If a 1kg English swallow is carrying a 1kg coconut and flies northward from rest to a final velocity of 10m/s in 7.5 seconds, what is its acceleration? A. What force was exerted by the bird during that acceleration? B. How far did the bird travel during that time?

Answer:

Force, [tex]F=8.71\times 10^{-4}\ N[/tex]

Explanation:

Given that,

Charges on pith balls, [tex]q_1=q_2=-28\ nC=-28\times 10^{-9}\ C[/tex]

Distance between balls, d = 9 cm = 0.09 m

Let F is the repulsive force between two pith balls. We know that the repulsive force between two charges is given by :

[tex]F=k\dfrac{q_1^2}{d^2}[/tex]

[tex]F=9\times 10^9\times \dfrac{(-28\times 10^{-9})^2}{(0.09)^2}[/tex]

F = 0.000871 N

or

[tex]F=8.71\times 10^{-4}\ N[/tex]

So, the repulsive force between the pith balls is [tex]8.71\times 10^{-4}\ N[/tex]. Hence, this is the required solution.

Answer: Option (b) is correct.

Explanation:

Since we know that,

P = VI

where;

P = power

V= Voltage

I = Current

Since it's given that,

P = 600W

I = 2.5 A

equating these values in the above equation, we get;

V = [tex]\frac{600}{2.5}[/tex]

V = 240 V

Answer:

The sled required 9.96 s to travel down the slope.

Explanation:

Please, see the figure for a description of the problem. In red are the x and y-components of the gravity force (Fg). Since the y-component of Fg (Fgy) is of equal magnitude as Fn but in the opposite direction, both forces get canceled.

Then, the forces that cause the acceleration of the sled are the force of the wind (Fw), the friction force (Ff) and the x-component of the gravity force (Fgx).

The sum of all these forces make the sled move. Finding the resulting force will allow us to find the acceleration of the sled and, with it, we can find the time the sled travel.

The magnitude of the friction force is calculated as follows:

Ff = μ · Fn

where :

μ = coefficient of kinetic friction

Fn =  normal force

The normal force has the same magnitude as the y-component of the gravity force:

Fgy = Fg · cos 30º = m · g · cos 30º

Where

m = mass

g = acceleration due to gravity

Then:

Fgy = m · g · cos 30º = 87.7 kg · 9.8 m/s² · cos 30º

Fgy = 744 N

Then, the magnitude of Fn is also 744 N and the friction force will be:

Ff = μ · Fn = 0.151 · 744 N = 112 N

The x-component of Fg, Fgx, is calculated as follows:

Fgx = Fg · sin 30º = m·g · sin 30º = 87.7 kg · 9.8 m/s² · sin 30º = 430 N

The resulting force, Fr, will be the sum of all these forces:

Fw + Fgx - Ff = Fr

(Notice that forces are vectors and the direction of the friction force is opposite to the other forces, then, it has to be of opposite sign).

Fr = 161 N + 430 N - 112 N = 479 N

With this resulting force, we can calculate the acceleration of the sled:

F = m·a

where:

F = force

m = mass of the object

a = acceleration

Then:

F/m = a

a = 479N/87.7 kg = 5.46 m/s²

The equation for the position of an accelerated object moving in a straight line is as follows:

x = x0 + v0 · t + 1/2 · a · t²

where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

Since the sled starts from rest and the origin of the reference system is located where the sled starts sliding, x0 and v0 = 0.

x = 1/2· a ·t²

Let´s find the time at which the position of the sled is 271 m:

271 m = 1/2 · 5.46 m/s² · t²

2 · 271 m / 5.46 m/s² = t²

t = 9.96 s

The sled required almost 10 s to travel down the slope.

