5) You pull a 10.0 kg wagon along a flat road. You exert a force of 80.0 N at an angle of 30.0 degrees above the horizontal while you move the wagon 10.0 m forward. The coefficient of friction between the wagon and road is 0.500. Calculate the work down by you and the work done by friction.

Answer:

B.-80.35 J

i dont know the calculation

Cohesive forces are the forces that draw molecules of the same type together. Adhesive forces are those that draw molecules of various types together.

Cohesive forces are the forces that draw molecules of the same type together. Adhesive forces are those that draw molecules of various types together.

The force that draws molecules of the same substance together is called the cohesive force. The force that holds molecules of various substances together is known as the adhesive force.

Between molecules of the same substance, there are cohesive forces. There is a natural tendency to resist separation due to these intermolecular forces between like elements. Conversely, adhesive forces draw disparate molecules together.

In physics, cohesion refers to the intermolecular attraction that exists between two adjacent parts of a substance, especially one that is solid or liquid. A piece of matter is held together by this force. Adhesion is a term for the intermolecular forces that act when two dissimilar substances come into contact.

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The water in the bucket containing the aluminium block is warmer than the bucket containing the copper block.

The specific heat is the amount of heat needed or required to elevate the temperature of 1 gram of a substance by 1° C.

At standard conditions;

We know that:

Therefore, since the specific heat of aluminium is higher than that of copper, the water bucket containing aluminium block will be warmer than the bucket containing the copper block.  

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Answer:

 |F| = 35 kN

Explanation:

a = F/m

100 km/hr(1000 m/km / 3600 s/hr) = 27.8 m/s

v² = u² + 2as

   a = (v² - u²) / 2s

F/m = (v² - u²) / 2s

   F = m(v² - u²) / 2s

   F = 5000(0² - 27.8²) / 2(55)

   F = - 35,072.9517...

Answer:

Explanation:

350 cm / 25 cm = 14 radians

14 rad / 2π rad/rev = 2.23 revolutions

How to find g (acceleration due to gravity)

We know,

Acceleration due to gravity (g)

[tex] = \frac{GM}{ {R}^{2} } [/tex]

where, G = Gravitational constant

[tex] = 6.67 \times {10}^{11} N {m}^{2}/k {g}^{2} \\ [/tex]

M = Mass of the earth

[tex] = 6 \times {10}^{24} \: kg[/tex]

R = Radius of the earth

[tex] = 6.4 \times {10}^{6} m[/tex]

Putting these values of G, M and R in the above formula, we get

[tex]g \: = \: \frac{6.67 \times {10}^{11} N {m}^{2}/k {g}^{2} \times \: 6 \times {10}^{24} \: kg }{(6.4 \times {10}^{6}m {)}^{2} } \\ = 9.8m/ {s}^{2} [/tex]

So, the value of acceleration due to gravity is

[tex]9.8m/s ^{2} [/tex]

Hope it helps.

Do comment if you have any query.

Answer:

c

Explanation:

The increase in the temperature of the soil when bag fall from the given height is 0.065 ⁰C.

The given parameters:

Apply the principle of conservation of energy as follows;

[tex]mgh = mC \Delta T\\\\gh = C \Delta T\\\\\Delta T= \frac{gh}{C}[/tex]

Convert the value of C in kcal/kg ⁰C  to  J/kg ⁰C

1 kcal = 4180 J

[tex]0.200 \ kcal/kg ^0 C= 0.2 \times 4180 (J/kg ^0C) = 836 \ J/kg^0 C[/tex]

The increase in the temperature of the soil is calculated as follows;

[tex]\Delta T= \frac{gh}{C}\\\\\Delta T= \frac{9.8 \times 5.5}{836} \\\\\Delta T= 0.065 \ ^0C[/tex]

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Answer:

i think the correct answer is B. cancel out

Answer:

a1 = -w^2 A sin w t        where w is the angular frequency

a2 =  - (2 w)^2 A sin 2 w t       where w2 = 2 w1

a2 / a1 = 4 sin 2 w t / sin  w t

Since sin 2 w t / sin w t = 2     sin w t will change twice as fast

a2 / a1 = 8      the max acceleration must change 8 tims as fast

Hi there!

We can begin by using a summation of torques, with the fulcrum being at the rope farthest from the person (the other end).

Thus:

F1 = weight of person

R1 = distance of person to fulcrum (farther end)

F2 = Tension of rope closest to person

R2 = distance of rope to fulcrum

∑τ = 0 = R2F2 - R1F1

R2F2 = R1F1

Plug in values:

2F2 = 1.5(700N)

2F2 = 1050 N

F2 = 525 N = Tension of closest rope

The shutter speed that should be used if the lens is changed to f/4.0 is; Choice D: 1/500 s.

