On a 100km track , a train travels the first 30km with a speed of 30km/h . How fast the train travel the next 70 km if the average speed for the entire journey is 40km/h?

Answer:

Explanation:

Given,

Force ( f ) = 30 N

Acceleration(a) = 7.6 m/s

Now, Let's find the mass of the ball

Using the Newton's second law of motion:

We get:

[tex]force \: = mass \: \times acceleration[/tex]

plug the value

[tex]30 \: = m \: \times 7.6[/tex]

Use the commutative property to reorder the terms

[tex] 30 = 7.6 \: m[/tex]

Swap the sides of the equation

[tex]7.6m = 30[/tex]

Divide both sides of the equation by 7.6

[tex] \frac{7.6 \: m}{7.6} = \frac{30}{7.6} [/tex]

Calculate

[tex]m = 3.94 \: kg[/tex]

Hope this helps..

Best regards!!

Answer:

[tex]\displaystyle \boxed{\mathrm{3.95 \: kg }}[/tex]

Explanation:

[tex]\mathrm{force \: (N) = mass \: (kg) \times acceleration \: (m/s^2)}[/tex]

[tex]\mathrm{force = 30N}[/tex]

[tex]\mathrm{acceleration = 7.6 \: m/s^2 }[/tex]

[tex]\mathrm{Find \: the \: mass.}[/tex]

[tex]\mathrm{30 = m \times 7.6}[/tex]

[tex]\displaystyle \mathrm{m =\frac{30}{7.6} }[/tex]

[tex]\displaystyle \mathrm{m = 3.947... }[/tex]

Answer:

1.01×10^20 molecules of ozone.

Explanation:

Data obtained from the question include:

Volume (V) = 2 L

Temperature (T) = 275 K

Pressure (P) = 1.89×10¯³ atm

Gas constant (R) = 0.0821 atm.L/Kmol

Number of mole (n) of ozone =.?

Using the ideal gas equation, we can obtain the number of mole of ozone as follow:

PV = nRT

1.89×10¯³ x 2 = n x 0.0821 x 275

Divide both side by 0.0821 x 275

n = (1.89×10¯³ x 2) /(0.0821 x 275)

n = 1.67×10¯⁴ mole.

Therefore the number of mole of ozone in 2 L of air is 1.67×10¯⁴ mole.

Finally, we shall determine the number of molecules present in 1.67×10¯⁴ mole of ozone.

This can be obtained as follow:

From Avogadro's hypothesis, 1 mole of any substance contains 6.02×10²³ molecules. This implies that 1 mole of ozone contains 6.02×10²³ molecules.

If 1 mole of ozone contains 6.02×10²³ molecules,

therefore, 1.67×10¯⁴ mole of ozone will contain = 1.67×10¯⁴ x 6.02×10²³ = 1.01×10^20 molecules.

Therefore, 1.01×10^20 molecules of ozone are present in 2 L of air.

Answer:

1) a    α,  m   I,  W=F.d    W =τ . θ,

2)  a = v²/r

Explanation:

1) The amounts of rotational and translational motion are related

acceleration is

        a = d²x / dt²

    linear displacement is equivalent to angular rotation, therefore angular acceleration is

      α = d²θ / dt²

force in linear motion is equivalent to moment in endowment motion

       F = m a

       τ = I α

the mass is the inertia of the translation, in rotational motion the moment of inertia is the rotational inertia

          I = m r²

Work is defined by W = F. d

in rotation it is defined by W = τ . θ

The linear momentum is p = mv

the angular momentum L = I w

momentum the linear motion is I = F dt

in the rotation it is I = τ dt

2) The velocity is a vector therefore it has modulus and direction, linear acceleration changes the modulus of velocity, whereas circular motion changes the direction (the other element of the vector).

      [tex]a_{c}[/tex]Ac = v²/r

Answer:

pressure=force/area

Answer:

A car driving in a straight line 20 m/s

Explanation:

ayepecks silly

Answer:

a)4500V

b)750V

Explanation:

Given:

Distance between the plate=

6.00 cm

We need to convert to m

Then the Distance between the plate=

0.06m

electric field strength between two parallel plates =

7.50 × 104 V/m .

Then E= 7.50 × 104 V/m .

