A man pushing a crate of mass

m = 92.0 kg

at a speed of

v = 0.845 m/s

encounters a rough horizontal surface of length

ℓ = 0.65 m

as in the figure below. If the coefficient of kinetic friction between the crate and rough surface is 0.351 and he exerts a constant horizontal force of 280 N on the crate.



(a) Find the magnitude and direction of the net force on the crate while it is on the rough surface.

magnitude_____N

What is the direction?

1. Opposite as the motion of the crate

2. Same as the motion of the crate



(b) Find the net work done on the crate while it is on the rough surface.

______J


(c) Find the speed of the crate when it reaches the end of the rough surface.

_______m/s

A. Synthesis, decomposition, single-replacement, combustion

List of reaction types are redox reactions:

"Redox reactions are oxidation-reduction chemical reactions in which the reactants undergo a change in their oxidation states." The term 'redox' is a short form of reduction-oxidation. All the redox reactions can be broken down into two different processes – a reduction process and an oxidation process.

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Answer: I think it C and B but I am really confident in C

Explanation:

Answer:

5.05225 moles

Explanation:

The computation of the number of moles of gas in the tank is shown below:

Given that

Volume = V = 50 L = 50.0 × 10^-3m^3

Pressure = P = 2.45 atm = 2.45 × 101325

Temperature = T = 22.5°C  = (22.5 + 273)k = 295.5 K

As we know thta the value of gas constant R is 8.314 J/mol.K

Now

PV = nRT

n = PV ÷ RT

= ((2.45 × 101325) (50.0 × 10^-3)) ÷ ((8.314) (295.5))

= 5.05225 moles

Answer:

yo imma so I dunno find out yourself

Explanation:

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Answer:

the required revolution per hour is 28.6849

Explanation:

Given the data in the question;

we know that the expression for the linear acceleration in terms of angular velocity is;

[tex]a_{c}[/tex] = rω²

ω² = [tex]a_{c}[/tex] / r

ω = √( [tex]a_{c}[/tex] / r )

where r is the radius of the cylinder

ω is the angular velocity

given that; the centripetal acceleration equal to the acceleration of gravity a [tex]a_{c}[/tex]  = g = 9.8 m/s²

so, given that, diameter = 4.86 miles = 4.86 × 1609 = 7819.74 m

Radius r = Diameter / 2 = 7819.74 m / 2 = 3909.87 m

so we substitute

ω = √( 9.8 m/s² / 3909.87 m )

ω = √0.002506477 s²  

ω = 0.0500647 ≈ 0.05 rad/s  

we know that; 1 rad/s = 9.5493 revolution per minute

ω = 0.05 × 9.5493 RPM

ω = 0.478082 RPM  

1 rpm = 60 rph  

so  

ω = 0.478082 × 60

ω = 28.6849  revolutions per hour  

Therefore, the required revolution per hour is 28.6849

The required revolution per hour is 28.6849

The expression for the linear acceleration with respect to the angular velocity is

= rω²

So,

ω² =  / r

ω = √(  / r )

Here r represent the radius of the cylinder

ω represent the angular velocity

Now

diameter = 4.86 miles = 4.86 × 1609 = 7819.74 m

And,

Radius r = Diameter / 2 = 7819.74 m / 2 = 3909.87 m

So,

ω = √( 9.8 m/s² / 3909.87 m )

ω = √0.002506477 s²  

ω = 0.0500647 ≈ 0.05 rad/s  

Now

1 rad/s = 9.5493 revolution per minute

So,

ω = 0.05 × 9.5493 RPM

ω = 0.478082 RPM  

Now

1 rpm = 60 rph  

so  

ω = 0.478082 × 60

ω = 28.6849  revolutions per hour  

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Answer:

A. 171.24 Ibs

Explanation:

To find the amount of salt in the tank,

Let Q = Amount of salt in the mixture

And let 100 + (3-2)t = 100 + t be the volume of mixture at anytime t.

