On page 14 of the call of the wild jack London, writes,in vague ways he remembered back to the youth of the breed ,this statement is example of

Answer:

coz teaching is their profession.

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Force is a push or a pull that changes or trends to change the state of rest or uniform motion of an object or changes the direction or shape of an object. It causes objects to accelerate. SI unit is Newton.

1) Can change the state of an object : For example, pushing a heavy stone in order to move it.

2) May change the speed of an object if it is already moving. For example, catching a ball hit by a batsman.

3) May change the direction of motion of an object.

Answer:

Solar radiation

Explanation:

Visible light, ultraviolet light, infrared, radio waves, X-rays, and gamma rays.

==>  Energy

==>  Radio noise, heat, visible light, ultraviolet radiation, X-rays, gamma rays

==>  They carry all these kinds of energy wherever they go.  Not only to the Earth.

Answer:

[tex]F = \frac{m {v}^{2} }{r} \\ = \frac{82 \times { (\frac{590 \times 1000}{3600} )}^{2} }{970} \\ = 2270.6 \: newtons[/tex]

Answer:

required vertical distance below the edge is 0.6648 m

Explanation:

Given the data in the question;

Horizontal speed of water falls v = 2.73 m/s

direction of water falls 52.9° below the horizontal

The vertical velocity must be such that;

tanθ = v[tex]_y[/tex] / v[tex]_x[/tex]

Now, vertical speed of water falls;

v[tex]_y[/tex] = v[tex]_x[/tex] × tanθ

we substitute

v[tex]_y[/tex] = 2.73 × tan(52.9°)

v[tex]_y[/tex] = 2.73 × 1.322237

v[tex]_y[/tex] = 3.6097

Now, at the top of falls, initial speed u = 0

v² - u² = 2as

s = ( v² - u² ) / 2as

we substitute

s = ( 0² - (3.6097)² ) / (2 × 9.8)

s = 13.029934 / 19.6

s = 0.6648 m

Therefore, required vertical distance below the edge is 0.6648 m

Answer:

ratio of the masses of the two enemies = 1.7

Explanation:

Applying the law of conservation of momentum,

Momentum fo Bonzo = Momentum of Ender

mv = m'v'................. Equation 1

Where m = mass of Bonzo, v = velocity of Bonzo, m' = mass of Ender, v' = velocity of Ender

m/m' = v'/v............... Equation 2

Where m/m' = ratio of the masses of the two enemies

Given: v = 2.0 m/s, v' = 3.4 m/s

Substitute into equation 2

m/m' = 3.4/2

m/m' = 1.7

Hnece, ratio of the masses of the two enemies = 1.7

Answer:

convex lens because it makes parallel light rays passing through it bends inwards and meets convex at a spot just beyond the lens known as the focal point

Answer:

Explanation:

[tex]\frac{dx}{dt} = 2.5 \: \frac{cm}{sec} [/tex]

Explanation:

The volume of a cube is V = x^3. Taking the time derivative of this expression, we get

[tex] \frac{dV}{dt} = 3 {x}^{2} \frac{dx}{dt} [/tex]

or

[tex]\frac{dx}{dt} = \frac{1}{3 {x}^{2}} \frac{dV}{dt} [/tex]

We know that dV/dt = 30 cm^3/sec so the value of dx/dt when x = 2 cm is

[tex]\frac{dx}{dt} = \frac{1}{3 {(2 \: cm)}^{2}}(30 \: \frac{ {cm}^{3} }{sec} ) = 2.5 \: \frac{cm}{sec} [/tex]

Answer:

Explanation:

[tex]V=x^3\\\\\frac{dV}{dt}=3x^2\frac{dx}{dt}\\\\30\frac{cm^3}{s}=3x^2\frac{dx}{dt}\\\\\frac{dx}{dt}=\frac{30\frac{cm^3}{s}}{3x^2}~at~x=2cm,~\frac{dx}{dt}=\frac{30\frac{cm^3}{s}}{3*(2cm)^2}=\frac{5}{2}\frac{cm}{s}[/tex]

Answer:

6.67×10^-11 Nm^2kg^_2

Answer:

by using p = mv equation we can find v,

6 = 1.3 v

4.615 = v

Answer:

No,it isn't concave. The correct answer is convex lens.

Explanation:

A lens is a piece of transparent material bound by two surfaces of which at least one is curved. A lens bound by two spherical surfaces bulging outwards is called a bi-convex lens or simply a convex lens. A single piece of glass that curves outward and converges the light incident on it is also called a convex lens.

Convex lens is the answer.

See the attached diagram.

Answer:

1.F: About 6*10^14 Hz

2.E: About 4*10^ -19 J

Explanation:

Frequency: We knew that the speed of a wave is its wavelength(λ)* frequency(f, in Hz).  By the wave-particle duality we know we can calculate the frequency of light in the same way. So, c=495nm *f, f=c/495nm=> (299,792,458 m/s) / (4.95*10^-7 m)

=6.05*10^14 /s

Energy: The energy photon contains can be calculate by this formula-- E=hf

f is the frequency and h is Planck's constant which is about 6.62 ×10^-34 *m^2*kg/s (after dimensional analysis ) =6.62*10^ -34 J*s.

