A 300 g bird flying along at 6.2 m/s sees a 10 g insect heading straight toward it with a speed of 35 m/s (as measured by an observer on the ground, not by the bird). The bird opens its mouth wide and enjoys a nice lunch.

Required:
What is the bird's speed immediately after swallowing?

Given that,

Sun exposure = 30%, 45%, 60%, 75%, 90%

Stem mass (g) = 275, 415, 563, 815, 954

Stem volume (ml) = 1100, 1215, 1425, 1610, 1742

(a). We need to convert the mass measurements to kilograms (kg) and the volume measurements to cubic meters

Using conversion of mass

[tex]1\ g=0.001\ kg[/tex]

Conservation of volume

[tex]1\ Lt=0.001\ m^3[/tex]

[tex]1\ mL=1\times10^{-6}\ m^3[/tex]

So, mass in kg

Stem mass (kg) = 0.275, 0.415, 0.563, 0.815, 0.954

Volume in m³,

Stem volume (m³) = 0.0011, 0.001215, 0.001425, 0.001610, 0.001742

(b). We need to calculate the density of the samples

Using formula of density

[tex]\rho=\dfrac{m}{V}[/tex]

Where, m = mass

V = volume

If the m = 0.275 kg and V = 0.0011 m³

Put the value into the formula

[tex]\rho=\dfrac{0.275}{0.0011}[/tex]

[tex]\rho=250\ kg/m^3[/tex]

If the m = 0.415 kg and V = 0.001215 m³

Put the value into the formula

[tex]\rho=\dfrac{0.415}{0.001215}[/tex]

[tex]\rho=341.56\ kg/m^3[/tex]

[tex]\rho=342\ kg/m^3[/tex]

If the m = 0.563 kg and V = 0.001425 m³

Put the value into the formula

[tex]\rho=\dfrac{0.563}{0.001425}[/tex]

[tex]\rho=395.08\ kg/m^3[/tex]

If the m = 0.815 kg and V = 0.001610 m³

Put the value into the formula

[tex]\rho=\dfrac{0.815}{0.001610}[/tex]

[tex]\rho=506.21\ kg/m^3[/tex]

If the m = 0.954 kg and V = 0.001742 m³

Put the value into the formula

[tex]\rho=\dfrac{0.954}{0.001742}[/tex]

[tex]\rho=547.6\ kg/m^3[/tex]

[tex]\rho=548\ kg/m^3[/tex]

(c). We need to convert the density values to scientific notation

In scientific notation

The densities are

[tex]\rho\ (kg/m^3)= 2.50\times10^{2}, 3.42\times10^{2}, 3.95\times10^{2}, 5.06\times10^{2}, 5.48\times10^{2}[/tex]

Hence, This is required solution.

Answer:

The fraction of the cooper's electrons that is removed is [tex]8.5222\times 10^{-11}[/tex].

Explanation:

An electron has a mass of [tex]9.1 \times 10^{-31}\,kg[/tex] and a charge of [tex]-1.6 \times 10^{-19}\,C[/tex]. Based on the Principle of Charge Conservation, [tex]-2.10\times 10^{-6}\,C[/tex] in electrons must be removed in order to create a positive net charge. The amount of removed electrons is found after dividing remove charge by the charge of a electron:

[tex]n_{R} = \frac{-2.10\times 10^{-6}\,C}{-1.6 \times 10^{-19}\,C}[/tex]

[tex]n_{R} = 1.3125 \times 10^{13}\,electrons[/tex]

The number of atoms in 56 gram cooper ball is determined by the Avogadro's Law:

[tex]n_A = \frac{m_{ball}}{M_{Cu}}\cdot N_{A}[/tex]

Where:

[tex]m_{ball}[/tex] - Mass of the ball, measured in kilograms.

[tex]M_{Cu}[/tex] - Atomic mass of cooper, measured in grams per mole.

[tex]N_{A}[/tex] - Avogradro's Number, measured in atoms per mole.