The relationship between the shutter speed and the focal ratio is an inverse relationship.

Ratio of Areas = 5.6²/4²

Since; T1A1 = T2A2

In essence; when the focal ratio is reduced as in this case; from f/5.6 to f/4.0; the shutter speed is increased.

Ultimately, the shutter speed of the camera in discuss increases to; 1/500 s.

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Answer:

25.0 cm3

Explanation:

The volume is 25.0 cm3 .

The motion of the rope which is perpendicular to the direction of the

propagation of the wave is a transverse wave motion.

Reasons:

The given function for the wave speed is presented as follows;

[tex]\displaystyle v = \sqrt{\frac{T}{\mu} } \\[/tex]

Where;

[tex]\displaystyle \mu = \frac{Mass \ of \ rope }{Length \ of \ rope}[/tex]

Taking the mass of the rope as, m = 2.00 kg

The length of the rope, L = 80.0 m

The mass hanging on the rope, M = 20.0 kg

We have;

T = 20.0 kg × 9.81 m/s² = 196.2 N

[tex]\displaystyle \mu = \frac{2.0 }{80.0} = \frac{2.0 }{80.0} = 0.025[/tex]

Therefore;

Taking the wavelength as, λ = 7.9 m, and the frequency as 20 Hz, we have;

v = f × λ

Therefore;

v = 7.9 Hz × 7.9 m = 62.41 m/s

Which gives;

[tex]\displaystyle 62.41 = \sqrt{\frac{T}{0.025} }[/tex]

T = 62.41² × 0.025 = 97.3752025

[tex]\displaystyle Mass, \ m = \mathbf{\frac{T}{g}}[/tex]

Where;

g = The acceleration due to gravity which is approximately 9.81 m/s²

[tex]\displaystyle Mass, \ m = \frac{97.3752025}{9.81} \approx 9.93[/tex]

Therefore;

The mass of the box, m ≈ 9.93 kg

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The parameters obtained from a similar question online are;

[tex]\displaystyle The \ equation \ applied, \, v = \sqrt{\frac{T}{\mu} } \\[/tex]

Length of the rope, L = 80.0 m

Mass of the rope, m = 2.0 kg

Frequency of a point on the rope, f = 20 Hz

Answer:

600

Explanation:

GPE=mgh (mass*gravitational force*height)

m=10 kg

g=10

h=6m

10*10*6= 600 Joules

Answer:

Our atmosphere has five different layers. They are:

1. Troposphere: This is the most important layer of the atmosphere with an average height of 13 km from the earth. It is in this layer that we find the air that we breathe. Almost all the weather phenomena such as rainfall, fog and hailstorm occur here.

2. Stratosphere: This layer extends up to a height of 50 km. It presents the most ideal condition for flying airplanes. It contains a layer of ozone gas which protects us from the harmful effect of the sun rays.

3. Mesosphere: This layer extends up to a height of 80 km. Meteorites bum up in this layer on entering from the space.

4. Thermosphere: In this layer, the temperature rises very rapidly with increasing height. The ionosphere is a part of this layer. It extends between 80-400 km. This layer helps in radio transmission. Radio waves transmitted from the earth the reflected back to the earth by this layer.

5. Exosphere: It is the uppermost layer where there is very thin air. Light gases such as helium and hydrogen float into space from here.

Answer:

its b it must be charge

Explanation:

Answer:

Explanation:

I'm not 10% sure but i think that if you were to do:

603 + 277 = 880

1000 - 880 = 120

So to keep it from moving, i think you would need to pull with a force of 120N

The force with which it should be pulled to keep the knot from moving is 663.58 N.

Newton's third law states that, for every action, there is an equal and opposite reaction.

Here,

Force exerted by Roberta, F₁ = 277 N

Force exerted by Michael, F₂ = 603 N

Since, Roberta is pulling to the west and Michael is pulling towards the south, the angle between the forces applied by them is 90°.

The force needed to keep the knot from moving is the resultant force of these two forces.

Therefore, the resultant force,

Fₙ = √F₁² + F₂²

Fₙ = √(277)² + (603)²

Fₙ = 663.58 N

Hence,

The force with which it should be pulled to keep the knot from moving is 663.58 N.

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Answer:

Explanation:

ASSUMING the 52° is the angle of incidence measured from the perpendicular to the surface

          n₁sinθ₁ = n₂sinθ₂

          1 sin52 = 1.33sinθ₂

                  θ₂ = arcsin(sin52 / 1.33)

                  θ₂ = 36°

as measured from the perpendicular to the surface

Answer:

Explanation:

v² = u² + 2as

s = (v² - u²)/2a

s = (0² - 75²) / (2(-14))

s = 200.8928

s = 200 m

Answer:

Identification

Explanation:

:-;..........

