A factory worker pushes a 32.0 kg crate a distance of 7.0 m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and floor is 0.26.

Required:
a. What magnitude of force must the worker apply?
b. How much work is done on the crate by this force?
c. How much work is done on the crate by friction?
d. How much work is done on the crate by the normal force? By gravity?
e. What is the total work done on the crate?

Explanation:

time=Distance/speed

t=32/8

t=4 seconds

Answer:

wool and fibers

Explanation:

Answer:

[tex]\mu=0.185[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=17kg[/tex]

Force [tex]F=33N[/tex]

Velocity [tex]v=1.6m/s[/tex]

Distance [tex]d= 9.8m[/tex]

Generally the equation for Work done is mathematically given by

 [tex]W=\triangle K.E+\triangle P.E[/tex]

Where

 [tex]\triangle K.E=(F-F_f)*2[/tex]

 [tex]F_f=F+\frac{\triangle K.E}{d}[/tex]

 [tex]F_f=33+\frac{0.5*17*1.6^2}{9.8}[/tex]

 [tex]F_f=30.8N[/tex]

Since

 [tex]f = \mu*m*g[/tex]

 [tex]\mu= 30.8/(m*g)[/tex]

 [tex]\mu= 30.8/(17*9.81)[/tex]

 [tex]\mu=0.185[/tex]

Two connection methods are used for brushless DC motors. One method is to connect the coils in a loop as we compared it with the rotor winding of DC motors in Fig. 2.27. This method is called a Δ (delta) connection.

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Answer:

a) [tex]T=0.01s[/tex]

b) [tex]T=0.001s[/tex]

c) [tex]T=0.00001s[/tex]

Explanation:

From the question we are told that:

Given Frequencies

a. 100 Hz,

b. 1 kHz,

c. 100 kHz.

Generally the equation for Waveform Period is mathematically given by

[tex]T=\frac{1}{f}[/tex]

Therefore

a)

For

[tex]T=100 Hz[/tex]

[tex]T=\frac{1}{100}[/tex]

[tex]T=0.01s[/tex]

b)

For

[tex]F=1kHz[/tex]

[tex]T=\frac{1}{1000}[/tex]

[tex]T=0.001s[/tex]

c)

For

[tex]F=100kHz[/tex]

[tex]T=\frac{1}{100*100}[/tex]

[tex]T=0.00001s[/tex]

Answer:

[tex] = \frac{30 \times {10}^{ - 3} }{1} \\ = 0.03 \: km \: per \: second[/tex]

Answer:

108 km/hr or 0.03 km/s

Explanation:

conversion factor for m/s to km/hr is 5/18

conversion factor for m/s to km/s is 1/1000

Answer: hello some data related to your question is missing attached below is the missing data and diagram related to the solution

answer:

P = 141.21 N

Explanation:

Given data:

Mass of crate = 50 kg

coefficient  of static friction ( μ ) = 0.25

Calculate minimum horizontal force ( P ) that holds the crate from sliding

∑fx = 0

     = P + Fcos θ - N*sinθ = 0

     = P + 0.25N cos 30° - Nsin30°  = 0

∴ P = 0.2835 N = 0

P - 0.2853 N = 0 ------- ( 1 )

∑fy = 0

     - 50g + Ncosθ + Fsinθ

     - 50*9.81 + Ncos30° + 0.25Nsin30°

∴ N = 494.942 N ----- ( 2 )

input 2 into 1

P - 0.2853 ( 494.942 ) = 0

P = 141.21 N

Answer:

B

Explanation:

The liquid evaporates in the solid is left in the dish..

Answer:

B.

Explanation:

sana makatulong sayo

Answer:

Mercury - 800°F (430°C) during the day, -290°F (-180°C) at night. Venus - 880°F (471°C) Earth - 61°F (16°C) Mars - minus 20°F (-28°C)30-Jan-2018

Answer:

Distance = speed * time

55*5

275 meters.

The train would have covered a distance of 275 m

We can define distance as to how much ground an object has covered despite its starting or ending point.

