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an object down, but this is not true. If you place a box of mass 8 kg on a moving horizontal conveyor belt, the friction force of the belt acting on the bottom of the box speeds up the box. At first there is some slipping, until the speed of the box catches up to the speed of the belt, which is 5 m/s. The coefficient of kinetic friction between box and belt is 0.6. (a) How much time does it take for the box to reach this final speed

Answers

Answer 1

Answer:

Answer:

Explanation:

ASSUMING the belt is horizontal

kinetic friction force is μmg = 0.6(8)(9.8) = 47.04 N

Horizontal acceleration is

a = F/m = 47.04 / 8 = 5.88 m/s²

t = v/a = 5.0 / 5.88 = 0.85034...

t = 0.85 s

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Two charged particles are located on the x axis. The first is a charge +Q at x = −a. The second is an unknown charge located at x = +3a. The net electric field these charges produce at the origin has a magnitude of 2keQ/a2 . Explain how many values are possible for the unknown charge and find the possible values.

Answers

Answer:

-9Q

Explanation:

Electric field at origin is:

$E=(2keQ)/(a^2)$

Electric field due to first charge at origin would be:

$E_1=(keQ)/(a^2)$

Electric field due to second charge would be:

$E_2=E-E_1\nE_2=(2keQ)/(a^2)-(keQ)/(a^2) = (keQ)/(a^2)$

If the second charge is Q', then $E_2$ should be:

$E_2=(keQ')/((3a)^2)=(keQ')/(9a^2)$

compare the above two values to find the possible values of Q':

$(|Q'|)/(9)=Q\n |Q'|=9Q$

The net electric field at origin is greater than the one due to first charge. It means the second charge adds on to the electric field at the origin. Thus, it should be a negative charge.

Thus, Q' = -9Q

One value is possible as the location of the second charge is given to be on the positive x-axis.

Final answer:

The possible values for the unknown charge are 1/9 of the magnitude of the known charge.

Explanation:

To find the possible values for the unknown charge, we need to use the principle of superposition. The net electric field at the origin is given by the sum of the electric fields due to each charge. We know that the magnitude of the net electric field is 2keQ/a^2, so we can set up the equation:

2keQ/a^2 = keQ/(-a)^2 - keq/(3a)^2

By solving this equation, we can find the possible values for the unknown charge. Simplifying the equation, we get:

2 = 1 - 1/9

1/9 = 1

After solving the equation, we find that the possible value for the unknown charge is 1/9 of the magnitude of the known charge.

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Dave rows a boat across a river at 4.0 m/s. the river flows at 6.0 m/s and is 360 m across.a. in what direction, relative to the shore, does dave’s boat go?
b. how long does it take dave to cross the river?
c. how far downstream is dave’s landing point?
d. how long would it take dave to cross the river if there were no current?

Answers

a) Let's call x the direction parallel to the river and y the direction perpendicular to the river.

Dave's velocity of 4.0 m/s corresponds to the velocity along y (across the river), while 6.0 m/s corresponds to the velocity of the boat along x. Therefore, the drection of Dave's boat is given by:

$\theta= arctan((v_y)/(v_x))=arctan((4.0 m/s)/(6.0 m/s))=arctan(0.67)=33.7^(\circ)$

relative to the direction of the river.

b) The distance Dave has to travel it S=360 m, along the y direction. Since the velocity along y is constant (4.0 m/s), this is a uniform motion, so the time taken to cross the river is given by

$t=(S_y)/(v_y)=(360 m)/(4.0 m/s)=90 s$

c) The boat takes 90 s in total to cross the river. The displacement along the y-direction, during this time, is 360 m. The displacement along the x-direction is

$S_x = v_x t =(6.0 m/s)(90 s)=540 m$

so, Dave's landing point is 540 m downstream.

d) If there were no current, Dave would still take 90 seconds to cross the river, because its velocity on the y-axis (4.0 m/s) does not change, so the problem would be solved exactly as done at point b).

Which statements explain why theories change over time? Check all that apply. Scientists change the definition of theory to have their ideas accepted. New information and technology may be developed that influence the theory. Theories change with each new generation of scientists. New experimental methods provide new information. Theories may or may not be supported by new information.