This question involves the concepts of density, volume, and mass.

(a) The volume occupied by this amount of water is "90 m³".

(b) The length of an edge of each small cube is "83 μm".

(c) The length of the edge of each small cube is "equal" to the diameter of a molecule.

(a)

The volume can be found using the following formula:

[tex]V = \frac{nM}{\rho}[/tex]

where,

V = volume occoupied = ?

n = no. of moles = 5 mol

M = molar mass = 18 g/mol

[tex]\rho[/tex] = density of water = 1 g/m³

Therefore,

[tex]V=\frac{(5\ mol)(18\ g/mol)}{1\ g/m^3}\\\\[/tex]

V = 90 m³

(b)

First, we will find the volume of an individual molecule:

[tex]V_i =\frac{V}{nN_A}[/tex]

where,

[tex]N_A[/tex] = Avogadro's number = 6.02 x 10²³ molecules/mol

Therefore,

[tex]V_i=\frac{90\ m^3}{5\ mol(6.02\ x\ 10^{23}\ molecules/mol)}[/tex]

Vi = 3 x 10⁻²³ m³

This volume can be given as a volume of the sphere:

[tex]V_i=\frac{4}{3}\pi r^3\\\\r=\sqrt[3]{\frac{3(3\ x\ 10^{-23}\ m^3)}{4\pi}}[/tex]

r = 4.15 x 10⁻⁷ m

Diameter = d = 2r = 2(4.15 x 10⁻⁷ m)

d = 8.3 x 10⁻⁷ m = 83 μm

since the cubes are adjacent to each other. Therefore, the diameter will be equal to the edge length.

Edge Length = L = d

L = 8.3 x 10⁻⁷ m = 83 μm

(c)

The edge length is equal to the diameter of the molecule.

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this is where the sun and moon line up where you asleep only a tiny bit of the sun it's pretty cool to see

Explanation:

a solar eclipse means when the moon goes infront of the sun and the earth turns dark

Answer:

Con el cielo del agua

Explanation:

espero que te ayude

The internal energy after heating in terms of U is 100U.

The given parameters;

Assuming a constant pressure, the internal energy of the ideal gas is equal to the change in the enthalpy of the ideal gas.

[tex]\Delta H = U \times \Delta T\\\\\Delta H = U (200 - 100)\\\\\Delta H = 100 U[/tex]

Thus, we can conclude that the internal energy after heating in terms of U is 100U.

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Answer:

bathing and wear clean clothes 2 pollute 3throwing waste in land air water pollution

Hi there!

We must begin by converting km/h to m/s using dimensional analysis:

[tex]\frac{62km}{1hr} * \frac{1hr}{3600sec}*\frac{1000m}{1km} = 17.22 m/s[/tex]

Now, we can use the kinematic equation below to find the required acceleration:

vf² = vi² + 2ad

We can assume the object starts from rest, so:

vf² = 2ad

(17.22)²/(2 · 75) = a

a = 1.978 m/s²

Now, we can begin looking at forces.

For an object moving down a ramp experiencing friction and an applied force, we have the forces:

Fκ  = μMgcosθ  = Force due to kinetic friction

Mgsinθ = Force due to gravity

A = Applied Force

We can write out the summation. Let down the incline be positive.

ΣF = A + Mgsinθ - μMgcosθ

Or:

ma = A + Mgsinθ - μMgcosθ

We can plug in the given values:

22(1.978) = A + 22(9.8sin(5)) - 0.10(22 · 9.8cos(5))

A = 46.203 N

Answer:

False..

Explanation:

An electoMagnets attract iron due to the influence of their magnetic field upon the iron. ...

Answer:

  1.5 m/s

Explanation:

Conservation of momentum means the momentum of the system before the collision is the same as after.

The before, after momentum of each ball is ...

  5 kg ball: (5 kg)(2 m/s), (5 kg)(-1 m/s)

  10 kg ball: (10 kg)(0 m/s), (10 kg)(v)

The sum of the "before" products is the same as the sum of the "after" products:

  (5 kg)(2 m/s) +0 = (5 kg)(-1 m/s) +(10 kg)v

  (10 +5) kg·m/s = (10 kg)·v . . . . . add (5 kg)(1 m/s) to both sides

  v = (15 kg·m/s)/(10 kg) = 1.5 m/s

The speed of the larger ball will be 1.5 m/s. Its direction of motion will be the opposite of that of the 5 kg ball after the collision.

Answer:

I don't see anything so you should repost