Rate of gain - Rate of loss = dQ / dt

Concentration of salt = Q / (100+t)

For the linear differential equation,

dQ / dt = 3(2) - 2 [Q/ (100 + t)]

dQ /dt + Q [2 / (100 + t)] = 6

The general solution of the linear differential equation is:

Q (i.f) = ∫ A(t) (i.f) dt + C

Therefore,

i.f = e ^ ∫ P(t) dt

And P(t) = 2 / (100 + t)

i.f = e ^ ∫ 2 / (100 + t)

  = e ^ 2㏑ (100 + t)

     = e ^ ㏑ (100 + t) ^2 = (100 + t) ^2

Q(100 + t) ^ 2 = ∫6 (100 + t) ^2 dt + C

 Q(100 + t) ^2 = 2(100 + t) ^ 3 + C

  When t = 0, Q = 50

Therefore,

50( 100) ^2 = 2(100) ^3 + C

 C = -1.5 * 10 ^6

therefore, when t = 30,

Q (100 + 30) ^2 = 2(100 + 30) ^3 - 1.5 * 10 ^6

 Q (400) ^2 = 2(130) ^3 - 1.5 * 10 ^6

    Q = 171.24 Ibs

The amount of salt in the tank at the end of 30 minutes is 171.24 lbs.

The given parameters:

The linear differential equation of the salt solution is calculated as follows;

[tex]\frac{dx}{dt} = Gain - loss[/tex]

where;

The salt concentration at time t, is calculated as follows;

[tex]\frac{dx}{dt} = 2(3) - 2(\frac{X}{100 + t} )\\\\\frac{dx}{dt} = 6 - 2(\frac{X}{100 + t} )\\\\\frac{dx}{dt} +2(\frac{X}{100 + t} ) =6[/tex]

Apply the general solution of linear differential equation as follows;

[tex]X(f) = \int\limits {At} \, dt \ + C\\\\f = e^{\int\limits {At} \, dt}\\\\ f = e^{\int\limits {\frac{2}{100 + t} } \, dt}\\\\f = e^{2 ln(100 + t)}\\\\f = (100 + t)^2[/tex]

[tex]X(100 + t)^2 = \int\limits {6(100 + t)^2} \, dt \ + \ C\\\\ X(100 + t)^2 = 2(100 + t)^3 + C[/tex]

When t = 0 and X = 50

[tex]50(100 + 0)^2 = 2(100+ 0)^3 + C\\\\C = -1.5 \times 10^6[/tex]

When t = 30 min, the concentration is calculated as;

[tex]X (100 + 30)^2 = 2(100 + 30)^3- 1.5 \times 10^6\\\\X(130)^2 = 2(130)^3 - 1.5\times 10^6\\\\X(130)^2 = 2894000\\\\X = \frac{2894000}{130^2} \\\\X = 171.24 \ lbs[/tex]

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Answer:

i think 692m/s2 is the correct answer

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!

Answer:

a) True

b) True

c) false

d) True

e) True

Explanation:

a) True

In forward bias, the resistance of the p–n junction reduces and hence the electric charges can flow easily.

b) True

In reverse bias condition, the electric current is due to the minority charge carriers (negative).

c) False

direction of diffusion current is in the direction of movement of positive charge i,e towards n side

d) True

Because the breakdown of charge carriers occur due to which the current increases rapidly

e) True

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:

letra A segundo o couculo a divisão e completa

Answer:

Their kinetic energies would be the same

Explanation:

This is because, since the force, F acting on them moves the same distance, d, the work done by the force is W = Fd.

Now, from work-kinetic energy principles,

W = ΔK where ΔK = change in kinetic energy of the carts.

Since the work-done is the same for both carts, their change in kinetic energies would also be the same.

Since they start from rest, ΔK = K' - K =  K' - 0 = K'

So, the kinetic energies of the carts would be the same

Answer:

Explanation:

For original spring , compression in spring due to a load of 1355 kg is

x = 12 - 8.55 = 3.45 cm = .0345 m

spring constant = W / x

= 1355 x 9.8 / .0345

= 384898.55 N /m

Spring constant of new spring

k = 384898.55 - 5655 = 379243.55 N /m

New compression for new spring

= W / k

= 1355 x 9.8 / 379243.55

= .035 m

= 3.50 cm

Difference of compression = 3.50 - 3.45

= .05 cm .

In later case , car will be more lowered by .05 cm .

