while riding an open freight elevator upwards at 5.1 m/s, a carpenter drops his hammer off the side. it hits the ground 2.4 s later. how far above the ground was the carpenter when he dropped the hammer?

Answer:

\(a_{c} = 50\ m/s^2\)

Explanation:

In uniform circular motion, the acceleration of the object is due to the changing direction of velocity of the object. this acceleration is referred to as centripetal acceleration. It can be calculated by following formula:

\(a_{c} = \frac{v^2}{r}\)

where,

ac = centripetal acceleration = ?

v = speed of the object = 15 m/s

r= radius of circular motion = distance of object from center of rotation = 4.5 m

Therefore, using these values in equation, we get:

\(a_{c} = \frac{(15\ m/s)^2}{4.5\ m}\\\\a_{c} = 50\ m/s^2\)

Answer:

true

Explanation:

Answer:

false

Explanation:

For the designing of differential amplifier following were found out :

the small-signal gain is zero.

the transconductance (gm) and output resistance (ro) of the NMOS transistors are  -640 * (W/L) μA/V and 1 / (8 * (W/L)) kΩ respectively.

the transconductance (gm) and output resistance (ro) of the PMOS transistors are -320 * (W/L) μA/V and  respectively.

NMOS transistor: Wn = 0.03 μm, L = 0.2 μm

PMOS transistor: Wp = 0.0075 μm, L = 0.2 μm

Bias current: Itail = 1 mA

Resistance: R = 0.3 kΩ

To design the differential amplifier according to the given specifications, we will follow these steps:

Step 1: Calculate the small-signal gain (Av)

Step 2: Determine the transconductance (gm) and output resistance (ro) of the NMOS transistors

Step 3: Determine the transconductance (gm) and output resistance (ro) of the PMOS transistors

Step 4: Calculate the tail current (Itail) based on the specified Iss

Step 5: Determine the resistance (R) value

Step 6: Calculate the width (Wp) of the PMOS transistor

Step 7: Calculate the width (Wn) of the NMOS transistors

Now let's go through each step in detail.

Step 1: Calculate the small-signal gain (Av)

Given: Av = 10, VOP = VON = 3.5V

Av = (vop - von) / (vip - vin)

10 = (3.5 - 3.5) / (0.1)

10 = 0 / 0.1

Since the numerator is zero, the small-signal gain is zero.

Step 2: Determine the transconductance (gm) and output resistance (ro) of the NMOS transistors

Given: unCox = 400 μA/V², Vtn = 0.8V, n = 0.02 V^(-1), L = 0.2 μm

gm = 2 * unCox * (W/L) * (Vgs - Vtn)

ro = 1 / (lambda * unCox * (W/L))

We need to design the amplifier for DC operation (Vin = Vbias), where the differential voltage (vgs = Vin - Vbias) should be zero to operate the transistors in the saturation region.

For the NMOS transistors:

Vgs = 0 (since Vin = Vbias)

gm = 2 * unCox * (W/L) * (Vgs - Vtn)

  = 2 * 400 μA/V² * (W/L) * (0 - 0.8)

  = -640 * (W/L) μA/V

ro = 1 / (lambda * unCox * (W/L))

  = 1 / (0.02 V^(-1) * 400 μA/V² * (W/L))

  = 1 / (8 * (W/L)) kΩ

Step 3: Determine the transconductance (gm) and output resistance (ro) of the PMOS transistors

Given: upCox = 200 μA/V², Vtp = -0.8V, p = 0.04 V^(-1), L = 0.2 μm

Similarly, for the PMOS transistors, we need to design the amplifier for DC operation (Vin = Vbias), where the differential voltage (vsg = Vbias - Vin) should be zero to operate the transistors in the saturation region.