Answer:

i)20369 photons

ii) 40 ps

Explanation:

Momentum of one Sodium atom:

[tex]P=m*v =600m/s*23amu*\frac{1 kg}{6.02*10^{23}amu}\\P=2.29*10^{-23}kgm/s[/tex]

In other to stop it, it must absorb the same momentum in photons:

[tex]P=2.29*10^{-23}kgm/s=n_{photons}*\frac{h_{planck}}{\lambda}\\=n*\frac{6.63*10^{-34}}{589*10^{-9}} \\==>n=20369 photons[/tex]

Now, for the minimun time, we use the speed of light and the wavelength. For the n photons:

[tex]t=n*T=n*\frac{\lambda}{c} =20369*\frac{589nm}{3*10^{8}m/s}=4*10^{-11} second=40 ps[/tex]

Answer:

Force between sun and earth will decrease by [tex]2.51^2=6.3[/tex] times

Explanation:

According to law of gravitation we know that force between two object is given by

[tex]F=\frac{GM_1M_2}{R^2}[/tex], here G is universal gravitational constant [tex]M_1[/tex] is mass of one object and [tex]M_2[/tex] is mass of other object , and R is the distance between them

From the relation we can see that force is inversely proportional to square of distance between them

So force between sun and earth will decrease by [tex]2.51^2=6.3[/tex] times

Final answer:

The strength of the gravitational force between the Earth and the Sun would decrease by a factor of approximately 6.30 if the distance between them were increased by a factor of 2.51, according to the inverse square law of universal gravitation.

Explanation:

If the distance between the Earth and the Sun were increased by a factor of 2.51, the strength of the force between them would change according to Newton's law of universal gravitation. This law states that the force is inversely proportional to the square of the distance between two masses. Therefore, if the distance increases by a factor of 2.51, the force of attraction would decrease by a factor of (2.51)2.

To calculate the specific factor by which the force decreases, we square 2.51 which gives us 6.3001. The gravitational force would thus be reduced by this factor. Since the law involves an inverse square relationship, as the distance increases by a factor of 2.51, the strength of the gravitational force decreases by a factor of approximately 6.30.

Answer:

The answer is [tex] 3.994 \times 10^{13}\ electrons[/tex]

Explanation:

The amount of negative charge that must be removed is

[tex]\Delta = Final\ charge - initial\ Charge = 2.6 - (-3.8) = 6.4 \ \mu C = 6.4 \times 10^{-6}\ C[/tex]

and the charge of one electron is

[tex]1 e = 1.60217662\times 10^{-19} \ C[/tex]

So the amount of electrons we need to remove is

[tex]x = \frac{6.4 \times 10^{-6}}{1.60217662\times 10^{-19}} \approx 3.994 \times 10^{13}\ electrons[/tex]

If the force exerted on an arrow by a bow is tripled, the speed of the arrow leaving the bow would also triple, assuming other factors remain constant. Therefore, the arrow would leave the bow with a speed of 75 m/s.

If the arrow starting from rest leaves the bow with a speed of 25 m/s due to a certain force, tripling the force would result in a tripling of the acceleration, assuming mass remains constant. From Newton's second law, we understand that Force equals mass times acceleration (F = ma). The acceleration of an object is directly proportional to the force applied, if mass is kept constant. Therefore, if we triple the force, the acceleration would also triple. With a tripling of acceleration, the final speed of the arrow would also triple, as speed is the product of acceleration and time. Therefore, if the force exerted on the arrow by the bow is tripled, the speed with which the arrow leaves the bow would be 75 m/s.

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The speed at which the arrow would leave the bow when the average force exerted on it is tripled is 38.40 m/s.

To find the speed at which the arrow would leave the bow if the average force exerted on it were tripled, we can use the principle of conservation of mechanical energy. The initial kinetic energy of the arrow can be calculated using the formula KE = 1/2 * m * v^2, where KE is the kinetic energy, m is the mass of the arrow, and v is the initial velocity. Since the mass of the arrow remains the same, tripling the force would cause the initial kinetic energy to triple as well. Therefore, the final velocity can be found using the formula KE = 1/2 * m * v^2 and solving for v.