Answer:

7.5 m/s

Explanation:

Unfortunately, I don't have an explanation but I guessed the correct answer.

Answer:

D. Weight

Explanation:

Hope that helps:)

Answer:

The potential between the plates will decrease.

Explanation:

An insulator is usually placed between the parallel plates and is also called a dielectric because it makes the amount of charge a capacitor can accommodate to increase at a particular potential difference.

Furthermore, the dielectric effect will make the electric field of the charged capacitor which is not connected to a source of supply to decrease.

Now, when the battery is removed, the charge Q remains constant and Capacity C will increase.

Formula for the potential difference is here;

V = Q/C

Since the numerator Q is constant and the denominator C increases, it means the potential difference V will decrease

Answer:

true

Explanation:

Answer:

true

Explanation:

please give brainlest need 1

Answer:

Velocity time graph

Explanation:

Draw on graph paper two straight lines originating at the same point and perpendicular to each other. This is the x-y axis. The x-axis is the horizontal line and the y-axis is the vertical line.

Mark appropriate equally-spaced time intervals on the x-axis so that you can easily graph the time values from the table.

Mark appropriate velocity increments on the y-axis so that you can easily graph the velocity values from the table. If you have negative velocity values, extend the y-axis downward.

Find the first time value from the table and locate it on the x-axis. Look at the corresponding velocity value and find it on the y-axis.

Put a dot where a straight line vertically drawn up through the x-axis value and a straight line horizontally drawn through the y-axis value intersect.

Plot in similar fashion for all other velocity-time pairs in your table.

Draw a straight line with a pencil, connecting each dot you have put down on the graph paper, going from left to right

Answer:

C= 50km/hr west

Explanation:

150/3= 50

Because it asks for velocity, make sure to include the direction as well.

Answer:

P = 84.1 MPa

Explanation:

       [tex]P = \rho * g * h (1)[/tex]

       where ρ = density of salt water (in Kg/m³),

       g = acceleration due to  gravity  (in m/s²)

       h = height of the column of water.

      [tex]P = \rho * g * h = 1023.6kg/m3*9.8m/s2*8380m = 84.1 MPa (2)[/tex]

Answer:

D

Explanation:

The net force is the combination of all forces acting on an object.

Answer:

it’s 2.5 m/s

Explanation:

i’m too lazy but trust

This question can be solved by using the law of conservation of momentum.

The final speed of the lighter 2 kg train is " 2.5 ".

When two moving objects collide with each other, the law of conservation of momentum can be applied to them as follows:

[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2[/tex]

where,

m₁ = mass of heavier train = 3 kg

m₂ = mass of lighter train = 2 kg

u₁ = initial speed of heavier train = 2.8

u₂ = initial speed of lighter train = 1.6

v₁ = final speed of heavier train = 2.2

v₂ = final speed of lighter train = ?

Therefore,

(3 kg)(2.8) + (2 kg)(1.6) = (3 kg)(2.2) + (2 kg)(v₂)

[tex]v_2 = \frac{5 kg}{2 kg}[/tex]

v₂ = 2.5

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The attached picture illustrates the law of conservation of momentum.

Answer:

a) attractiva, b) dF = [tex]k \frac{Q_1 \ dQ_2}{dx}[/tex], c)  F = [tex]k Q_1 \frac{Q_2}{d \ (d+L)}[/tex], d) F = -1.09 N

Explanation:

a) q1 is negative and the charge of the bar is positive therefore the force is attractive

b) For this exercise we use Coulomb's law, where we assume a card dQ₂ at a distance x

           dF = [tex]k \frac{Q_1 \ dQ_2}{dx}[/tex]

where k is a constant, Q₁ the charge at the origin, x the distance

c) To find the total force we must integrate from the beginning of the bar at x = d to the end point of the bar x = d + L

         ∫ dF = [tex]k \ Q_1 \int\limits^{d+L}_d {\frac{1}{x^2} } \, dQ_2[/tex]

as they indicate that the load on the bar is uniformly distributed, we use the concept of linear density

          λ = dQ₂ / dx

          DQ₂ = λ dx

we substitute

         F = [tex]k \ Q_1 \lambda \int\limits^{d+L}_d \, \frac{dx}{x^2}[/tex]

