carrier concentration for n type​

Answer:

44.73 MP/H or 71.98 KM/H

Explanation:

Answer:

This question is incomplete

Explanation:

This question is incomplete. However, the formula to be used here is

ω = 2π/T

Where ω is the angular frequency (in rad/s)

T is the period - the time taken for Block A to complete one oscillation and return to it's original position.

To solve for this period T, the formula below should be used

T = 2π√m/k

where m is the mass of the object (Block A) and k is the spring constant (281 J/m²)

Answer:  see attachment

Explanation:

Answer:

268.22m/s

Explanation:

Given;

    10mile/min to m/s

We need to convert between the two units;

    1 mile  = 1609.34m

     60s  = 1min

Now;

    10 x [tex]\frac{mile}{min}[/tex] x [tex]\frac{1min}{60s}[/tex] x [tex]\frac{1609.34m}{1mile}[/tex]

  = 268.22m/s

Answer:

(d) 2.93 x 10⁻³ N

Explanation:

Given;

current in the wire, I = 11.4 A

angle of inclination, θ = 11.4⁰

magnetic field on the wire, B = 11. 4  x 10⁻³

length of the wire, L = 11.4 cm = 0.114 m

The magnitude of magnetic force on the wire is given by;

F = BILsinθ

F = (11.4 x 10⁻³)(11.4)(0.114)(sin 11.4°)

F = 0.00293 N

F = 2.93 x 10⁻³ N

Therefore, the correct option is "D"

Answer:

C

Explanation:

Answer:

8.72*[tex]10^{-12}[/tex] N

Explanation:

force of attraction f = G m1m2/ r^2 = [tex]\frac{6.67*10^{-11}*7*5*2.70 }{.85*.85*1000}[/tex] = 8.72*[tex]10^{-12}[/tex] N

The gravitational force the bowling balls exert on a ping-pong is given as  8.72*[tex]10^{-12}[/tex] N.

A force is an effect that can alter an object's motion according to physics. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.

force of attraction

f = G [tex]m_1m_2[/tex]/r²

 = [tex]6.67*10^{-11}[/tex]*7*5*2.70/.85²*1000

 = 8.72*[tex]10^{-12}[/tex] N

The gravitational force the bowling balls exert on a ping-pong is given as  8.72*[tex]10^{-12}[/tex] N.

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Answer:

[tex]14400\ \text{N}[/tex], Attractive

[tex]3240\ \text{N}[/tex], Repulsive

Explanation:

[tex]q_1[/tex] = -20 μC

[tex]q_2[/tex] = 50 μC

r = Distance between the charges = 2.5 cm

k = Coulomb constant = [tex]9\times 10^9\ \text{Nm}^2/\text{C}^2[/tex]

Electrical force is given by

[tex]F=\dfrac{kq_1q_2}{r^2}\\\Rightarrow F=\dfrac{9\times 10^9\times (-20\times 10^{-6})\times (50\times 10^{-6})}{(2.5\times10^{-2})^2}\\\Rightarrow F=-14400\ \text{N}[/tex]

The magnitude of force each sphere will experience is [tex]14400\ \text{N}[/tex]

Since the charges have opposite charges they will attract each other.

Now the charges are brought into contact with each other so the resultant charge will be

[tex]q=\dfrac{q_1+q_2}{2}\\\Rightarrow q=\dfrac{-20+50}{2}\\\Rightarrow q=15\ \mu\text{C}[/tex]

[tex]F=\dfrac{kq^2}{r^2}\\\Rightarrow F=\dfrac{9\times 10^9\times (15\times 10^{-6})^2}{(2.5\times 10^{-2})^2}\\\Rightarrow F=3240\ \text{N}[/tex]

The magntude of the force the spheres experience will be [tex]3240\ \text{N}[/tex]

The spheres have the same charge now so they will repel each other.

Answer: 5.5m/s

Explanation:

vf=vi+at

vf= 4.0m/s + (0.50m/s^2)(3.0s)

The speed of the cart at the end of this 3.0 second interval is 5.5 meter per seconds.