Distance = speed * time

given

speed= 55 m/s

time = 5 sec

Distance = 55 * 5 = 275 m

The train would have covered a distance of 275 m

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Complete question is;

Use a variation model to solve for the unknown value.

The stopping distance of a car is directly proportional to the square of the speed of the car.

a. If a car traveling 50 mph has a stopping distance of 170 ft, find the stopping distance of a car that is traveling 70 mph.

b. If it takes 244.8 ft for a car to stop, how fast was it traveling before the brakes were applied?

Answer:

A) d = 333.2 ft

B) 60 mph

Explanation:

Let the stopping distance be d

Let the speed of the car be v

We are told that the stopping distance is directly proportional to the square of the speed of the car. Thus;

d ∝ v²

Therefore, d = kv²

Where k is constant of variation.

A) Speed is 50 mph and stopping distance of 170 ft.

v = 50 mph

d = 170 ft = 0.032197 miles

Thus,from d = kv², we have;

0.032197 = k(50²)

0.032197 = 2500k

k = 0.032197/2500

k = 0.0000128788

If the car is now travelling at 70 mph, then;

d = 0.0000128788 × 70²

d = 0.06310612 miles

Converting to ft gives;

d = 333.2 ft

B) stopping distance is now 244.8 ft

Converting to miles = 0.046363636 miles

Thus from d = kv², we have;

0.046363636 = 0.0000128788(v²)

v² = 0.046363636/0.0000128788

v² = 3599.99658

v = √3599.99658

v ≈ 60 mph

Answer:

Explanation:

From the given information:

the car's momentum = momentum of the truck

∴

(a) 816 kg × v = 2650 kg × 16.0 km/h

v = (2650 kg × 16.0 km/h) /  816 kg

v = 51.96 km/hr

(b) 816 kg × v = 9080 kg × 16.0 km/h

v = (9080 kg × 16.0 km/h) /  816 kg

v = 178.04 km/hr

Answer:

the largest distance we can measure is 10¹⁴ km

Explanation:

Given the data in the question;

Threshold hearing = 10⁻²⁰

smallest distance measured = 1 mm

Largest distance measured will be;

⇒ ( threshold hearing )⁻¹ × smallest distance

= ( 1 / 10⁻²⁰ ) × 1 mm

= 10²⁰ × 1mm

= 10²⁰ mm

we know that; 1000 mm = 10⁶ km

Largest distance = ( 10²⁰ / 10⁶ ) km

= 10¹⁴ km

Therefore, the largest distance we can measure is 10¹⁴ km

Explanation:

The weight of an object on the surface of the earth is equal to the gravitational force exerted by the earth on the object.

[tex]W=F_G[/tex]

[tex]mg = G \dfrac{mM}{R^2}[/tex]

which gives us an expression for the acceleration due to gravity g as

[tex]g = G\dfrac{M}{R^2}[/tex]

At a height h = R, the radius of a satellite's orbit is 2R. Then the acceleration due to gravity [tex]g_h[/tex] at this height is

[tex]mg_h = G \dfrac{mM}{(2R)^2}= G \dfrac{mM}{4R^2}[/tex]

Simplifying this, we get

[tex]g_h= G \dfrac{M}{4R^2} = \dfrac{1}{4} \left(G \dfrac{M}{R^2} \right) = \dfrac{1}{4}g[/tex]

Maximum horizontal force that can be applied on the box is 300.32 N.

Mass of the heaviest box that can be pushed on the ramp at constant speed is 105.16 pound.

Kinetic friction is defined as the opposing force exerted by the surface on an object in contact with it, when there is relative motion between the two surfaces.

Here,

Mass of the box, m = 150 lb = 68.1 kg

Coefficient of kinetic friction, μ = 0.45

Maximum horizontal force that can be applied on the box is the kinetic frictional force. Frictional force,

F(k) = μmg

F(k) = 0.45 x 68.1 x 9.8

F(k) = 300.32 N

Now, the box sits on a ramp inclined at 60°

Coefficient of kinetic friction, μ = 0.45

The net force here acting on the box placed in the ramp is due to the kinetic frictional force and the weight of the box.

