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4 years ago

Combining which of the following substances with germanium will cause the germanium to emit free electrons?

2 answers:

5 0

The appropriate response is Gallium. It is a concoction component with image Ga and nuclear number 31. It is in gathering 13 of the occasional table and subsequently has similitudes to alternate metals of the gathering, aluminum, indium, and thallium.

vladimir1956 [14]4 years ago

4 0

On page 94 it explains that The electrical resistance of germanium or silicon can be reduced to a very small fraction of its original value by combining either material with an extremely small percentage of a suitable impurity. In one type of combination, called negative,or N-type,the impurity causes the germanium or silicon to emit free electrons. Arsenic, bismuth, and antimony are suitable impurities for this purpose. In another type of combination, called positive,or P-type,the impurity causes the germanium or silicon to collect free electrons. Aluminum, indium, and gallium are suitable impurities to cause this effect.

It is asking what would cause it to emit free electrons not what causes it to collect free electrons. So the Answer is A. Bismuth

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A car travels a distance of 100 km. For the first 30 minutes it is driven at a constant speed of 80 km/hr. The motor begins to v

gregori [183]

Explanation:

First, we need to determine the distance traveled by the car in the first 30 minutes, $d_{\frac{1}{2}}$ .

Notice that the unit measurement for speed, in this case, is km/hr. Thus, a unit conversion of from minutes into hours is required before proceeding with the calculation, as shown below

$d_{\frac{1}{2}\text{h}} \ = \ \text{speed} \ \times \ \text{time taken} \\ \\ \\ d_{\frac{1}{2}\text{h}} \ = \ 80 \ \text{km h}^{-1} \ \times \ \left(\displaystyle\frac{30}{60} \ \text{h}\right) \\ \\ \\ d_{\frac{1}{2}\text{h}} \ = \ 80 \ \text{km h}^{-1} \ \times \ 0.5 \ \text{h} \\ \\ \\ d_{\frac{1}{2}\text{h}} \ = \ 40 \ \text{km}$

Now, it is known that the car traveled 40 km for the first 30 minutes. Hence, the remaining distance, $d_{\text{remain}}$ , in which the driver reduces the speed to 40km/hr is

$d_{\text{remain}} \ = \ 100 \ \text{km} \ - \ 40 \ \text{km} \\ \\ \\ d_{\text{remain}} \ = \ 60 \ \text{km}$ .

Subsequently, we would also like to know the time taken for the car to reach its destination, denoted by $t_{\text{remian}}$ .

$t_{\text{remain}} \ = \ \displaystyle\frac{\text{distance}}{\text{speed}} \\ \\ \\ t_{\text{remain}} \ = \ \displaystyle\frac{60 \ \text{km}}{40 \ \text{km hr}^{-1}} \\ \\ \\ t_{\text{remain}} \ = \ 1.5 \ \text{hours}$ .

Finally, with all the required values at hand, the average speed of the car for the entire trip is calculated as the ratio of the change in distance over the change in time.

$\text{speed} \ = \ \displaystyle\frac{\Delta d}{\Delta t} \\ \\ \\ \text{speed} \ = \ \displaystyle\frac{100 \ \text{km}}{(0.5 \ \text{hr} \ + \ 1.5 \ \text{hr})} \\ \\ \\ \text{speed} \ = \ \displaystyle\frac{100 \ \text{km}}{2 \ \text{hr}} \\ \\ \\ \text{speed} \ = \ 50 \ \text{km hr}^{-1}$

Therefore, the average speed of the car is 50 km/hr.