(a) What is the potential difference between the plates?

potential difference between the plates can be calculated using the formula below

Δ Vab=ED

Where E is the given electric field strength

D= The Distance between the plate

ΔVab=7.50 × 10⁴V/m ×

0.06m

= 4500V

(b) The plate with the lowest potential is taken to be zero volts. What is the potential 1.00 cm from that plate and 6.00 cm from the other?

the potential 1cm from the zero volt plate

Then the 1cm must be converted to m

= 0.01m

Let us say plate A as the plate at 0 volts:

The potential increases linearly going from plate A (0 V) to plate B (4500V).

Therefore,if the potential difference between A and B, separated by 6 cm, is 4500 V, then the potential difference between A and a point located at 1 cm from A is can be calculated also

If the plate with Lowest potential is taken to be zero then

=ΔVab=Vab-Vb=Va-0=Va=ED

Va=7.50 × 10⁴V/m × 0.01=750V

Answer:

210 m

Explanation:

The speed of train 2 relative to train 1 is 15 m/s + 20 m/s = 35 m/s.

It takes 6 seconds for the train to pass, so the length of the train is:

(35 m/s) (6 s) = 210 m

Answer:

Package's vertical displacement(s) = 22.12 meter

Explanation:

Given:

Speed of plane = 120 m/s

Total distance = 255 m

Find:

Package's vertical displacement(s)

Computation:

Time taken = Distance / Speed

Time taken = Total distance / Speed of plane

Time taken = 255 / 120

Time taken = 2.125 s

Acceleration due to gravity(g) = 9.8 m/s²

Initial velocity (u) = 0

So,

Package's vertical displacement(s) = ut + (1/2)gt²

Package's vertical displacement(s) = (0)(2.125) + (1/2)(9.8)(2.125)²

Package's vertical displacement(s) = 22.12 meter

Answer: -22.1

Explanation:

I just did the Khan Academy and that was the answer, not the one provided by that one person. :)))

Answer:

Explanation:

Load ( L ) = 500 N

Effort ( E ) = 200 N

Load distance ( LD ) = 3 m

Effort distance ( ED ) = ?

Now, Let's find the Effort distance ( ED )

We know that,

Output work = Input work

i.e L × LD = E × ED

plug the values

[tex]500 \times 3 = 200 \times ED[/tex]

multiply the numbers

[tex]1500 = 200 \times ED[/tex]

Swipe the sides of the equation

[tex]200 \: ED \: = 500[/tex]

Divide both sides of the equation by 200

[tex] \frac{200 \: ED}{200} = \frac{500}{200} [/tex]

Calculate

[tex]ED\: = 2.5 \: m[/tex]

Hope this helps..

best regards!!

ANSWER

C 11.5 m/s

EXPLANATION

Answer:

11.5m/s south

Explanation:

Online classes

Answer: 0.0180701 s

Explanation:

Given the following :

Length of string (L) = 10 m

Weight of string (W) = 0.32 N

Weight attached to lower end = 1kN = 1×10^3

Using the relation:

Time (t) = √ (weight of string * Length) / weight attached to lower end * acceleration due to gravity

g = acceleration due to gravity = 9.8m/s^2

Weight of string = 0.32N

Time(t) = √ (0.32 * 10) / [(1*10^3) * (9.8)]

Time = √3.2 / 9800

= √0.0003265

= 0.0180701s

Answer:

Explanation:

Reddening of sun's rays at sunset and sunrise is due to scattering of light . The white light consisting of seven colours coming from the sun are scattered in different directions when they fall on the air particles present in atmosphere . Red coloured light scatters least and it travels straight forward to the viewer on the earth . On the other hand other colours scatter most and therefore they go out of area of vision for the viewer on the earth . Since only red colour reaches the eye of the viewer , sun's ray appear red . This happens during sunrise and sunset . It is so because during this period , sun rays travel far greater distance through  atmosphere , so scattering is most pronounced .

Answer:

The actual height of the tree is 28 m

Explanation:

The given information are;

The length of the shadow of an upright meter rule = 25 cm

The actual height of the meter rule = 100 cm

The length of the shadow of the tree = 7 m

The actual  height of the tree  = h

We have

[tex]\dfrac{The \ length \ of \ the \ shadow \ of \ an \ upright \ metre \ rule}{The \ actual \ height \ of \ the \ metre \ rule} = \dfrac{The \ length \ of \ the \ shadow \ of \ the \ tree}{The \ actual \ height \ of \ the \ tree}[/tex]Which gives;

[tex]\dfrac{25 \ cm}{100 \ cm} = \dfrac{7 \ m}{The \ actual \ height \ of \ the \ tree}[/tex]

Therefore;

[tex]The \ actual \ height \ of \ the \ tree = 7 \ m \times \dfrac{100 \ cm}{25\ cm} = 7 \ m \times 4 = 28 \ m[/tex]

That is the actual height of the tree = 28 m.