So,

Frictional force, F(k)' = μmgcosθ

F(k)' = 0.45 x M x 9.8 x cos 60

F(k)' = 2.2M

Weight of the box acting horizontally,

W = Mgsinθ

W = M x 9.8 x sin60

W = 8.5M

Therefore, net force,

Fn = W - F(k)'

Fn = 8.5M - 2.2M

Fn = 6.3M

The total force acting on the box is

F = F(k) - Fn

ma = 300.32 - 6.3M

Since, the box is moving with constant speed, the acceleration, a = 0

Therefore,

300.32 - 6.3M = 0

6.3M = 300.32

M = 300.32/6.3

M = 47.7 kg = 105.16 pound

Hence,

Maximum horizontal force that can be applied on the box is 300.32 N.

Mass of the heaviest box that can be pushed on the ramp at constant speed is 105.16 pound.

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Answer:

The surface tension is 190.2 N/m.

Explanation:

Initial radius, r = 4 cm

final radius, r' = 6 cm

Work doen, W = 15 J

Let the surface tension is T.

The work  done is given by

W = Surface Tension x change in surface area

[tex]15 = T \times 4\pi^2(r'^2 - r^2)\\\\15 = T \times 4 \times 3.14\times 3.14 (0.06^2- 0.04^2)\\\\15 = T\times 0.0788\\\\T = 190.2 N/m[/tex]

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]

Answer:

belpw

Explanation:

The distance prior to the sliding friction dispersing her energy would be:

- The distance will remain unaffected by the sliding friction i.e. 354m

As we know, When Sliding friction dissolves her energy, leading her Kinetic Energy to turn 0 on coming to the state of rest. So,

[tex]1/2 mv^2 - 1/2 mu^2 = -W[/tex]            (∵ Work in -ve denotes it is done opposite to friction)

Given that,

m(mass) [tex]= 50 kg[/tex]

v(velocity) [tex]= 30 km/hr[/tex] or [tex]8.33 m/s[/tex]

The coefficient of Kinetic Friction [tex]= 0.01[/tex]

g(gravitational force) [tex]= 9.8 m/s^2[/tex]

Initial Velocity(u) [tex]= 30[/tex] × [tex]1000/3600 m/s[/tex]

[tex]= 8.33 m/s[/tex]

Now by employing the provided values,

[tex]F =[/tex] μ[tex]mg[/tex]

[tex]= (0.01) (50) (9.8)[/tex]

[tex]= 4.9[/tex]

∵ [tex]F = 4.9 N[/tex]

By using the above expression, we will find the distance;

[tex]1/2 mv^2 - 1/2 mu^2 = -W[/tex]

⇒ [tex]1/2 (50) (0)^2 - 1/2 (50) (8.33)^2 = -4.9(S)[/tex]

⇒ [tex]1734.7225 = 4.9S[/tex]

⇒ [tex]S = 1734.7225/4.9[/tex]

∵ [tex]S = 354 m[/tex]

Because [tex]1/2 mv^2 - 1/2 mu^2 = -W[/tex]  [tex]= -[/tex] μmgS

⇒ [tex]S = (u^2 - v^2)[/tex]/2μ[tex]g[/tex]

Thus, the distance will remain unaffected by the sliding friction i.e. 354m

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Answer:

the answer is c which is a+2 charge

Explanation:

Beryllium is in group 2A. It's nearest noble gas is Helium, which is 2 elements behind Beryllium. ThBeryllium wants to lose two electrons. When it does that, Beryllium will have a positive chargeof two, and it will be stated as B-e two plus.

The Beryllium (Be) has an atomic number of 4 and belongs to Group-2 elements. The Beryllium will form a divalent cation (+2). Thus, option C is correct.