For the PMOS transistors:

Vsg = 0 (since Vin = Vbias)

gm = 2 * upCox * (W/L) * (Vtp - Vsg)

  = 2 * 200 μA/V² * (W/L) * (-0.8 - 0)

  = -320 * (W/L) μA/V

ro = 1 / (lambda * upCox * (W/L))

  = 1 / (0.04 V^(-1) * 200 μA/V² *

  = 1 / (5 * (W/L)) kΩ

Step 4: Calculate the tail current (Itail) based on the specified Iss

Given: Iss = 2 mA

Itail = Iss / 2

= 2 mA / 2

= 1 mA

Step 5: Determine the resistance (R) value

Given: Vs = 0.3 V, VDD = 5 V

We can calculate the resistance (R) value using Ohm's Law:

Vs = Itail * R

0.3 V = 1 mA * R

R = 0.3 kΩ

Step 6: Calculate the width (Wp) of the PMOS transistor

To calculate Wp, we'll use the equation for the tail current:

Itail = 2 * upCox * (Wp/L) * (VDD - Vtp)^2

1 mA = 2 * 200 μA/V² * (Wp/0.2 μm) * (5 V + 0.8 V)^2

1 mA = 2 * 200 μA/V² * (Wp/0.2 μm) * (5.8 V)^2

Solving for Wp:

Wp = (1 mA * 0.2 μm) / (2 * 200 μA/V² * (5.8 V)^2)

Wp = 0.01 μm / (2 * 200 μA/V² * 33.64 V^2)

Wp ≈ 0.0075 μm

Step 7: Calculate the width (Wn) of the NMOS transistors

Given: Wn = 4 * Wp

Wn = 4 * 0.0075 μm

Wn = 0.03 μm

So, the design parameters for the differential amplifier are as follows:

the small-signal gain is zero.

the transconductance (gm) and output resistance (ro) of the NMOS transistors are  -640 * (W/L) μA/V and 1 / (8 * (W/L)) kΩ respectively.

the transconductance (gm) and output resistance (ro) of the PMOS transistors are -320 * (W/L) μA/V and  respectively.

NMOS transistor: Wn = 0.03 μm, L = 0.2 μm

PMOS transistor: Wp = 0.0075 μm, L = 0.2 μm

Bias current: Itail = 1 mA

Resistance: R = 0.3 kΩ

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The most effective insulating material is polystyrene.

The material which isolates the body and doesn't allow heat and mass to flow out of the control system.

The insulating material should have the lowest thermal conductivity.

Polystyrene foam has low thermal conductivity which makes it a great insulator to heat.

Aluminum foil can be an effective insulating material because it doesn't flow heat out into the environment.

Cotton wool is effective, but not it is not fire resistant.

Thus, the best effective insulating material is polystyrene.

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The change in his internal energy for all these activities performed by the  65-kg man is -435 kJ.

To calculate the change in his internal energy for all these activities, we use the formula; ΔE = (P)(Δt)

where P = power expended, and Δt = time

The formula states that the internal energy change is proportional to the time spent and the power expended by an individual during the activity. Using the data in the table, the metabolic rates for light activity and moderate activities are 4.2 kJ/min and 11.1 kJ/min, respectively, assuming that the man's mass is 65 kg.

Arranging the given data from the problem in the table below.

Activity Power, P (kJ/min)

Time, Δt (h)

Energy, ΔE (kJ)

Sleeping0.89.80

Light Chores4.22.5(4.2)(150)

Walking Slowly4.21.4(4.2)(84)

Jogging Moderately11.10.2(11.1)(12)

Sum 435. Note that the negative sign indicates that the man's internal energy decreased by 435 kJ during the activities.