Given the following data:

To find the speed of the cart at the end of this 3.0 second interval, we would use the first equation of motion;

[tex]V = U + at[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]V = 4 + 0.5(3)\\\\V = 4 + 1.5[/tex]

Final velocity, V = 5.5 m/s.

Therefore, the speed of the cart at the end of this 3.0 second interval is 5.5 meter per seconds.

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Answer:

Explanation:

initial velocity u = 32.7 m /s

final velocity v = 50.3 m /s

displacement s = 44500 m

acceleration a = ?

v² = u² + 2 a s

50.3² = 32.7² + 2 x a x 44500

2530.09 = 1069.29 + 89000a

a .016 m /s²

time taken t = ?

v = u + at

50.3 = 32.7 + .016 t

t = 1100 s

Answer:

attract

Explanation:

that is the answer

Answer:

Attract.

Explanation:

I took the quiz.

Answer: Titanium (Ti)

Explanation:

First, each element has a unique atomic number Z, that is equal to the number of protons in the nucleus.

The halogen in period 3 is chlorine (Ch)

Chlorine's atomic number is Z = 17, this means that it has 17 protons.

Now we want to find a transition metal that has 5 more protons, then this transition metal has Z = 17 + 5 = 22

Now we can look at the periodic table and find the element with Z = 22, and if this is in the d-block, then this will be a transition metal.

The element with Z = 22 is titanium (Ti)

Answer:

Option A

Explanation:

Ast the force is equal and the diayance is equal the beam is also balanced

Answer:

Less precipitation, droughts9: How might agriculture in southern Europe change by the end of the century if conditions follow the RCP8.

Explanation:

Precipitation and droughts are the specific changes in two climate variables that are expected to lead to major decreases in soil moisture.

Drought is defined as a period of protracted water scarcity, whether it is due to atmospheric surface water, or groundwater constraints.

Droughts can last months or years, although they can be proclaimed in as little as 15 days.

It has the potential to have a significant influence on the afflicted region's ecology and agriculture as well as harm the local economy.

Precipitation and droughts are the specific changes in two climate variables that are expected to lead to major decreases in soil moisture in southern Africa and the Mediterranean region.

Hence Precipitation and droughts are the specific changes in two climate variables.

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Answer:10.m and 10. M

Explanation:

A displacement vector with a magnitude of 20. m could have perpendicular components with magnitudes of C. 12 m and 16 m.

A displacement vector with a magnitude of 20. meters can be decomposed in 2 perpendicular components.

They would form a right triangle, in which the displacement vector would be the hypotenuse (a) and the components would be the legs (b, c).

Given the magnitude of the legs, we can calculate the magnitude of the hypotenuse using the Pythagorean theorem.

[tex]c = \sqrt{a^{2} + b^{2} }[/tex]

Let's use this formula to calculate the displacement vector for each pair of legs.

[tex]c = \sqrt{a^{2} + b^{2} } = \sqrt{(10.m)^{2} + (10.m)^{2} }= 14.1m[/tex]

[tex]c = \sqrt{a^{2} + b^{2} } = \sqrt{(12m)^{2} + (8.0m)^{2} }= 14.4m[/tex]

[tex]c = \sqrt{a^{2} + b^{2} } = \sqrt{(12m)^{2} + (16m)^{2} }= 20m[/tex]

[tex]c = \sqrt{a^{2} + b^{2} } = \sqrt{(16m)^{2} + (8.0m)^{2} }= 17.9m[/tex]

A displacement vector with a magnitude of 20. m could have perpendicular components with magnitudes of C. 12 m and 16 m.

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Answer:

a

Explanation:

..................

Answer:

 x = 10.75 m

Explanation:

For this problem we will solve it in two parts, the first using energy and the second with kinematics

Let's use the energy work relationship to find the velocity of the block as it exits the ramp

       W = [tex]Em_{f}[/tex] - Em₀

Starting point. Higher

       Em₀ = U = m g h

the height from the edge of the ramp of the graph has a value

        h = 9-3 = 6 m

Final point. At the bottom of the ramp

       Em_{f} = K = ½ m v²

Friction force work

      W = - fr  d

The friction force has the formula

      fr = μ N

On the ramp, we can use Newton's second law

         N - W cos θ = 0

         N = W cos θ

where the angle is obtained from the graph

         tan θ = (9-3) / (0.5-4) = -6 / 3.5

         θ = tan⁻¹ (-1,714)