In an atom, the number of electrons equals the number of protons. If the electrons are removed from the atom or the electrons are added to the atom, the atom has an excessive positive or negative charge.

This excessive of electrons or lack of electrons forms Ions. The excess of electrons has a negative charge or anions and the lack of electrons has a positive charge or cations.

Beryllium has 4 electrons. Two electrons are occupied in the valence shell of beryllium. Group 2 elements always form the positive ions or cations, to become stable ions.

The outermost shell of beryllium has two electrons. In order to form a stable ion, beryllium should lose its two electrons or gain six electrons. Beryllium belongs to the Group-2 element, it always loses two electrons and forms Be²⁺, to form a stable ion.

Hence, Beryllium forms an ion with a +2 charge. Thus, the correct option is C.

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Answer:

Answer:

A). The paper airplane’s motion is due to horizontal inertia and the vertical pull of gravity.

Explanation:

Edge.

Answer:

The motion of the paper airplane  is best explained by horizontal inertia and vertical pull of gravity.

Explanation:

Inertia is the property by which the body wants to remain in its position unless any external for is applied. Here horizontal inertia is inertia of motion which is acting horizontally .

While vertical pull is due to the earth .

Hence horizontal inertia and vertical pull best explain the projectile motion of paper airplane.

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Answer:

The range of the photon energies is between:

2.652 x 10⁻²⁵ J    to    4.973 x 10⁻²⁵ J

Explanation:

The energy of a photon is calculated using the following equation;

E = hf

where;

h is Planck's constant = 6.63 x 10⁻³⁴ Js

f is frequency of the photon

[tex]E = h \frac{c}{\lambda} \\\\where;\\\\\lambda \ is \ the \ wavelength\\\\c \ is \ the \ speed \ of \ light \ = 3\times 10^8 \ m/s\\\\When \ \lambda = 400 \ mm = 400 \ \times 10^{-3} \ m\\\\E = \frac{(6.63 \times 10^{-34})(3\times 10^8)}{400 \times 10^{-3}} \\\\E = 4.973 \times 10^{-25} \ J[/tex]

[tex]When \ \lambda = 750 \ mm = 750 \ \times 10^{-3} \ m\\\\E = \frac{(6.63 \times 10^{-34})(3\times 10^8)}{750 \times 10^{-3}} \\\\E = 2.652 \times 10^{-25} \ J[/tex]

The range of the photon energies is between:

2.652 x 10⁻²⁵ J    to    4.973 x 10⁻²⁵ J

Work done by  man will be A. 640 J

The work-energy theorem states that the net work done by the forces on an object equals the change in its kinetic energy.

according to work energy theorem

Work done = final Kinetic energy - initial kinetic energy

                   = KE (final) - KE (initial )

                   = 1/2 m ([tex]v^{2}[/tex])  - 1/2 m ([tex]u^{2}[/tex])

                  = 1/2 m ([tex]v^{2}[/tex] - [tex]u^{2}[/tex])

                  = 1/2 * 500 * ( [tex]2^{2}[/tex] - [tex]1.2^{2}[/tex])

                  = 250 * 2.56 = 640 J

correct answer is A. 640 J

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Answer:

ΔV = 2 10¹ V

Explanation:

The calculation of the uncertainty or error in an expression is given by

         ΔV = [tex]\frac{dV}{di}[/tex]  |Δi| + [tex]\frac{dV}{dR}[/tex]  |ΔR |

         V = i R

let's make the derivatives

        [tex]\frac{dV}{di}[/tex] = R

        [tex]\frac{dV}{dR}[/tex] = i

we substitute

         ΔV = R | Δi | + i | ΔR |

in the exercise give the values

         i = (5.9 ± 0.4) A

         R = (42.7 ± 0.6) Ω

we calculate

          ΔV = 42.7  0.4 + 5.9  0.6

          ΔV = 20.6 V

          ΔV = 2 10¹ V

the voltage is

         V = i R

         V = 5.9  42.7

          V = 251.9 V

the result is

         V = (25 ± 2) 10¹ V

Answer:

1. Friction enables us to walk freely.

2. It helps to support ladder against wall.

3. It becomes possible to transfer one form of energy to another.

4. Objects can be piled up without slipping.

Answer:

a. 1 Newton = 100000 Dyne

b. a represents acceleration.