If [tex]m_{ball} = 56\,g[/tex], [tex]M_{Cu} = 63.5\,\frac{g}{mol}[/tex] and [tex]N_{A} = 6.022\times 10^{23}\,\frac{atoms}{mol}[/tex], the number of atoms is:

[tex]n_{A} = \left(\frac{56\,g}{63.5\,\frac{g}{mol} } \right)\cdot \left(6.022\times 10^{23}\,\frac{atoms}{mol} \right)[/tex]

[tex]n_{A} = 5.3107\times 10^{23}\,atoms[/tex]

As there are 29 protons per each atom of cooper, there are 29 electrons per atom. Hence, the number of electrons in cooper is:

[tex]n_{E} = \left(29\,\frac{electrons}{atom} \right)\cdot (5.3107\times 10^{23}\,atoms)[/tex]

[tex]n_{E} = 1.5401\times 10^{23}\,electrons[/tex]

The fraction of the cooper's electrons that is removed is the ratio of removed electrons to total amount of electrons when net charge is zero:

[tex]x = \frac{n_{R}}{n_{E}}[/tex]

[tex]x = \frac{1.3125\times 10^{13}\,electrons}{1.5401\times 10^{23}\,electrons}[/tex]

[tex]x = 8.5222 \times 10^{-11}[/tex]

The fraction of the cooper's electrons that is removed is [tex]8.5222\times 10^{-11}[/tex].

Answer:

The radius of the sphere is 4.05 m

Explanation:

Given;

potential at surface, [tex]V_s[/tex] = 450 V

potential at radial distance, [tex]V_r[/tex] = 150

radial distance, l = 8.1 m

Apply Coulomb's law of electrostatic force;

[tex]V = \frac{KQ}{r} \\\\V_s = \frac{KQ}{r} \\\\V_r = \frac{KQ}{r+ l}[/tex]

[tex]450 = \frac{KQ}{r} ------equation (i)\\\\150 = \frac{KQ}{r+8.1} ------equation (ii)\\\\divide \ equation (i)\ by \ equation \ (ii)\\\\\frac{450}{150} = (\frac{KQ}{r} )*(\frac{r+8.1}{KQ} )\\\\3 = \frac{r+8.1}{r} \\\\3r = r + 8.1\\\\2r = 8.1\\\\r = \frac{8.1}{2} \\\\r = 4.05 \ m[/tex]

Therefore, the radius of the sphere is 4.05 m

Answer:

t = 0.414s

Explanation:

In order to calculate the time that the ball takes to reach home plate, you assume that the speed of the ball is constant, and you use the following formula:

[tex]t=\frac{d}{v}[/tex]         (1)

d: distance to the plate = 18.4m

v: speed of the ball = 160.0km/h

You first convert the units of the sped of the ball to appropriate units (m/s)

[tex]160.0\frac{km}{h}*\frac{1h}{3600s}*\frac{1000m}{1km}=44.44\frac{m}{s}[/tex]

Then, you replace the values of the speed v and distance s in the equation (1):

[tex]t=\frac{18.4m}{44.44m/s}=0.414s[/tex]

THe ball takes 0.414s to reach the home plate

Answer:

The distance is  [tex]d = 1.5 *10^{15} \ km[/tex]

Explanation:

From the question we are told that

        The smallest shift is [tex]d = 0.2 \ grid \ units[/tex]

Generally a grid unit is  [tex]\frac{1}{10}[/tex] of  an arcsec

  This implies that  0.2 grid unit is  [tex]k = \frac{0.2}{10} = 0.02 \ arc sec[/tex]

The maximum distance at which a star can be located and still have a measurable parallax is mathematically represented as

           [tex]d = \frac{1}{k}[/tex]

substituting values

           [tex]d = \frac{1}{0.02}[/tex]

           [tex]d = 50 \ parsec[/tex]

Note  [tex]1 \ parsec \ \to 3.26 \ light \ year \ \to 3.086*10^{13} \ km[/tex]

So  [tex]d = 50 * 3.08 *10^{13}[/tex]

     [tex]d = 1.5 *10^{15} \ km[/tex]

Answer:

x = 46.54m

Explanation:

In order to find the length of the incline you use the following formula:

[tex]x=v_ot+\frac{1}{2}at^2[/tex]      (1)

vo: initial speed of the soccer ball = 0 m/s

t: time

a: acceleration

You first use the the fact that the ball traveled 22 m in 2.2 s. Whit this information you can calculate the acceleration a from the equation (1):

[tex]22m=\frac{1}{2}a(2.2s)^2\\\\a=9.09\frac{m}{s^2}[/tex]      (2)

Next, you calculate the distance traveled by the ball for t = 3.2 s (one second later respect to t = 2.2s). The values of the distance calculated is the lenght of the incline:

[tex]x=\frac{1}{2}(9.09m/s^2)(3.2s)^2=46.54m[/tex]       (3)

The length of the incline is 46.54 m

Answer:

a) v₁ = 3.92 m / s , b)     ΔP =  = 9.0 10⁴ Pa, c)  t = 0.0297 s  

Explanation:

