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3 years ago

10

Un avión tarda en llegar a su destino 12 horas. Si recorrió una distancia de 10.700 kilómetros. Calcular su velocidad y expesarl

a en Km/h y en m/s

1 answer:

elixir [45]3 years ago

5 0

Given that,

Distance, d = 10700 km

Time taken by the airplane to complete the destination, t = 12 hours

We need to find the speed of the airplane. It is equal to the total distance covered divided by total time taken. So,

$v=\dfrac{d}{t}\\\\v=\dfrac{10700\ km}{12\ h}\\\\v=891.66\ km/h$

We know that,

1 km = 1000 m

1 h = 3600 s

So,

$v=891.66\ km/h =\dfrac{891.66\times 1000\ m}{3600\ s}\\\\v=247.68\ m/s$

So, the speed of the airplane is 891.66 km/h or 247.68 m/s.

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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

A steel ball bearing with a radius of 1.5 cm forms an image of an object that has been placed 1.1 cm away from the bearing’s sur

Nonamiya [84]

Answer:

Check the explanation

Explanation:

given

R = 1.5 cm

object distance, u = 1.1 cm

focal length of the ball, f = -R/2

= -1.5/2

= -0.75 cm

let v is the image distance

use, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/(-0.75) - 1/(1.1)

v = -0.446 cm <<<<<---------------Answer

magnification, m = -v/u

= -(-0.446)/1.1

= 0.405 <<<<<<<<<---------------Answer

The image is virtual

The image is upright

given

R = 1.5 cm

object distance, u = 1.1 cm

focal length of the ball, f = -R/2

= -1.5/2

= -0.75 cm

let v is the image distance

use, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/(-0.75) - 1/(1.1)

v = -0.446 cm <<<<<---------------Answer

magnification, m = -v/u

= -(-0.446)/1.1

= 0.405 <<<<<<<<<---------------Answer

Kindly check the diagram in the attached image below.

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3 years ago

The total negative charge on the electrons in 1mol of helium (atomic number 2, molar mass 4) is ________?

DaniilM [7]

Answer:

$1.92\times 10^5 C$

Explanation:

We are given that

Atomic number=2

We have to find the total negative charge on the electrons in one mole of Helium.

We know that atomic number=Proton number

Proton number=Number of electrons=2

Number of electrons in Helium=2

1 mole of Helium= $6.02\times 10^{23}$ atoms

We know that q=ne

Where n =Number of fundamental units

e=Charge on electron

1 e= $1.6\times 10^{-19}C$

Using the formula

$q=2\times 1.6\times 10^{-19}=3.2\times 10^{-19}C$

Total negative charge in 1 mole= $3.2\times 10^{-19}\times 6.02\times 10^{23}=1.92\times 10^5C$

Hence, the total negative charge on the electrons in 1 mole of Helium= $1.92\times 10^5 C$

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Answer:

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