Experimenting with free fall, Mariana observes that her baseball takes 1.5 s to travel the last 30m before hitting the ground. From what height above the ground did Mariana drop the ball?

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:

b. Red light is transmitted through and reflected by the glass

Explanation:

Give me brainliest plz!

All of the wavelengths of the red light are absorbed when it passes through a translucent red glass window, but the red light is transmitted and reflected.

Therefore, the correct answer is option b) Red light is transmitted through and reflected by the glass.

To learn more about light refer to:

https://brainly.com/question/12002703

#SPJ2

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:

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:

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

I inferred you've referring to a close electrical circuit.

Answer:

The circuit is complete.

Explanation:

A closed electrical circuit is indeed a complete circuit. Also, it allows charges to flow, the bulbs in the circuit will shine and it is not broken.

It is termed closed circuit because there is no brokage in the series of electrical wires or the switch; which may prevent the free flow of current or charges. Thus, a feature that marks closed circuits is that they are complete.

Answer:

The circuit is complete.

Explanation:

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:

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:

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.

Answer:

1525 meters above ground

Explanation:

So to do this you will need to write this in slope intercept form or [tex]y=mx+b[/tex]. So 650 would be the b, 175 would be the m, and the x would be 5 so the equation would be [tex]y=175(5)+650[/tex] so if you solve or simplify the equation you will get 1525 meters above the ground and that would be our final answer.

Answer:

A car driving in a straight line 20 m/s

Explanation:

ayepecks silly

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:

Given,

Mass ( m ) = 10 kg

Force ( f ) = 2 Newtons

Acceleration ( a ) = ?

Now, let's find the acceleration :

We know that,

[tex]f = ma[/tex]

Plug the values

[tex]2 = 10a[/tex]

Swap the sides of the equation

[tex]10a = 2[/tex]

Divide both sides of the equation by 10

[tex] \frac{10a}{10} = \frac{2}{10} [/tex]

Calculate

[tex]a = 0.2 \: {metre \: per \: second \: }^{2} [/tex]

Hope this helps...

Best regards!!

Answer:

personally I'd say C by do not know if that is the exact answer

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:

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:

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:

Kinetic energy.

Explanation:

Answer:

pressure=force/area

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

C 11.5 m/s

EXPLANATION

Answer:

11.5m/s south

Explanation:

Online classes

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:

they get closer together

Explanation:

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:

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:

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:

a commutator

Explanation:

Answer: 43.2 J

Explanation:

Work = change in KE

initial KE = 0

final KE = 1/2mv^2 = 1/2(0.370 kg)(15.2778 m/s)^2 = 43.2 J

i'm not sure about sig figs though

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:

(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]