Explanation:

It is given that,

Object distance from the mirror, u = -30 cm

Focal length of the mirror, f = +15 cm

Size of the object, h = 7.5 cm

We need to find the image distance and the size of the image.

Mirror's formula, [tex]\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}[/tex]

v is image distance

[tex]\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{(15)}-\dfrac{1}{(-30)}\\\\v=10\ cm[/tex]

Let h' is the size of the image. So,

[tex]\dfrac{h'}{h}=\dfrac{-v}{u}\\\\h'=\dfrac{-vh}{u}\\\\h'=\dfrac{-10\times 7.5}{-30}\\\\h'=2.5\ cm[/tex]

So, the image is located at a distance of 10 cm and the size of the image is 2.5 cm.

ANSWER:

0.4081%

Explanation:

Difference=24.6-24.5=0.1

Relative error = 0.1/24.5*100=0.4081%

Relative error is equal to the = difference between both the values/The true value *100

Answer:

V = 268.8 m/s

Explanation:

The speed of a wave in general is given by the following formula:

V = fλ

where,

V = Speed of that wave

f = Frequency of the wave

λ = wavelength of the wave

In this case we have a sound wave, travelling across carbon dioxide. The properties of sound wave are as follows:

V = Speed of Sound in Carbon dioxide = ?

f = frequency of sound wave = 140 Hz

λ = wavelength of sound wave = 1.92 m

Therefore,

V = (140 Hz)(1.92 m)

V = 268.8 m/s

Answer:

Kinetic energy.

Explanation:

Answer:

(a) By small angle approximation, we have;

F = -2×T×Δy/l

(b) [tex]The \ frequency \ of \ oscillation, \ f = \dfrac{1}{2\cdot \pi }\cdot\sqrt{\dfrac{2 \cdot T}{l \cdot M} }[/tex]

Explanation:

(a) The diagram shows the mass, M, being restored by two equal tension, T acting on the elastic strings l, such the restoring force, F acts along the path of motion of the mass, with distance Δy

Therefore, the component of the tension T that form part of the restoring force is given as follows;

Let the angle between the line representing the extension of the elastic strings T and the initial position of the string = ∅

Then we have;

String force, [tex]F_{string}[/tex] = T×sin∅ + T×cos∅ + T×sin∅ - T×cos∅  = 2×T×sin∅

Whereby the angle is small, we have;

sin∅ ≈ tan∅ = Δy/l

Which gives;

[tex]F_{string}[/tex] = 2×T×sin∅ = 2×T×Δy/l (for small angles)

Restoring force F = [tex]-F_{string}[/tex] = -2×T×Δy/l

F = -2×T×Δy/l

(b) Given that the the tensions do not change appreciably as the mass, M, oscillates from Δy we have;

By Hooke's law, F = -k×x

Whereby Δy corresponds to the maximum displacement of the mass, M from the rest position, which gives;

Which gives;

F = M×a = -k×Δy

a =  -k×Δy/M

d²(Δy)/dt² =  -k×Δy/M

When we put angular frequency as follows;

ω² = k/M

We get;

d²(Δy)/dt² =  -ω²×Δy

Which gives;

Δy(t) = A×cos(ωt + Ф)

The angular frequency is thus, ω = √(k/M)

Period of oscillation = 2·π/ω = 2·π/√(k/M)

The frequency of oscillation, f = 1/T = √(k/M)/(2·π)

Where:

k = 2·T/l, we have;

f = √(k/M)/(2·π) = √(2·T/l)/m)/(2·π)

The frequency of oscillation is given as follows;

[tex]f = \dfrac{1}{2\cdot \pi }\cdot\sqrt{\dfrac{2 \cdot T}{l \cdot M} }[/tex]

Answer:

D. Mars is a long distance away

Answer:

The teacher should swim in a direction 29.24° North of East

Explanation:

Given that the there is a water  current across the lake, and the physics teacher intends to swim directly across the lake, the direction the physics teacher will have to swim is found as follows;

The speed of the water current is given by the speed of the floating leaf traveling with the water current  

Distance traveled by the leaf = 4.2 m South

Time of travel of the leaf = 5.0 s

Speed of leaf = 4.2/5 = 0.84 m/s = Speed of the water current

Swimming peed of the teacher, v = 1.9 m/s

To swim directly across the lake, the teacher has to swim slightly in the opposite direction of the water current, the y-component of the teacher's swimming speed should be equal to and opposite that of the speed of the water current.