         θ = -59.7º

the distance d is

         d = √ (Δx² + Δy²)

         d = √ [(0.5-4)² + (9-3)²]

         d = 6.95 m

for which the work is

       W = - μ mg cos 59.7 d

we substitute

        W = Em_{f} -Em₀

        - μ mg cos 59.7 d = ½ m v² - m g h

In the graph o text the value of the friction coefficient is not observed, suppose that it is μvery = 0.2

        - μ g cos 59.7 d = ½ v² - g h

         v² = 2g (h - very d coss 59.7)

let's calculate

         v² = 2 9.8 (6 - 0.2 6.95 cos 59.7)

         v = √ 103.8546

         v = 10.19 m / s

in the same direction as the ramp

in the second part we use projectile launch kinematics

let's look for the components of velocity

         v₀ₓ = vo cos -59.7

         [tex]v_{oy}[/tex] = vo sin (-59,7)

         v₀ₓ = 10.19 cos (-59.7) = 5.14 m / s

         v_{oy} = 10.19 if (-59.7) = -8.798 m / s

Let's find the time to get to the floor (y = o)

          y = y₀ + v_{oy} t - ½ g t²

to de groph y₀=3 m

          0 = 3 - 8.798 t - ½ 9.8 t²

          t² - 1.796 t - 0.612 = 0

we solve the quadratic equation

          t = [1.796 ±√(1.796² + 4 0.612)] / 2

          t = [1,795 ± 2,382] / 2

          t₁ = 2.09 s

          t₂ = -0.29 s

since time must be a positive quantity the correct value is t = 2.09 s

we calculate the horizontal displacement

          x = v₀ₓ t

          x = 5.14 2.09

          x = 10.75 m

The motion of the box, after it exits the incline is the motion and trajectory

of a projectile.

Horizontal distance from the right-hand edge of the incline to the point of

contact with the floor is approximately 1.24613 m.

Reasons:

Mass of the block,  m = 3.5 kg

Coefficient of kinetic friction, μ = 1.2

Location of the = 0.5 m from the origin

Required:

Horizontal distance between the block's point of contact with the floor and

the bottom right-hand edge of the incline.

Solution:

Let θ represent the angle the incline make with the horizontal.

The normal reaction of the incline on the block, [tex]F_N[/tex] = m·g·cos(θ)

Work done on friction = [tex]F_N[/tex]×μ×Length of the incline, L

Rise of the incline = 10 - 3 = 7

Run of the incline = 4

L = √(6.125² + 3.5²) = [tex]\dfrac{7 \times \sqrt{65} }{8}[/tex]

Let ΔP.E.₁  represent the potential energy transferred to kinetic energy

and work along the incline, we have;

Energy of the block at the bottom of the incline, M.E.₂, is found as follows;

K.E.₂ = mgh - m·g·μ·cos(θ)·L

[tex]K.E. =\frac{1}{2} \times 3.5 \times v^2 = 3.5 \times 9.81 \times 6.125 - 3.5 \times 9.81 \times 1.2 \times \dfrac{4}{\sqrt{65} } \times \dfrac{7 \times \sqrt{65} }{8}[/tex]

v ≈ 6.1456 m/s

The vertical component of the velocity is therefore;

[tex]v_y = v \cdot sin(\theta)[/tex]

[tex]v_y = 6.1456 \times \dfrac{7}{\sqrt{65} } \approx 5.33588[/tex]

From the equation, h = u·t + 0.5·g·t² derived from Newton's Laws of motion, we have;

ΔP.E.₁ = 3.5×9.81×7

3 = 5.33588·t + 0.5×9.81·t²

Factorizing, the above quadratic equation, we get;

The time it takes the block to reach the floor, t ≈ 0.40869 seconds

Horizontal component of the velocity is [tex]v_x \approx 6.1456 \times \dfrac{4}{\sqrt{65} } \approx 3.04908[/tex]

The horizontal distance, x = vₓ × t

∴ x = 3.04908 × 0.40869 ≈ 1.08194

Horizontal distance from the right-hand edge of the incline to the point of

contact with the floor, x ≈ 1.24613 m.