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Answer:

can someone help me with physics to

Answer:

equation D=mass/volume

D=m/v

example sample : D=m/v= 250/100 = 2.5 g/cm³

1- D=m/v = 454/2274 = 0.2 g/cm³

2- 1 liter = 1000 cm³

D=m/v=1000/1000= 1 g/cm³

3 - d=m/v

0.6 g/cm³=m / 1.2 cm³

m=0.6*1.2 = 0.72 g

4: a:171/15  = 11.4 g/cm^3     lead

   b:148/40 = 3.7 g/cm^3     Aluminum

   c:475/250 =1.9 g/cm^3   Bone

   d: 680/1000 = 0.68 g/cm^3   Gasoline (lithium)

5: the density of a substance equal to the mass of the substance divide by volume

The value of charge q on an equipotential surface that surrounds a point charge is calculated to be 12.52× 10⁻⁹ C.

The expression to find out electric potential at a distance r is given by,

v = k q /r

where,

v is electric potential

q is charge

r is distance

k is coulomb's constant (9 × 10⁹ Nm²/C²)

Electric potential is given as 490 V.

Area is given as 1.1 m². The expression for area is A = 4 π r².

Making r as subject, we have,

Radius r = √(A/4π) = √(1.1/4π) = √0.087 = 0.23 m

To find out charge, let us make q as subject,

q = v r / k = (490 × 0.23)/(9 × 10⁹) = 12.52× 10⁻⁹ C

Thus, the charge on an equipotential surface that surrounds a point charge is calculated to be 12.52× 10⁻⁹ C.

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A basic harmonic oscillator has a 0.42 m amplitude and a 3.8 sec period. The maximum acceleration in the simple harmonic motion will be 1.144 rad/s².

Given that,

Amplitude of simple harmonic motion = 0.42 m

Time period of oscillation T = 3.8 sec

We have to find the maximum acceleration, but before that we have to find the angular frequency

The relation for angular frequency is

ω = (2*π)/T = (2* 3.14)/3.8 = 1.65 rad/sec

Maximum acceleration is given by, α = ω²* A = 1.65²* 0.42 = 1.144 rad/s²

Thus, maximum acceleration in the simple harmonic motion will be 1.144 rad/s².

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The following are the values:-

(a)Water content (W%)  = 13.99%

(b) Total unit weight (Υ\(_{t}\)) = 1778 kg/m³

(c) Dry unit weight (Y\(_{d}\)) = 1559.75 kg/m³

(d) Void ratio (e)  = 0.718

(e) Porosity (n) = 0.417

Volume of mold (V\(_{t}\)) = 0.4 m³

Dry mass of sand (W\(_{s}\)) = 623.9 kg

Wet mass of sand (W\(_{t}\)) = 711.2 kg

Specific Gravity (G\(_{s}\)) = 2.68

Water content (W%) = \(\frac{W_{w} }{W_{s} }\) × \(100\)

Water content (W%) = \(\frac{711.2 - 623.9}{623.9}\) × \(100\)

Water content (W%) = 13.99%

Total unit weight (Υ\(_{t}\)) = \(\frac{W_{t} }{V_{t} }=\frac{711.2}{0.4}\)

Total unit weight (Υ\(_{t}\)) = 1778 kg/m³

Dry unit weight (Y\(_{d}\)) = \(\frac{W_{s} }{V_{t} }=\frac{623.9}{0.4}\)

Dry unit weight (Y\(_{d}\)) = 1559.75 kg/m³

Solid unit weight (Y\(_{s}\)) = \(\frac{W_{s} }{V_{s} }\)

Or

G\(_{s}\) = \(\frac{Y_{s} }{Y_{w} }\) ⇒ Y\(_{s}\) = G\(_{s}\)Y\(_{w}\)

Y\(_{s}\) = 2.68 × \(\frac{10^{-3} }{(10^{-2})^{3} }\frac{kg}{m}\) = 2680 kg/m³

Now,

Void Ratio (e) = \(\frac{Y_{s} }{Yd}-1\)

Void Ratio (e) = \(\frac{2680}{1559.72}-1\)

Void Ratio (e) = 0.718

Porosity (n) = \(\frac{e}{1+e}\)

Porosity (n) = \(\frac{0.718}{1+0.718}\)

Porosity (n) = 0.417

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A laboratory sample of sand is formed inside a mold of 0.4 cubic meters. It took 711.2 kg of wet sand (dry mass of sand = 623.9 kg) to fill the mold. Assuming the specific gravity of the solid is 2.68, compute the:

(a)Water content (W%)  

(b) Total unit weight (Υ\(_{t}\))

(c) Dry unit weight (Y\(_{d}\))

(d) Void ratio (e)

(e) Porosity (n)

Answer:

I believe D

Explanation:

You need to have a more accurate reading and you want to test it multiple times throughout the week though to get a base resting rate.