Let's use an example to illustrate this. Suppose the mass of the arrow is 0.5 kg. Initially, the arrow leaves the bow with a speed of 25 m/s. The initial kinetic energy is given by KE = 1/2 * 0.5 kg * (25 m/s)^2 = 156.25 J. If the average force is tripled, the new kinetic energy will be 3 times the initial kinetic energy, which is 3 * 156.25 J = 468.75 J. Plugging this value into the kinetic energy formula and solving for v, we get v = √(2 * 468.75 J / 0.5 kg) = 38.40 m/s.

Therefore, if the average force exerted on the arrow by the bow were tripled, the arrow would leave the bow with a speed of 38.40 m/s.

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Answer:

637 km/h, not reasonable

Explanation:

Total average speed = 91 km/h

Average speed of first half, v1 = 49 km/h

Let the average speed for the next half is v2.

Let the total distance is 2s.

Time taken for first half of the journey, t1 = s / v1 = s / 49

Time take for the next half of the journey, t2 = s / v2

The average speed is defined as the total distance traveled to the total time taken

[tex]91=\frac{2s}{\frac{s}{49}+\frac{s}{v_{2}}}[/tex]

[tex]\frac{s}{49}+\frac{s}{v_{2}}=\frac{2s}{91}[/tex]

[tex]\frac{1}{v_{2}}=\frac{2}{91}-\frac{1}{49}[/tex]

[tex]\frac{1}{v_{2}}=\frac{7}{4459}

v2 = 637 km/h

This is very high speed which is not reasonable.

Answer:

a) 2.41 km

b) 38.8°

Questions c and d are illegible.

Explanation:

We can express the displacements as vectors with origin on the point he started (0, 0).

When he traveled south he moved to (-3, 0).

When he moved east he moved to (-3, x)

The magnitude of the total displacement is found with Pythagoras theorem:

d^2 = dx^2 + dy^2

Rearranging:

dy^2 = d^2 - dx^2

[tex]dy = \sqrt{d^2 - dx^2}[/tex]

[tex]dy = \sqrt{3.85^2 - 3^2}  = 2.41 km[/tex]

The angle of the displacement vector is:

cos(a) = dx/d

a = arccos(dx/d)

a = arccos(3/3.85) = 38.8°

Final answer:

The density of the object is 2.40 g/cm³ with an absolute uncertainty of 0.01 g/cm³, which corresponds to a relative uncertainty of approximately 0.411%.

Explanation:

To calculate the density of an object, you divide the mass by the volume. Given that the mass is 53.5 ± 0.1 g and the volume is 22.30 ± 0.05 cm³, the density can be computed as follows:

Density = Mass/Volume = 53.5 g / 22.30 cm³ = 2.3991 g/cm³
However, when it comes to reporting this value, we need to match the significant figures to the least number in any of the measurements used, which in this case would be three significant figures. Thus, the density is reported as 2.40 g/cm³.

To calculate the uncertainty in density, you combine the relative uncertainties of mass and volume by simply adding them because we're dividing the two quantities. The relative uncertainty of mass is (0.1 g / 53.5 g) * 100% ≈ 0.187%, and the relative uncertainty of volume is (0.05 cm³ / 22.30 cm³) * 100% ≈ 0.224%. Adding these gives us the total uncertainty:

Total relative uncertainty = 0.187% + 0.224% ≈ 0.411%

To find the absolute uncertainty in density, you multiply the total relative uncertainty by the calculated density:

Absolute uncertainty = 0.411% * 2.3991 g/cm³ ≈ 0.00986 g/cm³

Rounded to match the significant figures of the calculated density, this would be 0.01 g/cm³. So, the density reported with its absolute uncertainty is 2.40 ± 0.01 g/cm³.

Answer:

The spot of light hits the bottom of the pool at 4.634 m from the wall beneath the watchman's feet.

Explanation:

Use the diagram attached below this answer to see the notation we will use.

