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3 years ago

Which question about the discovery of the earth revolving around the sun could be best answered using scientific inquiry

1 answer:

8 0

AAaaahh yes, T'was that a lovely chap by the name of Copernicus, has discovered us a little secret, well. why dont you share?

Copernicus: well uh, i'm a bit shy

ME: who a gives a flying *bleep* about your feelings man! just man up an discover the earth's rotation already!

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True or false. there would be no like in earth without the sun

alisha [4.7K]

hello!

True

have a goed day

6 0

3 years ago

Electromagnetic radiation that has a short wavelength will have a ________ frequency.

ArbitrLikvidat [17]

High frequency , it is because wavelength is inversely proportional to frequency

8 0

3 years ago

A small box of mass m1 is sitting on a board of mass m2 and length L. The board rests on a frictionless horizontal surface. The

Nadusha1986 [10]

Answer:

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is $\left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}$

Explanation:

The Newton’s second law states that the net force on an object is the product of mass of the object and final acceleration of the object. The expression of newton’s second law is,

$\sum {F = ma}$

Here, is the sum of all the forces on the object, mm is mass of the object, and aa is the acceleration of the object.

The expression for static friction over a horizontal surface is,

$F_{\rm{f}}} \leq {\mu _{\rm{s}}}mg$

Here, ${\mu _{\rm{s}}}$ is the coefficient of static friction, mm is mass of the object, and $g$ is the acceleration due to gravity.

Use the expression of static friction and solve for maximum static friction for box of mass ${m_1}$

Substitute for in the expression of maximum static friction ${F_{\rm{f}}} = {\mu _{\rm{s}}}mg$

${F_{\rm{f}}} = {\mu _{\rm{s}}}{m_1}g$

Use the Newton’s second law for small box and solve for minimum acceleration aa to pull the box out.

Substitute for , [/tex]{m_1}[/tex] for in the equation .

${F_{\rm{f}}} = {m_1}a$

Substitute ${\mu _{\rm{s}}}{m_1}g$ for ${F_{\rm{f}}}$ in the equation ${F_{\rm{f}}} = {m_1}a$

${\mu _{\rm{s}}}{m_1}g = {m_1}a$

Rearrange for a.

$a = {\mu _{\rm{s}}}g$

The minimum acceleration of the system of two masses at which box starts sliding can be calculated by equating the pseudo force on the mass with the maximum static friction force.

The pseudo force acts on in the direction opposite to the motion of the board and the static friction force on this mass acts in the direction opposite to the pseudo force. If these two forces are cancelled each other (balanced), then the box starts sliding.

Use the Newton’s second law for the system of box and the board.

Substitute for for in the equation .

${F_{\min }} = \left( {{m_1} + {m_2}} \right)a$

Substitute for in the above equation .

${F_{\min }} = \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g$

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is $\left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g$

There is no friction between the board and the surface. So, the force required to accelerate the system with the minimum acceleration to slide the box over the board is equal to total mass of the board and box multiplied by the acceleration of the system.

5 0

3 years ago

Which of the following situations would cause the greatest decrease in the motion of molecules in a system?

Alina [70]

A. is the right answer since work is negative and Q which is heat in negative also

7 0

2 years ago

Read 2 more answers

Calculate the radius of the orbit of a proton moving at 2.2x10^6 m/s in a magnetic field 0.7 T where v and B are perpendicular.

Juliette [100K]

Answer:

3.28 cm

Explanation:

To solve this problem, you need to know that a magnetic field B perpendicular to the movement of a proton that moves at a velocity v will cause a Force F experimented by the particle that is orthogonal to both the velocity and the magnetic Field. When a particle experiments a Force orthogonal to its velocity, the path it will follow will be circular. The radius of said circle can be calculated using the expression:

r = $\frac{mv}{qB}$

Where m is the mass of the particle, v is its velocity, q is its charge and B is the magnitude of the magnetic field.

The mass and charge of a proton are:

m = 1.67 * 10^-27 kg

q = 1.6 * 10^-19 C

So, we get that the radius r will be:

r = $\frac{1.67 * 10^-27 kg * 2.2*10^6 m/s}{1.6 * 10^-19 C* 0.7 T}$ = 0.0328 m, or 3.28 cm.

8 0

3 years ago