I hope this is correct good luck!

She left you.. for another pokemon.

O_o - SikeBrehh 9

Answer:

cuz charmander is better lol

Explanation:

The acceleration due to gravity is dependent on the gravitational force between the two bodies, which is given by Newton's Law of Universal Gravitation:

F = (G x m1 x m2) / r^2

where F is the gravitational force, G is the gravitational constant, m1 and m2 are the masses of the two bodies, and r is the distance between their centers of mass.

When an object is falling towards the Earth, the gravitational force acting on it is given by:

F = m x g

where m is the mass of the object and g is the acceleration due to gravity. The two equations can be equated to give:

m x g = (G x m x M) / r^2

where M is the mass of the Earth.

Rearranging the equation for g, we get:

g = (G x M) / r^2

This equation shows that the acceleration due to gravity is dependent only on the mass of the Earth and the distance between the center of the Earth and the falling object. It does not depend on the mass of the falling object. Therefore, the acceleration due to gravity is independent of the mass of the falling body.

(a) Total, 910.2 coulombs of charge pass through any cross section of the wire's width. (b) Total, approximately 5.69 × 10²¹ electrons pass through any cross section of the wire's width.

To calculate the number of coulombs and electrons that pass through a cross-section of the wire during the given time, we can use the formula;

Charge (Q)=Current (I) × Time (t)

Given;

Current (I) = 4.9 A

Time (t) = 3.1 min

First, convert the time from minutes to seconds;

t = 3.1 min × 60 s/min

t = 186 s

To find the number of coulombs passing through the wire, we can use the formula;

Q = I × t

Substituting the given values, we have;

Q = 4.9 A × 186 s

Calculating the product, we find;

Q ≈ 910.2 C

Therefore, approximately 910.2 coulombs of charge pass through any cross section of the wire's width.

To find the number of electrons passing through the wire, we can use the relationship between charge and elementary charge (e);

Number of electrons = Charge (Q) / Elementary charge (e)

The elementary charge is approximately 1.6 × 10⁻¹⁹ C.

Substituting the values, we get;

Number of electrons = 910.2 C / (1.6 × 10⁻¹⁹ C)

Calculating the division, we find;

Number of electrons ≈ 5.69 × 10²¹ electrons

Therefore, approximately 5.69 × 10²¹ electrons pass through any cross section of the wire's width.

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The force between the rods are -3.486 x 10⁻⁴ N, the two rods are oppositely charged, the force is an attractive force.

Force is an influence on an object or system that will cause it to undergo acceleration, change in direction or shape, or be deformed. It is the result of an interaction between two bodies, and it is measured as a vector quantity with direction and magnitude. Force can be classified into contact forces, such as a push or pull, and non-contact forces, such as gravitation.

The force between two oppositely charged rods can be calculated using Coulomb's law. Coulomb's law states: F = k(q₁×q₂)/r²

where F is the force, k is a constant, q₁ and q₂ are the charges of the two rods (in coulombs), and r is the distance between the two rods (in meters).

Therefore, the force between the two rods in your example can be calculated as follows:

F = (8.99 x 10⁹ Nm²/C²)(7.5 x 10⁻⁷ C)( -5.1 x 10⁻⁵ C) / (0.09 m)²

F = -3.486 x 10⁻⁴ N

Since the two rods are oppositely charged, the force is an attractive force.