For this case, we're trying to find x and we have:

h=1.3 m

b=2.1 m

a=2.7 m

We also know Snell's law for refraction:

[tex]n_{1} sin\theta_{1}=n_{2}sin\theta_{2}[/tex]

n is the refractive index of each substance (in this case, air and water), which are:

[tex]n_{air}=1[/tex]

[tex]n_{water}=1.33[/tex]

Triangle theory says that [tex]\theta_{1}=\beta[/tex] and:

[tex]tan\beta=\frac{a}{h}[/tex]

[tex]\beta=arc tan(\frac{a}{h})=arctan(\frac{2.7m}{1.3m})=64.29[/tex]

Using Snell's law:

[tex]\theta_{2}=arcsin(\frac{n_{1}sin\theta{1}}{n_{2}})=arcsin(\frac{1sin(64.29)}{1.33})=42.644[/tex]

Using triangle theory:

[tex]tan\theta_{2}=\frac{(x-a)}{b}[/tex]

[tex]x=b*tan\theta_{2}+a=2.1m*tan(42.644)+2.7m=4.634m[/tex]

Answer:

The distance of spot of light from his feet equals 3.425 meters.

Explanation:

The situation is represented in the attached figure below

The angle of incidence is computed as

[tex]\theta _i=tan^{-1}(\frac{1.3}{2.7})\\\\\therefore \theta _i=25.71^{o}[/tex]

Now by Snell's law we have

[tex]n_{i}sin(\theta _i)=n_{r}sin(\theta _r)[/tex]

where

[tex]n_{i},n_{r}[/tex] are the refractive indices of the incident and the refracting medium respectively

[tex]\theta _i,\theta _r[/tex] are the angle of incidence and the angle of refraction respectively

Thus using the Snell's relation we have

[tex]1.0\times sin(25.71)=1.33\times sin(\theta _r)\\\\\therefore sin(\theta _r)=\frac{sin(25.71}{1.33}=0.326\\\\\therefore \theta _r=sin^{-1}(0.326)=19.04^{o}[/tex]

from the attached figure we can see

[tex]tan(\theta _r)=\frac{L_{2}}{H}=\frac{L_{2}}{2.1}\\\\\therefore L_{2}=2.1\times tan(19.04)=0.725m[/tex]

Thus distance of spot on the pool bed from his feet equals [tex]2.7+0.725=3.425m[/tex]

Answer:

a) 0.2399 mi³

b) 440.8 × 10³ Pounds

Explanation:

Given:

Volume of cumulus cloud, V = 1 km³

Liquid water content = 0.2 g/m³

Now,

a) 1 km = [tex]\frac{\textup{1 miles}}{\textup{1.6093}}[/tex]

thus,

1 km³ = [tex](\frac{\textup{1 miles}}{\textup{1.6093}})^3[/tex]

1 km³ =  0.2399 mi³

Hence, volume of cloud in cubic miles is 0.2399 mi³

b)

Liquid water content = 0.2 g/m³

Now,

1 Km = 1000 m

thus,

1 km³ = 1000³ m³

Therefore,

Liquid water content in 1 Km³ of cloud = 0.2 g/m³ × 1000³ m³

= 200 × 10⁶ gram

or

= 200 × 10³ Kg

also,

1 kilogram =  2.204 pounds

Therefore,

200 × 10³ Kg = 200 × 10³ × 2.204 pounds = 440.8 × 10³ Pounds

The volume of a cumulus cloud occupying one cubic kilometer is equal to 0.386102 cubic miles. The water inside this cloud, with a liquid water content of 0.2 g/m³, weighs approximately 441 pounds.

The subject of this problem is to find the volume of a cloud in a different unit of measure and the weight of the water within it. Volume conversion from cubic kilometers to cubic miles and weight calculation of water in pounds are required.

To find the volume of the cloud in cubic miles, we use the conversion factor that 1 cubic kilometer is equal to about 0.386102 cubic miles. Hence, the volume of 1 cubic kilometer in cubic miles is:

1 cubic kilometer * 0.386102 cubic miles/cubic kilometer = 0.386102 cubic miles.