So, the energy of a blue photon is (6.05*10^14)*(6.62*10^-34)=40.051*10^-20=  4.051*10^-19 J

Explanation:

Let's set the x-axis to be parallel to the and positive up the plane. Likewise, the y-axis will be positive upwards and perpendicular to the plane. As the problem stated, we are going to assume that m1 will move downwards so its acceleration is negative while m2 moves up so its acceleration is positive. There are two weight components pointing down the plane, [tex]m_1g \sin \theta[/tex] and [tex]m_2g \sin \theta[/tex] and two others pointing up the plane, the two tensions T along the strings. There is a normal force N pointing up from the plane and two pointing down, [tex]m_1g \sin \theta[/tex] and [tex]m_2g \sin \theta[/tex]. Now let's apply Newton's 2nd law to this problem:

x-axis:

[tex]m1:\:\:\:\displaystyle \sum_i F_i = T - m_1g \sin \theta = - m_1a\:\:\:\:(1)[/tex]

[tex]m2:\:\:\:\displaystyle \sum_i F_i = T - m_2g \sin \theta = m_2a\:\:\:\:(2)[/tex]

y-axis:

[tex]\:\:\:\displaystyle \sum_i F_i = N - m_1g \cos \theta - m_2g \cos \theta = 0[/tex]

Use Eqn 1 to solve for T,

[tex]T = m_1(g \sin \theta - a)[/tex]

Substitute this expression for T into Eqn 2,

[tex]m_1g \sin \theta - m_1a - m_2g \sin \theta = m_2a[/tex]

Collecting all similar terms, we get

[tex](m_1 + m_2)a = (m_1 - m_2)g \sin \theta[/tex]

or

[tex]a = \left(\dfrac{m_1 - m_2}{m_1 + m_2} \right)g \sin \theta[/tex]

Answer:

3.6 KJ

Explanation: Given that a 70-kg boy is surfing and catches a wave which gives him an initial speed of 1.6 m/s. He then drops through a height of 1.60 m, and ends with a speed of 8.5 m/s. How much nonconservative work (in kJ) was done on the boy

The workdone = the energy.

There are two different energies in the scenario - the potential energy (P.E ) and the kinetic energy ( K.E )

P.E = mgh

P.E = 70 × 9.8 × 1.6

P.E = 1097.6 J

P.E = 1.098 KJ

K.E = 1/2mv^2

K.E = 1/2 × 70 × 8.5^2

K.E = 2528.75 J

K.E = 2.529 KJ

The non conservative workdone = K.E + P.E

Work done = 1.098 + 2.529

Work done = 3.63 KJ

Therefore, the non conservative workdone is 3.6 KJ approximately

Answer:

B

Explanation:

Answer:

B

Explanation:

Bc i say

Answer:

a. [tex]1000\ kg(\frac{1000\ g}{1\ kg}) = 1\ x\ 10^{6}\ g[/tex]

b. [tex]50\ m (\frac{100\ cm}{1\ m} ) = 5000\ cm[/tex]

c. "Nano" is 10⁻⁹ , so there are 10⁻⁹ meter in a nm (OR)  "Nano" is 10⁻⁹ , so there are 10⁹ nm in a meter.

d. micro is 10⁻⁶, so 1kg is 10⁹ ug​  

Explanation:

a.

The conversion factor is written inverted. The correct statement will be:

[tex]1000\ kg(\frac{1000\ g}{1\ kg}) = 1\ x\ 10^{6}\ g[/tex]

b.

The values in the conversion factor used are wrong. The correct statement will be:

[tex]50\ m (\frac{100\ cm}{1\ m} ) = 5000\ cm[/tex]

c.

Change of units is the mistake here. The correct statement will be:

"Nano" is 10⁻⁹ , so there are 10⁻⁹ meter in a nm (OR)  "Nano" is 10⁻⁹ , so there are 10⁹ nm in a meter.

d.

the conversion will be as follows:

[tex]1\ kg(\frac{1000\ g}{1\ kg})(\frac{1\ \mu g}{10^{-6}\ g}) = 10^9 \mu g[/tex]

therefore, the correct statement will be:

micro is 10⁻⁶, so 1kg is 10⁹ ug​

Answer:

Person's body pulling on the earth

Explanation:

The weight of an object is the attraction of a person's body toward the Earth. The weight of an object is given by :

Weight (W) = mass (m) × acceleration due to gravity (g)

We know that, for an action there is an equal and opposite reaction.

So, there must be a reaction force for the weight of an object. The reaction force must be the person's body pulling on the earth.