This is a fluid mechanics exercise

a) let's use the continuity equation

let's use index 1 for the hose and index 2 for the nozzle

        A₁ v₁ = A₂v₂

in area of ​​a circle is

       A = π r² = π d² / 4

we substitute in the continuity equation

        π d₁² / 4 v₁ = π d₂² / 4 v₂

        d₁² v₁ = d₂² v₂

the speed of the water in the hose is v1

       v₁ = v₂ d₂² / d₁²

       v₁ = 14 (1.8 / 3.4)²

        v₁ = 3.92 m / s

b) they ask us for the pressure difference, for this we use Bernoulli's equation

       P₁ + ½ ρ v₁² + m g y₁ = P₂ + ½ ρ v₂² + mg y2

as the hose is horizontal y₁ = y₂

       P₁ - P₂ = ½ ρ (v₂² - v₁²)

      ΔP = ½ 1000 (14² - 3.92²)

       ΔP = 90316.8 Pa = 9.0 10⁴ Pa

c) how long does a tub take to flat

the continuity equation is equal to the system flow

        Q = A₁v₁

        Q = V t

where V is the volume, let's equalize the equations

         V t = A₁ v₁

         t = A₁ v₁ / V

A₁ = π d₁² / 4

let's reduce it to SI units

         V = 120 l (1 m³ / 1000 l) = 0.120 m³

          d1 = 3.4 cm (1 m / 100cm) = 3.4 10⁻² m

let's substitute and calculate

         t = π d₁²/4   v1 / V

         t = π (3.4 10⁻²)²/4 3.92 / 0.120

         t = 0.0297 s

Answer:

The found acceleration in terms of h and t is:

[tex]a=\frac{h}{5(t_1)^2}[/tex]

Explanation:

(The complete question is given in the attached picture. We need to find the acceleration in terms of h and t in this question)

We are given 3 stages of movement of elevator. We'll first model them each of the stage one by one to find the height covered in each stage. After that we'll find the total height covered by adding heights covered in each stage, and equate it to Total height h. From that we can find the formula for acceleration.

Constant acceleration, starts from rest.

Distance = [tex]y = \frac{1}{2}a(t_1)^2[/tex]

Velocity = [tex]v_1=at_1[/tex]

Constant velocity where

Velocity = [tex]v_o=v_1=at_1[/tex]

Distance =

Constant deceleration where

Velocity = [tex]v_0=v_1=at_1[/tex]

Distance =

[tex]y_3=v_1t_3-\frac{1}{2}a(t_3)^2\\\text{Where}~t_3=t_1\\y_3=v_1t_1-\frac{1}{2}a(t_1)^2\\\text{Where}~ v_1t_1=a(t_1)^2\\y_3=a(t_1)^2-\frac{1}{2}a(t_1)^2\\\text{Subtracting both terms:}\\y_3=\frac{1}{2}a(t_1)^2[/tex]

Total height = y₁ + y₂ + y₃

Total height = [tex]\frac{1}{2}a(t_1)^2+4a(t_1)^2+\frac{1}{2}a(t_1)^2 = 5a(t_1)^2[/tex]

Find acceleration by rearranging the found equation of total height.

Total Height = h

h = 5a(t₁)²

[tex]a=\frac{h}{5(t_1)^2}[/tex]

Answer:

P = 5.97 × 10^(5) Pa

Explanation:

We are given;

Mass of balloon;m_b = 4.3 kg

Radius;r = 1.54 m

Temperature;T = 289 K

Density;ρ = 1.19 kg/m³

We know that, density = mass/volume

So, mass = Volume x Density

We also know that Force = mg

Thus;

F = mg = Vρg

Where m = mass of balloon(m_b) + mass of helium (m_he)

So,

(m_b + m_he)g = Vρg

g will cancel out to give;

(m_b + m_he) = Vρ - - - eq1

Since a sphere shaped balloon, Volume(V) = (4/3)πr³

V = (4/3)π(1.54)³

V = 15.3 m³

Plugging relevant values into equation 1,we have;

(3 + m_he) = 15.3 × 1.19

m_he = 18.207 - 3

m_he = 15.207 kg = 15207 g

Molecular weight of helium gas is 4 g/mol

Thus, Number of moles of helium gas is ; no. of moles = 15207/4 ≈ 3802 moles

From ideal gas equation, we know that;

P = nRT/V

Where,

P is absolute pressure

n is number of moles

R is the gas constant and has a value lf 8.314 J/mol.k

T is temperature

V is volume

Plugging in the relevant values, we have;

P = (3802 × 8.314 × 289)/15.3

P = 597074.53 Pa

P = 5.97 × 10^(5) Pa