Answers

The theories change over time because new information and technology may be developed that influence the theory. Therefore, option B is correct.

Why theories change over time ?

As new information and perspectives become available, established theories may be changed or refuted. If a new or modified theory explains all the old theory did and then some, scientists are likely to accept it.

When fresh data emerges that refutes an existing theory, scientific theories can be revised or replaced. All scientific theories are founded on observable, testable data, and they can be changed when new information is found that contradicts the present idea.

When certain features of a hypothesis are refuted by fresh experimental data, a theory might be altered in science.

Thus, option B is correct.

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Final answer:

Theories change over time due to new information, technology, and the perspectives of new generations of scientists.

Explanation:

Theories change over time for several reasons. First, new information and technology may be developed that influence the theory. For example, advancements in experimental methods can provide new information and lead to the refinement or modification of theories. Additionally, theories may change with each new generation of scientists as they bring new perspectives and ideas. It is important to note that theories may or may not be supported by new information, and scientists do not change the definition of theory to have their ideas accepted.

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Calculate the ionization potential for C+5 ( 5 electrons removed for the C atom) and in addition compute the wavelength of the transition from n=3 to n= 2.

Answers

Answer:

Ionization potential of C⁺⁵ is 489.6 eV.

Wavelength of the transition from n=3 to n=2 is 1.83 x 10⁻⁸ m.

Explanation:

The ionization potential of hydrogen like atoms is given by the relation :

$E = (13.6Z^(2) )/(n^(2) ) eV$ .....(1)

Here E is ionization potential, Z is atomic number and n is the principal quantum number which represents the state of the atom.

In this problem, the ionization potential of Carbon atom is to determine.

So, substitute 6 for Z and 1 for n in the equation (1).

$E = (13.6*(6)^(2) )/(1^(2) )$

E = 489.6 eV

The wavelength (λ) of the photon due to the transition of electrons in Hydrogen like atom is given by the relation :

$(1)/(\lambda) =RZ^(2)[(1)/(n_(1) ^(2))-(1)/(n_(2) ^(2) )]$ ......(2)

R is Rydberg constant, n₁ and n₂ are the transition states of the atom.

Substitute 6 for Z, 2 for n₁, 3 for n₂ and 1.09 x 10⁷ m⁻¹ for R in equation (2).

$(1)/(\lambda) =1.09*10^(7) *6^(2)[(1)/(2 ^(2))-(1)/(3 ^(2) )]$

$(1)/(\lambda)$ = 5.45 x 10⁷

λ = 1.83 x 10⁻⁸ m

Which are electromagnetic waves? check all that apply.earthquake waves
infrared waves
ocean waves
radio waves
untraviolet waves

Answers

Since electromagnetic waves do not require a medium for their transmission, the electromagnetic waves are radio waves, ultraviolet waves and infrared waves.

What are electromagnetic waves?

Electromagnetic waves or radiations are waves which occur as a result of the interaction between the electric and magnetic fields.

Electromagnetic waves do not require a material medium for their transmission and as such can travel through a vacuum.

Some examples of electromagnetic waves are radio waves, ultraviolet waves, microwaves, infrared waves etc.

Therefore, the electromagnetic waves are radio waves, ultraviolet waves and infrared waves.

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The electromagnetic waves are:
Radio waves
Ultraviolet waves
And Infrared waves
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What are the density, specific gravity and mass of the air in a room whose dimensions are 4 m * 6 m * 8 m at 100 kPa and 25 C.

Answers

Answer:

Density = 1.1839 kg/m³

Mass = 227.3088 kg

Specific Gravity = 0.00118746 kg/m³

Explanation:

Room dimensions are 4 m, 6 m & 8 m. Thus, volume = 4 × 6 × 8 = 192 m³

Now, from tables, density of air at 25°C is 1.1839 kg/m³

Now formula for density is;

ρ = mass(m)/volume(v)

Plugging in the relevant values to give;

1.1839 = m/192

m = 227.3088 kg

Formula for specific gravity of air is;

S.G_air = density of air/density of water

From tables, density of water at 25°C is 997 kg/m³

S.G_air = 1.1839/997 = 0.00118746 kg/m³

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