Answer: hello your question is incomplete below is the missing part

A 69-kg petty thief wants to escape from a third-story jail window. Unfortunately, a makeshift rope made of sheets tied together can support a mass of only 58 kg.

answer:

To 2 significant Figures = 1.6 m/s^2

Explanation:

Calculate the magnitude of minimum acceleration at which the thief can descend

we apply the relation below

Mg - T = Ma  --- ( 1 )

M = 69kg

g = 9.81

T = 58 * 9.81

a = ? ( magnitude of minimum acceleration)

From equation 1

a = [ ( 69 * 9.81 ) - ( 58 * 9.81 ) ] / 69

  = 1.5639 m/s^2

To 2 significant Figures = 1.6 m/s^2

answer 8N 8N

la tensión del ladrillo inferior es 8N ya que los ladrillos son idénticos, decido 24 por 3 para darme 8

la tensión entre el ladrillo superior y el medio también es 8N

Answer:

The moon is actually quite dim.

Explanation:

compared to other astronomical bodies. The moon only seems bright in the night sky because it is so close to the earth and because the trees, houses, and fields around you are so dark at night. In fact, the moon is one of the least reflective objects in the solar system.

Answer:

It reflects the light send from the sun.

Explanation:

If the moon is between you and the sun, you will see the back of it which doesnt reflect light.

Complete question:

a metal of work function 1.8eV is illuminated by light of wavelength 3.0x*10-7 m,calculate the maximum kinetic energy of the emitted photons.

Answer:

the maximum kinetic energy of the emitted photons is 3.742 x 10⁻¹⁹ J

Explanation:

Given;

work function, Ф = 1.8 eV

wavelength of the light, λ = 3 x 10⁻⁷ m

The maximum kinetic energy of the emitted photons is calculated from photoelectric equation.

[tex]E = K.E_{max} + \phi\\\\KE_{max} = E- \phi\\\\where;\\\\E \ is \ the \ energy \ of \ the \ incident \ light\\\\E = hf = h \frac{c}{\lambda} \\\\where;\\\\c \ is \ speed \ of \ light = 3 \times 10^8 \ m/s\\\\h \ is \ Planck's \ constant = 6.626 \times 10^{-34} \ Js\\\\E = \frac{ (6.626 \times 10^{-34})\times ( 3 \times 10^8)}{3\times 10^{-7}} \\\\E = 6.626 \times 10^{-19} \ J[/tex]

[tex]K.E_{max} = E - \phi\\\\K.E_{max} = 6.626\times 10^{-19} \ J \ - \ (1.8 \times 1.602 \times 10^{-19} \ J)\\\\K.E_{max} = 6.626\times 10^{-19} \ J \ - \ 2.884 \times 10^{-19} \ J\\\\K.E_{max} =3.742 \times 10^{-19} \ J[/tex]

Therefore, the maximum kinetic energy of the emitted photons is 3.742 x 10⁻¹⁹ J

Solution :

Given :

Mass of the baseball, m = 200 g

Velocity of the baseball, u = -30 m/s

Mass of the baseball after struck by the bat, M = 900 g

Velocity of the baseball after struck by the bat, v = 47 m/s

According to the conservation of momentum,

[tex]Mv+mu=Mv_1+mv_2[/tex]

(900 x 47) + (200 x -30)  = (900 x [tex]v_1[/tex]) + (200 x [tex]v_2[/tex])

36300 =  (900 x [tex]v_1[/tex]) + (200 x [tex]v_2[/tex])

[tex]9v_1 + 2v_2 = 363[/tex] ..............(i)

[tex]9v_1 = 363 - 2v_2[/tex]

[tex]v_1=\frac{363 - 2v_2}{9}[/tex]

The mathematical expression for the conservation of kinetic energy is

[tex]\frac{1}{2}Mv^2+\frac{1}{2}mu^2 = \frac{1}{2}Mv_1^2+\frac{1}{2}mv_2^2[/tex]

[tex]\frac{1}{2}(900)(47)^2+\frac{1}{2}(200)(-30)^2 = \frac{1}{2}(900)v_1^2+\frac{1}{2}(200)v_2^2[/tex]    ................(ii)

[tex]$(9)(14)^2+(2)(-30)^2 = (9)v_1^2+2v_2^2$[/tex]  

[tex]21681 = 9v_1^2+2v_2^2[/tex]

Substituting (i) in (ii)

[tex]21681= 9\left( \frac{363-2v_2}{9}\right)^2+2v_2^2[/tex]

[tex](363-2v_2)^2+18v_2^2=195129[/tex]

[tex](363)^2+18v_2^2-2(363)(2v_2)+(363)^2-195129=0[/tex]

[tex]22v_2^2-145v_2-63360=0[/tex]

Solving the equation, we get

[tex]v_2=96 \ m/s, -30 \ m/s[/tex]

The negative velocity is neglected.

Therefore, substituting 96 m/s for [tex]v_2[/tex] in (i), we get

[tex]v_1=\frac{363-(2 \times 96)}{9}[/tex]

     = 19

Thus, only impulse of importance is used to find final velocity.

Answer:

t1= 8.40/10 =.84 s

t2 = 5.60/10 = .56s

t3= 1.4/10 = .14s

total time = 1.54 sec