         F = k Q1 λ ([tex]-\frac{1}{x}[/tex])  

we evaluate the integral

        F = k Q₁ λ [tex](- \frac{1}{d+L} + \frac{1}{d} )[/tex]

        F = k Q₁ λ  [tex]( \frac{L}{d \ (d+L)})[/tex]

we change the linear density by its value

      λ = Q2 / L

       F = [tex]k Q_1 \frac{Q_2}{d \ (d+L)}[/tex]

d) we calculate the magnitude of F

       F =9 10⁹ (-4.2 10⁻⁶)   [tex]\frac{10.4 10x^{-6} }{0.45 ( 0.45 +0.35)}[/tex]

       F = -1.09 N

the sign indicates that the force is attractive

Answer:

a)Toward the rod

b)|dF| = k|Q1|Q2(dx/L)/x^2

c)|F| = k|Q1|Q2/(d(d+L))

d)Plug in for answer c and solve

Explanation:

A)

Q1 is negative and Q2 is positive so it is an attractive force to  where the rod is located.

B)

The formula for Force due to electric charges is F=kQ1Q2/r^2

In this case, Q2 is distrusted through the length of the rod as opposed to a single point charge. As such Q2 is actually Q2*dx/L as dx is a small portion of the full length, L.

The radius between Q1 and Q2 depends on the section of the rod taken so r will be the variable x distance from Q1.

The force is only from a small portion of the rod so more accurately, we are finding |dF| as opposed to the full force, F, caused by the whole rod.

The final formula is |dF| = k|Q1|Q2(dx/L)/x^2

C)

Integrating with respect to the only changing variable, x, which spans the length of the rod, from radius = d to d+L we get this:

F = integral from d to d+L of k|Q1|Q2(dx/L)/x^2

factor out constants

F = kQ1Q2/L * integral d to d+L(1/x^2)dx

F = kQ1Q2/L * (-1/x)| from d to d+L

F = kQ1Q2/L * (-1/d+L - -1/d)

F = kQ1Q2/L * (-d/(d(d+L)) + (d+L)/(d(d+L))

F = kQ1Q2/L * (L)/(d(d+L))

F = kQ1Q2/(d(d+L))

D)

Plug in the given values into c and you have your answer.

Answer:

a

Explanation:

Answer:

The electric potential at the center of the meter stick is 54 KV.

Explanation:

Electric potential (V) is given as:

i.e V = [tex]\frac{kq}{r}[/tex]

Where: k is the Coulomb constant, q is the charge and r is the distance.

Given: q = 3.0 μC = 3.0 x [tex]10^{-6}[/tex] C, r = 0.5 m

So that,

V = [tex]\frac{9*10^{9}*3.0*10^{-6} }{0.5}[/tex]

   = [tex]\frac{2.7*10^{4} }{0.5}[/tex]

V = 54000

  = 54 000 volts

The electric potential at the center of the meter stick is 54 KV.

Answer:

The drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

Explanation:

We can find the drift speed by using the following equation:

[tex] v = \frac{I}{nqA} [/tex]

Where:

I: is the current = 4.50 A

n: is the number of electrons

q: is the modulus of the electron's charge = 1.6x10⁻¹⁹ C

A: is the cross-sectional area = 2.20x10⁻⁶ m²

We need to find the number of electrons:

[tex] n = \frac{6.022\cdot 10^{23} atoms}{1 mol}*\frac{1 mol}{26.982 g}*\frac{2.70 g}{1 cm^{3}}*\frac{(100 cm)^{3}}{1 m^{3}} = 6.03 \cdot 10^{28} atom/m^{3} [/tex]                  

Now, we can find the drift speed:

[tex]v = \frac{I}{nqA} = \frac{4.50 A}{6.03 \cdot 10^{28} atom/m^{3}*1.6 \cdot 10^{-19} C*2.20 \cdot 10^{-6} m^{2}} = 2.12 \cdot 10^{-4} m/s[/tex]              

Therefore, the drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

I hope it helps you!      

Answer:

D

Explanation:

If you need an explanation feel free to ask.