Y-component of v = v×sin(θ), where θ is the angle of the direction, the teacher should swim

Therefore;

1.9 × sin(θ) = 0.84

sin(θ) = 0.84/1.9 = 0.44

θ = 26.24°

That is the teacher should swim in a direction 29.24° North of East.

To cross the lake the teacher has to swim in a direction 29.24° North of the East

The speed of the water current can be derived from the speed of the floating leaf :

The distance traveled by the leaf L = 4.2 m South

Time taken T = 5s

So, the speed of the leaf is:

u = 4.2/5

u = 0.84 m/s South

So, the speed of the current is 0.84 m/s South

Now, it is given that the speed of the teacher is, v = 1.9 m/s East

To cross the lake the speed of the teacher must be in a Northeast direction so that the North component of the speed of the teacher cancels out the speed of the current which is directed towards the South.

Let, the speed of the teacher makes an angle of θ from the EAST.

So, the North component is given by:

v(north) = vsinθ

it must be equal to the speed of the current:

vsinθ = u

1.9 × sinθ = 0.84

sinθ = 0.84/1.9

sinθ = 0.44

θ = 26.24°

The teacher should swim in a direction 29.24° North of East.

Learn more about vector components:

https://brainly.com/question/1686398?referrer=searchResults

Question

The large-scale distribution of galaxies in the universe reveals

A) a smooth, continuous, and homogenous arrangement of clusters

B) large voids, with most of the galaxies lying in filaments and sheets a large supercluster at the center of the universe

C) a central void with walls of galaxies at the edge of the universe

Group of answer choices

Answer:

The correct answer is B)

Explanation:

The universe is arranged in a filamentary structure. Filamentary structures are very large. They are the largest kind of structures in the universe and comprise mostly of galaxies that are held together by gravity.

The structures found within Galaxy filaments have thread-like qualities spanning 52 to 78.7 megaparsecs h⁻¹ in lenght.

Other phenomena associated with the nature fo the universe is the existence of void spaces.

Cheers!

Answer:

Explanation:

just use the gravational force equation which is G x m of earth x m of object divided by r squared (which is radius of earth)

Answer:

1666.7 ETB (birr)

833.3 ETB (birr)

2083.3 ETB (birr)

Explanation:

The first worker

5000*1/3=1666.7

The second worker

5000*1/6=833.3

The third worker

5000*5/12=2083.3

Hope this helps :) ❤❤❤

Answer:

Q1 acceleration = 16m/s²

Q2 Net force = 9N North

Explanation:

Q1 Using the formula F=ma

Q2 R = F1 + F2

Answer:

W =1562.53 N

Explanation:

It is given that,

Radius of the aluminium ball, r = 24 cm = 0.24 m

The density of Aluminium, [tex]d=2698.4\ kg/m^3[/tex]

We need to find the thrust and the force. The mass of the liquid displaced is given by :

[tex]m=dV[/tex]

V is volume

Weight of the displaced liquid

W = mg

[tex]W=dVg[/tex]

So,

[tex]W=dg\times \dfrac{4}{3}\pi r^3\\\\W=2698.4\times 10\times \dfrac{4}{3}\times \pi \times (0.24)^3\\\\W=1562.53\ N[/tex]

So, the thrust and the force is 1562.53 N.

Answer:

a commutator

Explanation:

Answer:

Explanation:

For entropy change the formula is

ΔS = ΔQ / T

ΔQ = Δ H

ΔS = Δ H / T

Given

Δ H = + 162.8 kJ

We can take equilibrium temperature as average temperature of the whole process

So, T = 273 + 87 = 360 K

ΔS = Δ H / T

=  162.8 kJ  / 360

= +  0.508 kJ / K .

