength and frequency

             v = λ f

              λ = v / f

              λ = 330/1000

              λ = 0.330 m

c) The precision in the measurement of the wavelength refers to the error or uncertainty in the measurement, if the measurement is direct with a tape measure the precision is the appreciation of the tape measure, in general it is 0.1 cm

d) the accuracy of the frequency in general the frequency is calculated from the measurements of period T

                f = 1 / T

The precision of the period is data by the chronometer used, in general time a press of 0.01s, by the response time of the people

               Δf = df / dT    ΔT

               Δf = 1 / T² ΔT

                Δf = 1 / (0.001)² 0.01

                 Δf = 104

As we can see, a much more precise system is needed to reduce the error

Answer:

h^2 - ( 2t_o v_s + 2v_s^2 /g) h + v_s^2  \  t_o^2 =0

The correct result is that of a positive height

Explanation:

For this exercise we use the kinematic relations, let's start by finding the time it takes for the sound to reach the man

             v_s = y / t

             t = [tex]\frac{y}{ v_s}[/tex]

this height is y = h

             t =  \frac{h}{ v_s}

the man has a response time of t = t₀, therefore

time to move is

             t' =  t - t₀

the initial height of flower pot is

           y = y₀ + v₀ t' - ½ g  t'²

when it reaches the floor the height is zero y = 0 and as the pot is dropped its initial velocity is zero v₀ = 0

            0 = y₀ +0 - ½ g (t -t₀)²

if the initial height is i = h,

            h = ½ g ([tex]\frac{h}{v_s}[/tex] - t₀)²2

            [tex]\frac{2}{g} h[/tex] = [tex]\frac{h^2}{v_s^2}[/tex] - [tex]\frac{2t_o }{v_s} h[/tex] + t₀²

            [tex]\frac{h^2}{v_s^2} - ( \frac{2t_o}{v_s} + \frac{2}{g} ) h + t_o^2 = 0[/tex]h2 / vs2 - (2nd / vs + 2 / g) h + to2 - = 0

            [tex]h^2 - ( 2t_o v_s + 2v_s^2 /g) h + v_s^2 \ t_o^2 =0[/tex]

To know the height, you must solve the second degree equation, it is much easier with numerical values.

             The correct result is that of a positive height

Answer:

-3.77 m/s

3.85 m/s

Explanation:

given that

Frequency at stationary = 180 Hz

Beat frequency = 2 Hz

Using Doppler effect, we know that

f' = f[(v ± v0) / (v ± vs)], where

v = speed of sound, 343 m/s

v0 = speed of the observer, 0

vs = speed of light, ?

f = stationary frequency, 180 Hz

f' = stationary ± beat frequency, 180 ± 2

Applying the formula, we have

f' = f[(v ± v0) / (v ± vs)]

182 = 180 [(343 + 0) / (343 + vs)]

182/180 = 343 / 343 + vs

343 + vs = 343 * 180/182

343 + vs = 339.23

vs = 339.23 - 343

vs = -3.77 m/s

Again, using

f' = f[(v ± v0) / (v ± vs)]

178 = 180 [(343 + 0) / (343 + vs)]

178/180 = 343 / 343 + vs

343 + vs = 343 * 180/178

343+ vs = 346.85

vs = 346.85 - 343

vs = 3.85 m/s

Answer: it is thin near the center and thick at the edges

Explanation: took the test on Plato :)

Answer:

The minimum coefficient of friction required is 0.35.  

Explanation:

The minimum coefficient of friction required to keep the crate from sliding can be found as follows:

[tex] -F_{f} + F = 0 [/tex]      

[tex] -F_{f} + ma = 0 [/tex]      

[tex] \mu mg = ma [/tex]

[tex] \mu = \frac{a}{g} [/tex]

Where:

μ: is the coefficient of friction

m: is the mass of the crate

g: is the gravity

a: is the acceleration of the truck

The acceleration of the truck can be found by using the following equation:

[tex] v_{f}^{2} = v_{0}^{2} + 2ad [/tex]

[tex] a = \frac{v_{f}^{2} - v_{0}^{2}}{2d} [/tex]

Where:  

d: is the distance traveled = 46.1 m

[tex] v_{f}[/tex]: is the final speed of the truck = 0 (it stops)      

[tex]v_{0}[/tex]: is the initial speed of the truck = 17.9 m/s

[tex] a = \frac{-(17.9 m/s)^{2}}{2*46.1 m} = -3.48 m/s^{2} [/tex]        

If we take the reference system on the crate, the force will be positive since the crate will feel the movement in the positive direction.  

[tex] \mu = \frac{a}{g} [/tex]  

[tex] \mu = \frac{3.48 m/s^{2}}{9.81 m/s^{2}} [/tex]

[tex] \mu = 0.35 [/tex]

Therefore, the minimum coefficient of friction required is 0.35.  

I hope it helps you!