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Answer: intertia

Explanation:

Answer:

The value  is  [tex]\theta =407.3 \ radian[/tex]

Explanation:

From the question we are told that

    The angular acceleration is  [tex]\alpha = (7t + 8) \ rad/ s^2[/tex]

    The first time is  [tex]t_1 = 1.00 \ s[/tex]

    The second time [tex]t_2 = 6.10 \ s[/tex]

Generally the angular velocity is mathematically represented as

     [tex]w = \int\limits {\alpha } \, dt[/tex]

=>  [tex]w = \int\limits {7t + 8 } \, dt[/tex]

=>  [tex]w =\frac{ 7t^2}{2} + 8 t[/tex]

Generally the angular displacement  is mathematically represented as

[tex]\theta = \int\limits^{t_2}_{t_1} { w} \, dt[/tex]

=>  [tex]\theta = \int\limits^{t_2}_{t_1} { \frac{7t^2}{2} + 8t } \, dt[/tex]

=>  [tex]\theta = { \frac{7t^3}{6} + \frac{8t^2}{2} } | \left \ t_2} \atop {t_1}} \right.[/tex]

=> [tex]\theta = { \frac{7t^3}{6} + 4t^2} } | \left \ 6.10} \atop {1}} \right.[/tex]

=> [tex]\theta =[ { \frac{7}{6}[6.10 ]^3 + 4[6.10]^2} } ] -[ { \frac{7}{6}[1 ]^3 + 4[1]^2} } ][/tex]

=> [tex]\theta =407.3 \ radian[/tex]

Answer:

Energy can flow into and out of the system but matter cannot.

Explanation:

In a closed system, energy can flow in and out of the system but matter cannot.

For an isolated system, energy and matter cannot flow out of the system.

For open systems, energy and matter can flow out of the system.

Such systems are used for certain thermodynamics experiment.

We know that Period of a wave is the inverse of its Frequency

So,  Period = 1 / Frequency

From the above, we can say that Period is inversely proportional to Frequency and hence, any change in Frequency will be the inverse change in the period

Therefore, we can say that if the frequency is increased by 5 times, the period will increase by 1/5 times

Answer:

Explanation:

The mass of the object can be found by using the formula

[tex]m = \frac{f}{a} \\ [/tex]

f is the force

a is the acceleration

From the question we have

[tex]m = \frac{1200}{8} \\ [/tex]

We have the final answer as

Hope this helps you

Answer: When two objects in contact with each other are at different temperatures, they are said to be in thermal equilibrium.

Explanation: . When two objects not in contact with each other are at the same pressure, they are said to be in thermal equilibrium.

Answer:

29223.6J

Explanation:

Given parameters:

Mass of Piano = 852kg

Height of lifting  = 3.5m

Unknown:

Gravitational potential energy  = ?

Solution:

The gravitational potential energy of a body can be expressed as the energy due to the position of a body;

  G.P.E  = mgh

m is the mass

g is the acceleration due to gravity

h is the height

 Now insert the given parameters and solve;

   G.P.E  = 852 x 9.8 x 3.5 = 29223.6J

Answer:

The answer is D

Explanation:

The object with the heavy mass will reach first because the cause it is heaver so it will go faster

Answer:

bcuz ov yo fat mamma

Explanation:

Answer:

Water pollution

Explanation:

Answer:

The range of the projectile is 60 m

Explanation:

Horizontal Motion

When an object is thrown horizontally with a speed vo from a height h, it describes a curved path ruled exclusively by gravity until it eventually hits the ground.

The horizontal component of the velocity is always constant because no acceleration acts in that direction, thus:

[tex]v_x=v_o[/tex]

The vertical component of the velocity changes in time because gravity makes the object fall at increasing speed given by:

[tex]v_y=g.t[/tex]

The horizontal distance is calculated as a constant speed motion:

[tex]x = v_x.t[/tex]

Knowing the crossbow is fired horizontally at vo=vx=15 m/s and it takes t=4 s to hit the ground, thus the range of the projectile is:

x = 15*4 = 60

The range of the projectile is 60 m