When the magnitude of the heat transfer for the process is 162.8 kJ, Then the entropy change for the mixing process, in kJ/K is = + 0.508 kJ / K

For The entropy change, the formula is

Then ΔS = ΔQ / T

After that ΔQ = Δ H

Then ΔS = Δ H / T

Given as per question are:

Then Δ H = + 162.8 kJ

Now We can take equilibrium temperature as average temperature of the whole process are:

So, T is = 273 + 87 = 360 K

Then ΔS = Δ H / T

After that = 162.8 kJ / 360

Therefore, = + 0.508 kJ / K.

Find more information about Entropy change here:

https://brainly.com/question/17241209

Answer:

λ₀ = 2 d n

Explanation:

A soap film is a layer where the lus is reflected on the surface and on the inside of the film, these two reflected rays can interfere with each other either constructively or destructively.

Let's analyze the general conditions of this interference,

* When the ray of light reaches the surface of the film it is reflected, as the index of refraction of the air is less than the index of the film, the reflected ray has a phase change of 180º

* When the ray penetrates the film, its wavelength changes due to the refractive index of the film.

          λ = λ₀ / n

where lick is the wavelength in the vacuum or air and n index of refraction of the film, in general this interference is observed perpendicular to the film, so the sine veils 1. the expression for constructive interference taking in what previous remains

         2d = (m + ½) λ

the expression for destructive interference remains

         2d = m λ

          2d = m λ₀ / n

When the film is placed on a glass plate whose index of refraction is greater than the index of refraction of the film, in the reflection in the lower part of the film another phase difference of 180º is created, for which we have a difference of total phase of 180 +180 = 360º, which is equivalent to no phase difference, therefore the two previous equations are interchanged.

Therefore where we had destructive interference now a cosntructive interference happens we can see the reflected light.

Find us the wavelength that this constructive interference creates

           2d n = m λ₀

           λ₀ = 2 d n / m

To find the minimum wavelength, suppose we observe the first interference pattern m = 1

         λ₀ = 2 d n

where d is the thickness of the film and n the index of refraction of the same

Answer:

3.94

Explanation:

divide total mass by the number of blocks since they are identical

Answer:

3.94

Explanation:

You want to find the mass of one block. Since we know there is 8 blocks with the same mass, you can divide the total mass by 8 since the mass is equally distributed within the 8 blocks

Answer:

88 %

Explanation:

The rate of heat loss by a conducting material of thermal conductivity K, cross-sectional area,A and thickness d with a temperature gradient ΔT is given by

P = KAΔT/d

The total heat lost by the styrofoam wall is P₁ = K₁A₁ΔT₁/d₁ where K₁ =thermal conductivity of styrofoam wall 0.033 W/m-K, A₁ = area of styrofoam wall = 17 m², ΔT₁ = temperature gradient between inside and outside of the wall and d₁ = thickness of styrofoam wall = 0.20 m

The total heat lost by the glass window is P₂ = K₂A₂ΔT₂/d₂ where K₂ =thermal conductivity of glass window pane wall 0.96 W/m-K, A₂ = area of glass window pane = 0.15 m², ΔT₂ = temperature gradient between inside and outside of the window and d₂ = thickness of glass window pane = 7 mm = 0.007 m

The total heat lost is P = P₁ + P₂ = K₁A₁ΔT₁/d₁ + K₂A₂ΔT₂/d₂

Now, since the temperatures of both inside and outside of both window and wall are the same, ΔT₁ = ΔT₂ = ΔT

So, P = K₁A₁ΔT/d₁ + K₂A₂ΔT/d₂

Since P₂ = K₂A₂ΔT₂/d₂ = K₂A₂ΔT/d₂is the heat lost by the window, the fraction of the heat lost by the window from the total heat lost is

P₂/P = K₂A₂ΔT/d₂ ÷ (K₁A₁ΔT/d₁ + K₂A₂ΔT/d₂)

= 1/(K₁A₁ΔT/d₁÷K₂A₂ΔT/d₂ + 1)

= 1/(K₁A₁d₂÷K₂A₂d₁ + 1)

= 1/[(0.033 W/m-K × 17 m² × 0.007 m ÷ 0.96 W/m-K × 0.15 m² × 0.20 m) + 1]

= 1/(0.003927/0.0288 + 1)

= 1/(0.1364 + 1)

= 1/1.1364

= 0.88.

The percentage is thus P₂/P × 100 % = 0.88 × 100 % = 88 %

The percentage of heat lost by window of the total heat is 88 %