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3 years ago

20. Using the picture, how many neutrons does lithium have?

Physics

1 answer:

Igoryamba3 years ago

8 0

Answer:

No. of Neutrons = 3

Explanation:

The atomic number of Lithium is given as 3 in the symbol while the mass number is given as 5.941 which is approximately equal to 6.

Mass Number = No. of Protons + No. of Neutrons = 6

Atomic Number = Number of Electrons = No. of Protons = 3

Therefore,

Mass Number - Atomic Number = (No. of Protons + No. of Neutrons) - No. of Protons

Mass Number - Atomic Number = No. of Neutrons

No. of Neutrons = 6 - 3

No. of Neutrons = 3

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Write relationship between hertz and megahertz

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Explanation:

1 mega Hertz = 1000000 hertz

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3 years ago

A wave is produced from a vibrating string on a violin. Two observers are standing 20 m and 40 m away from the musician. The obs

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Well the one that is closer can see and hear more

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3 years ago

At t=0, a block A of mass 8 kg and block B of mass 16 kg are both at position x=0 . Block A is at rest, and block B is moving at

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The center of mass of the two objects is the average position of the parts of the two object system

The center of mass of block A, and block B after displacement of block B is at 20 m from block A

Reason:

The given parameters are;

The position of block A and block B at t = 0 is x = 0

The mass of block A, m₁ = 8 kg

Mass of block B, m₂ = 16 kg

Speed of block A = 0 m/s

Speed of block B, v₂ = 10 m/s

Location of the center of mass of the two object at t = 3 s; Required

Solution;

The location of block A, after 3 s is x₁ = 0 (block A is at rest)

The location of block B, = v₂ × t

The location of block B, after 3 s is x₂ = 10 m/s × 3 s = 30 m

The center of mass of two masses are given as follows;

$x_{cm} = \dfrac{m_1 \cdot x_1 +m_2\cdot x_2}{m_1 + m_2}$

$x_{cm} = \dfrac{8 \times0 + 16 \times 30}{8 + 16} = 20$

The center of mass of the two objects is at at the position x = 20 m (from block A)

Learn more about the center of mass here:

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4 0

2 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.

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3 years ago

An airplane, starting from rest, moves down the runway at constant acceleration for 18 s and then takes off at a speed of 60 m/s

ruslelena [56]

Answer:

a =3.33 m/s²

Explanation:

given,

initial speed of Plane, u = 0 m/s

final speed of plane, v = 60 m/s

time of the acceleration, t = 18 s

average acceleration of the plane, a = ?