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Answer:

Explanation:

A simple pendulum consists of a mass (usually represented as a small object or bob) attached to a string or rod of negligible mass. The mass is free to swing back and forth under the influence of gravity.

In the figure, the point of suspension is denoted by "O," and the mass (bob) is represented by the small circle. The string or rod is represented by the vertical line connecting the point of suspension to the bob.

Amplitude:

The amplitude of a pendulum refers to the maximum displacement or swing of the bob from its equilibrium position. In the figure, the amplitude can be represented by the angle formed between the vertical position and the position of the bob when it swings to its maximum distance on one side. It is usually denoted by the symbol "A."

Effective Length:

The effective length of a pendulum refers to the distance from the point of suspension to the center of mass of the bob. It represents the distance over which the mass swings back and forth. In the figure, the effective length can be measured as the length of the string or rod from the point of suspension to the center of the bob. It is usually denoted by the symbol "L."

It is important to note that the amplitude and effective length of a simple pendulum affect its period of oscillation (the time taken for one complete swing). The relationship between these parameters and the period can be described by mathematical formulas.

Overall, the simple pendulum is a fundamental concept in physics and provides a simplified model for understanding oscillatory motion and the principles of periodic motion.

Wind speeds of above 56 kilometers per hour (35 miles per hour) are common during blizzards. These winds reduce visibility by causing a lot of snow to blow around in the air and close to the ground.

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Answer:

Answer: C. 3000kg

Explanation:

Hydraulic Jack

It's a device that is used to lift heavy loads by applying force via a hydraulic cylinder.

Since the internal pressure on the liquid of the jack is constant, the following relation applies:

\(\displaystyle \frac{F_1}{A_1}=\frac{F_2}{A_2}\)

Where

F1  = Force applied in one end of the jack

A1  = Area of the cylinder of the same end

F2  = Force applied in the other end of the jack

A2  = Area of the cylinder of the other end

If we assume the force is being applied on side 1, then the force on side 2 is calculated by solving for F2:

\(\displaystyle F_2=\frac{F_1.A_2}{A_1}\)

We have F1=1000N, A1= 48~cm^2, A2= 1440~cm^2, thus:

\(\displaystyle F_2=\frac{1000.1440}{48}\)

\(F_2 = 30000 ~N\)

Since we need to know the mass that can be lifted by that force, we use the formula:

W = m.g

Where W is the weight of the mass and g= 10~m/s^2. Since the weight is equal to the force exerted by the jack:

\(\displaystyle m=\frac{W}{g}=\frac{30000}{10}\)

m = 3000 Kg

Answer: C. 3000kg

Answer:

24° north of east

Explanation:

tan(x)=90/200 ie x=arctan(90/200)=24°

So the plane took off 24° east.

Answer:

24° north of east

Explanation:

Hope this will help

Answer:

2m/s²

Explanation:

F = 12 N

m = 3kg

a = ?

F = ma

a = F/m = 12/3 = 4 m/s²

when m = 6kg

a = F/m = 12/6

a = 2 m/s²

Answer:

This has been explained below.

Explanation:

Sound waves move along the auditory canal and when moving they hit the tympanic membrane making it to vibrate. this vibration would make the 3 ossicles to move. the tympanic membrane is sticked to the auditory ossicles and the stapes are joined to the oval window. as movements occur in the oval window, there would be motions happening in the cochlea

Wavelength of the incident photon = \(\lambda = 0.025nm = 0.025 \times 10^{-9}m\)

\(\lambda ^{'}-\lambda = \frac{h}{m_{e}c} \times (1- cos\theta )\)

Using the above formula we get Shifting wavelength \(\lambda = 0.027 \times 10^{-9}m\)

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Question:

A photon moving in the +x-direction, scatters off a free stationary electron. The wavelength of the incident photon is 0.0250 nm. After the collision, the electron moves at an angle α below the +x-axis, while the photon moves at an angle θ = 83.3° above the +x-axis. (Assume that the electron is traveling slow enough that the non-relativistic relationship between momentum and velocity can be used.)