8 0

3 years ago

Lance Armstrong bikes at a constant speed up the Col d’Izoard, a famous mountain pass. Assume his teammates do such a good job r

svetoff [14.1K]

Work done against gravity to climb upwards is always stored in the form of gravitational potential energy

so we can say

$W = mgh$

here h = vertical height raised

so here we know that

$h = 14.1 sin7.3 km$

here we have

$h = 1.79 km$

now from above equation

$W = (83 kg)(9.81 m/s^2)(1.79 \times 10^3 m)$

$W = 1.46 \times 10^6 J$

so work done will be given by above value

7 0

3 years ago

If the radius of an atom is 60 pm and the radius of the Earth is 6000 km, by how many orders of

kherson [118]

Answer:

1 x 10¹⁷

Explanation:

Given data:

Radius of the earth = 6000km

Radius of an atom = 60pm

Now, how many orders is the radius of the earth larger than an atom

Solution:

To solve this problem, let us express both quantity as the same unit;

1000m = 1km

6000km = 6000 x 10³m = 6 x 10⁶m

60pm;

1 x 10⁻¹²m = 1pm

60pm = 60 x 1 x 10⁻¹²m = 6 x 10⁻¹¹m

Now;

The order: $\frac{6 x 10^{6} }{6 x 10^{-11} }$ = 1 x 10¹⁷

6 0

3 years ago

Find the moment of inertia about each of the following axes for a rod that is 0.360 {cm} in diameter and 1.70 {m} long, with a m

mamaluj [8]

The complete question is;

Find the moment of inertia about each of the following axes for a rod that is 0.36 cm in diameter and 1.70m long, with a mass of 5.00 × 10 ^(−2) kg.

A) About an axis perpendicular to the rod and passing through its center in kg.m²

B) About an axis perpendicular to the rod and passing through one end in kg.m²

C) About an axis along the length of the rod in kg.m²

Answer:

A) I = 0.012 kg.m²

B) I = 0.048 kg.m²

C) I = 8.1 × 10^(-8) kg.m²

Explanation:

We are given;

Diameter = 0.36 cm = 0.36 × 10^(−2) m

Length; L = 1.7m

Mass;m = 5 × 10^(−2) kg

A) For an axis perpendicular to the rod and passing through its center, the formula for the moment of inertia is;

I = mL²/12

I = (5 × 10^(−2) × 1.7²)/12

I = 0.012 kg.m²

B) For an axis perpendicular to the rod and passing through one end, the formula for the moment of inertia is;

I = mL²/3

So,

I = (5 × 10^(−2) × 1.7²)/3

I = 0.048 kg.m²

C) For an axis along the length of the rod, the formula for the moment of inertia is; I = mr²/2

We have diameter = 0.36 × 10^(−2) m, thus radius;r = (0.36 × 10^(−2))/2 = 0.18 × 10^(−2) m

I = (5 × 10^(−2) × (0.18 × 10^(−2))^2)/2

I = 8.1 × 10^(-8) kg.m²

3 0

3 years ago

An electron moving in a direction perpendicular to a uniform magnetic field at a speed of 1.6 107 m/s undergoes an acceleration

umka2103 [35]

Answer:

B = 0.024T positive z-direction

Explanation:

In this case you consider that the direction of the motion of the electron, and the direction of the magnetic field are perpendicular.

The magnitude of the magnetic force exerted on the electron is given by the following formula:

$F=qvB$ (1)

q: charge of the electron = 1.6*10^-19 C

v: speed of the electron = 1.6*10^7 m/s

B: magnitude of the magnetic field = ?

By the Newton second law you also have that the magnetic force is equal to:

$F=qvB=ma$ (2)

m: mass of the electron = 9.1*10^-31 kg

a: acceleration of the electron = 7.0*10^16 m/s^2

You solve for B from the equation (2):

$B=\frac{ma}{qv}\\\\B=\frac{(9.1*10^{-31}kg)(7.0*10^{16}m/s^2)}{(1.6*10^{-19}C)(1.6*10^7m/s)}\\\\B=0.024T$

The direction of the magnetic field is found by using the right hand rule.

The electron moves upward (+^j). To obtain a magnetic forces points to the positive x-direction (+^i), the direction of the magnetic field has to be to the positive z-direction (^k). In fact, you have:

-^j X ^i = ^k

Where the minus sign of the ^j is because of the negative charge of the electron.

Then, the magnitude of the magnetic field is 0.024T and its direction is in the positive z-direction

8 0

3 years ago