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3 years ago

Which element is a halogen and has a larger average atomic mass than arsenic (As)?.

Physics

2 answers:

DaniilM [7]3 years ago

3 0

The element would be iodine

Alla [95]3 years ago

3 0

Explanation:

Arsenic is a group 15 element and lies in period 4. Whereas elements of halogen belongs to group 17.

Elements of group 17 are F, Cl, Br, I. It is known that on moving down the group there will be increase in size of atom because of increase in number of electrons.

Therefore, we can conclude that iodine is the halogen which has a larger average atomic mass than arsenic.

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Assuming the starting height is 0.0 m, calculate the potential energy of the cart after it has been elevated to a height of 0.5

Bogdan [553]

The potential energy is most often referred to as the "energy at rest" and is dependent on the elevation of an object. This can be calculated through the equation,

E = mgh

where E is the potential energy, m is the mass, g is the acceleration due to gravity, and h is the height. In this item, we are not given with the mass of the cart so we assume it to be m. The force is therefore,

E = m(9.8 m/s²)(0.5 m) = 4.9m

Hence, the potential energy is equal to 4.9m.

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4 years ago

As shown in the figure below, a bullet is fired at and passes through a piece of target paper suspended by a massless string. Th

NikAS [45]

Answer:

M = 0.730*m

V = 0.663*v

Explanation:

Data Given:

$v_{bullet, initial} = v\\v_{bullet, final} = 0.516*v\\v_{paper, initial} = 0\\v_{paper, final} = V\\mass_{bullet} = m\\mass_{paper} = M\\Loss Ek = 0.413 Ek$

Conservation of Momentum:

$P_{initial} = P_{final}\\m*v_{i} = m*0.516v_{i} + M*V\\0.484m*v_{i} = M*V .... Eq1$

Energy Balance:

$\frac{1}{2}*m*v^2_{i} = \frac{1}{2}*m*(0.516v_{i})^2 + \frac{1}{2}*M*V^2 + 0.413*\frac{1}{2}*m*v^2_{i}\\\\0.320744*m*v^2_{i} = M*V^2\\\\M = \frac{0.320744*m*v^2_{i} }{V^2} ....... Eq 2$

Substitute Eq 2 into Eq 1

$0.484*m*v_{i} = \frac{0.320744*m*v^2_{i} }{V^2} *V \\0.484 = 0.320744*\frac{v_{i} }{V} \\\\V = 0.663*v_{i}$

Using Eq 1

$0.484m*v_{i} = M* 0.663v_{i}\\\\M = 0.730*m$

7 0

3 years ago

The equivalent capacitance of three capacitors, each of capacitance C connected at series

katrin2010 [14]

Answer:

In Series:

1/C = 1/c1 + 1/c2 + 1/c3

Explanation:

For equivalent,

1/C = 3/c

Ce = c/3

5 0

3 years ago

This is written question

Vinil7 [7]

Answer:

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Explanation:

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3 years ago

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adelina 88 [10]

Answer:

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Explanation:

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3 years ago