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3 years ago

Question 1(Multiple Choice Worth 2 points)

Physics

1 answer:

tia_tia [17]3 years ago

3 0

Answer:

Explanation:

the reason for each column is to show the electronic configuration of the elements in that column.

Hope This Helps ▼・ᴥ・▼

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Q 28.7: A strip 1.2 mm wide is moving at a speed of 25 cm/s through a uniform magnetic field of 5.6 T. What is the maximum Hall

Zinaida [17]

Answer:the maximum Hall voltage across the strip= 0.00168 V.

Explanation:

The Hall Voltage is calculated using

Vh= B x v x w

Where

B is the magnitude of the magnetic field, 5.6 T

v is the speed/ velocity of the strip, = 25 cm/s to m/s becomes 25/100=0.25m/s

and w is the width of the strip= 1.2 mm to meters becomes 1.2 mm /1000= 0.0012m

Solving

Vh= 5.6T x 0.25m/s x 0.0012m

=0.00168T.m²/s

=0.00168Wb/s

=0.00168V

Therefore, the maximum Hall voltage across the strip=0.00168V

3 0

3 years ago

Consider the following neutral electron configurations in which n has a constant value. Which configuration would belong to the

grandymaker [24]

Answer:

he configuration with the highest electronic affinity is 2s2 2p5

Explanation:

Electronic affinity is the variation of energy when we add an electron to a neutral atom to form an ion

When an electron is added, it must occupy a space is the sub-level of the atom, giving more stability when it approaches the configuration of a complete shell with eight electrons (noble gas), so the affinity must increase when moving in a period Group VIII noble gases)

Let's examine the given settings

In this case, when adding an electron, 2s2 is very far from a complete level configuration, so its affinity must be small.

2s2 2p2 when adding an electro the one has a little more affinity, but is still a long way from a full shell, it would be missing 3 electrons

2s2 2sp5 this is the atom with the highest electronic affinity, since i = that when adding an electron the ion has the configuration of a noble gas. This is the most stable on the list

2s2 2p6 already has a full shell making it very difficult to insert an electron into this atom.

In summary, the configuration with the highest electronic affinity is 2s2 2p5

3 0

3 years ago

Read 2 more answers

While finding the spring constant, if X1 = 12 cm, X2 = 15 cm, and hanging mass = 22 grams, the value of spring constant K would

Pavel [41]

Answer:

If x₁=12 cm then k=1.7985 N/m

If x₂=15 cm then k=1.4388 N/m

Explanation:

Hanging mass= 22 g=0.022 kg

Acceleration due to gravity g=9.81 m/s²

If x₁=displacement= 12 cm=0.12 m

k= spring constant

$F=ma\\\Rightarrow F=0.022\times 9.81\\\Rightarrow F=0.21582\ N$

$\text {For spring}\\F=kx\\\Rightarrow 0.21582=k\times 0.012\\\Rightarrow k=1.7985\ N/m\\$

∴k = 1.7985 N/m

If x₂=15 cm=0.15 m

Force of the hanging mass is same however the spring constant will change

$\text {For spring}\\F=kx\\\Rightarrow 0.21582=k\times 0.015\\\Rightarrow k=1.4388\ N/m\\$

∴k = 1.4388 N/m

As the mass is not changing the spring constant has to change. That means that here there are two spring one with k=1.7985 N/m and the other with k= 1.4388 N/m

4 0

4 years ago

What happens if the breakdown voltage is exceeded.

s2008m [1.1K]

Answer:

this may help

Explanation:

maybe it will cause the eletric cause a fire that will mean it will spread till the fire men should come

7 0

3 years ago

When a parachutist jumps from an airplane, he eventually reaches a constant speed, called the terminal speed. Once he has reache

Masteriza [31]

Answer:

b) True. the force of air drag on him is equal to his weight.

Explanation:

Let us propose the solution of the problem in order to analyze the given statements.

The problem must be solved with Newton's second law.

When he jumps off the plane

fr - w = ma

Where the friction force has some form of type.

fr = G v + H v²

Let's replace

(G v + H v²) - mg = m dv / dt

We can see that the friction force increases as the speed increases

At the equilibrium point

fr - w = 0

fr = mg

(G v + H v2) = mg

For low speeds the quadratic depended is not important, so we can reduce the equation to

G v = mg

v = mg / G

This is the terminal speed.

Now let's analyze the claims

a) False is g between the friction force constant

b) True.

c) False. It is equal to the weight

d) False. In the terminal speed the acceleration is zero

e) False. The friction force is equal to the weight

3 0

3 years ago