(a) Find the shifting Wavelength of the incident photon

The separation of the lenses is 110 cm.

The separation of the lenses in an astronomical telescope can be calculated using the lens formula, which states that:

1/f = 1/v - 1/u,

where f is the focal length of the lens, v is the image distance, and u is the object distance.

In this case, the objective lens has a focal length of 100 cm, and the eyepiece lens has a focal length of 10 cm.
For the objective lens, u1 is the distance between the object being observed and the lens, and v1 is the distance between the lens and the image formed by the objective lens.
For the eyepiece lens, u2 is the distance between the image formed by the objective lens and the eyepiece lens, and v2 is the distance between the eyepiece lens and the final image formed.

Since the telescope is in normal adjustment, the final image is formed at infinity, so v2 is equal to infinity.

Using the lens formula for the objective lens, we can write:
1/f1 = 1/v1 - 1/u1.

Using the lens formula for the eyepiece lens, we can write:
1/f2 = 1/v2 - 1/u2.

Substituting v2 = infinity, the equation becomes:

1/f2 = 0 - 1/u2.

1/f2 = -1/u2.

Since the telescope is in normal adjustment, the final image formed is at infinity. Therefore, u2 is equal to the distance between the eyepiece lens and the image formed by the objective lens, which is equal to v1.
So, u2 = v1.

1/f2 = -1/v1.

1/10 = -1/v1.
v1 = -10 cm.
Since the distance between the object and the objective lens is equal to the distance between the objective lens and the image formed (u1 = v1), the separation between the lenses is the sum of the focal lengths of the two lenses:

separation = focal length of objective lens + focal length of eyepiece lens.

Substituting the values:

separation = 100 cm + 10 cm = 110 cm.

Therefore, the separation of the lenses in the astronomical telescope is 110 cm.

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The hydrostatic pressure at a given depth in a large tank of liquid is a function of depth and liquid density. Therefore, choices a and c are both valid.

The hydrostatic pressure at a given depth in a liquid is determined by the weight of the liquid above that depth.

As the depth increases, the weight of the liquid above it increases, resulting in an increase in pressure.

The pressure at a given depth can be calculated using the following formula:

P = ρgh

where P is the pressure, ρ is the density of the liquid, g is the acceleration due to gravity, and h is the depth.

As we can see from the formula, the pressure is directly proportional to the depth and the density of the liquid. The surface area of the tank does not affect the hydrostatic pressure at a given depth.

Therefore, choices a and c are both valid as the hydrostatic pressure at a given depth is a function of depth and liquid density.
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Answer:

In the given function b(t), the variable t represents the number of years after 2008. The   domain of this function is . The range more than 258.7 would not make sense. The graph of the function is always continuous

The given function b(t) shows the population of bobcats in northern Arizona since 2008. Therefore the variable t represents the number of years after 2008.

Since t represents the number of years after 2008, which is either positive or zero, therefore the domain of this function is .

The given function is a quadratic function and the coefficient of  is negative, so it is a downward parabola. The range above the y-coordinate of the vertex doesn't make any sense.

The vertex of parabola is defined by .

Vertex of the given function is (4.2,258.7).

Thus, the range more than 258.7 would not make sense for this function.

Since the given function is a polynomial function and polynomial functions are always continuous, therefore the graph of the function is always continuous

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The names of meteor showers (Geminids, Leonids, Perseids, Quadrantids) are names of meteor radiant points which are in the apparent direction of the luminous tails of individual meteors seen all over the sky.

The Quadrantids, for example, appear to radiate from the constellation Boötes.

The names of meteor showers (Geminids, Leonids, Perseids, Quadrantids) are names of radiant points which are in the apparent paths of the luminous tails of individual meteors seen all over the sky.

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