Explanation:

Newton is the standard unit (S.I) of measurement of force. Converting 1 Newton to dyne we have;

1 Newton = 10⁵ Dyne

1 Newton = 100000 Dyne

Newton's Second Law of Motion states that the acceleration of a physical object is directly proportional to the net force acting on the physical object and inversely proportional to its mass.

Mathematically, it is given by the formula;

Force = mass * acceleration

[tex] F = ma[/tex]

Hence, we can deduce that a represents the acceleration of an object and it's measured in meters per seconds square.

Answer:

The ratio of the resistances of second coil to the first coil is the ratio of square of radius of the first coil to the square of radius of  second coil.

And

The ratio of the resistances of fourth coil to the third coil is the ratio of square of radius of the third coil to the square of radius of  fourth coil.

Explanation:

The resistance of the coil is directly proportional to the length of the coil and inversely proportional to the area of coil and hence inversely proportional to the square of radius of the coil.

So, the ratio of the resistances of second coil to the first coil is the ratio of square of radius of the first coil to the square of radius of  second coil.

And

The ratio of the resistances of fourth coil to the third coil is the ratio of square of radius of the third coil to the square of radius of  fourth coil.

Answer:

the correct answer is C   v = 60 cm / s

Explanation:

The speed of a wave is related to the frequency and the wavelength

         v = λ f

They indicate that the object performs 20 oscillations every second, this is the frequency

         f = 20 Hz

the wavelength is the distance until the wave repeats, the distance between two consecutive peaks corresponds to the wavelength

         λ = 3 cm = 0.03 m

let's calculate

       v = 20 0.03

       v = 0.6 m / s

       v = 60 cm / s

the correct answer is C

Complete question is;

A projectile of mass m is fired horizontally with an initial speed of v0 from a height of h above a flat, desert surface. Neglecting air friction, at the instant before the projectile hits the ground, find the following in terms of m, v0, h and g:

(a) the work done by the force of gravity on the projectile,

(b) the change in kinetic energy of the projectile since it was fired, and

(c) the final kinetic energy of the projectile.

(d) Are any of the answers changed if the initial angle is changed?

Answer:

A) W = mgh

B) ΔKE = mgh

C) K2 = mgh + ½mv_o²

D) No they wouldn't change

Explanation:

We are expressing in terms of m, v0​, h, and g. They are;

m is mass

v0 is initial velocity

h is height of projectile fired

g is acceleration due to gravity

A) Now, the formula for workdone by force of gravity on projectile is;

W = F × h

Now, Force(F) can be expressed as mg since it is force of gravity.

Thus; W = mgh

Now, there is no mention of any angles of being fired because we are just told it was fired horizontally.

Therefore, even if the angle is changed, workdone will not change because the equation doesn't depend on the angle.

B) Change in kinetic energy is simply;

ΔKE = K2 - K1

Where K2 is final kinetic energy and K1 is initial kinetic energy.

However, from conservation of energy, we now that change in kinetic energy = change in potential energy.

Thus;

ΔKE = ΔPE

ΔPE = U2 - U1

U2 is final potential energy = mgh

U1 is initial potential energy = mg(0) = 0. 0 was used as h because at initial point no height had been covered.

Thus;

ΔKE = ΔPE = mgh

Again like a above, the change in kinetic energy will not change because the equation doesn't depend on the angle.

C) As seen in B above,

ΔKE = ΔPE

Thus;

½mv² - ½mv_o² = mgh

Where final kinetic energy, K2 = ½mv²

And initial kinetic energy = ½mv_o²

Thus;

K2 = mgh + ½mv_o²

Similar to a and B above, this will not change even if initial angle is changed

D) All of the answers wouldn't change because their equations don't depend on the angle.