Next, we calculate the weight of the water in pounds, considering that the liquid water content is 0.2 grams per cubic meter and there are 1,000,000 cubic meters in a cubic kilometer:

0.2 g/m³ * 1,000,000 m³/km³ = 200,000 grams of water in the cloud.

Now, convert grams to pounds using the conversion factor 453.59237 grams per pound:

200,000 grams * (1 pound / 453.59237 grams) = 441 pounds.

Therefore, the water weight in the cloud is 441 pounds.

Answer:d

Explanation:

Given

Magnitude of Vector A is 3 m

and Magnitude of vector B is 4 m

So the maximum value of resultant can be 7 m when both are at an angle of [tex]0^{\circ}[/tex]

and its minimum value can be 1 m when both are at angle of [tex]180 ^{\circ}[/tex]

So the resultant must lie between 1 m to 7 m  

The length of the sum of the vectors A and B with no direction given must be some value from 1 m to 7 m. Thus, the correct option is D.

Vector addition is the property or operation of addition of two or more vectors together into a vector sum. The sign used with the vectors depends upon the direction in which they move. If the vectors are moving in the same direction then they will be added and if they are moving in opposite directions then they are substracted.

Given, Magnitude of Vector A is 3 m and Magnitude of vector B is 4 m.

So, in this case the maximum value of resultant vector can be 7 m when both are moving in the same direction with an angle of 0° and it has minimum value which can be 1 m when both are at angle of 180° in opposite direction.

So, the resultant vector must lie between 1 m to 7 m.  

Therefore, the correct option is D.

Learn more about Vector sum here:

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The police car takes approximately 30 seconds to reach the speeder, requires an acceleration of about 1.67 m/s², and its velocity at the moment it reaches the speeder is approximately 50 m/s.

This problem is a two-body pursuit scenario. The speeder's motion can be described by x = Ut, while the police car's motion is represented by the equation x = 1/2at². The speeder is moving at a constant speed of 120 km/h which is equal to 33.33 m/s.

(a) Time for the police car to overtake the speeder:
To find the time, we equate the two equations (since they both cover the same distance of 750m) and solve for 't'. Thus, 750m = 33.33m/s * t = 1/2 * a * t². By solving this equation, we get two values of 't', out of which the realistic answer is t = 30 seconds.

(b) Required acceleration of the police car:
Plugging the time into the equation for the police car, we get 750m = 1/2 * a * (30s)². Solving for 'a', we find the required acceleration to be approximately 1.67 m/s².

(c) Velocity of the police car at the moment it reaches the speeder:
Since the police car starts from rest and maintains a constant acceleration, the final velocity can be calculated using the equation v = at. Hence, at the moment it reaches the speeder, the police car's velocity would be approximately 50 m/s.

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Answer:

[tex]q=1.18*10^{-6}C}[/tex]

Explanation:

The potential V due to a charge q,  at a distance r, is:

[tex]V=k\frac{q}{r}[/tex]

k=8.99×109 N·m^2/C^2      :Coulomb constant

We solve to find q:

[tex]q=\frac{V*r}{k}=\frac{6.25*10^{2}*17}{8.99*10^{9}}=1.18*10^{-6}C[/tex]

Answer:

Explanation:

The pressure of the gas can be found out as follows

Gas law formula is as follows

PV = nRT

P = nRT / V

= 32.4 X 8.31 X ( 273 + 459.38 ) / 51 X 10⁻⁶

3866.45 X 10⁶ Pa.

The first change is isobaric therefore

V₁ / T₁ = V₂ / T₂

V₂ = V₁ X T₂/ T₁

= 51 X 10⁻⁶ X ( 855.6 +273) / (459.38 +273)

= 78 X 10⁻⁶

78 cm³

After the first operation , the pressure of the gas remains at

= 3866.45 X 10⁶ Pa.

Now volume of the gas changes from 78 cm³ to 25.3 cm³ isothermally so

P₁V₁ = P₂V₂

P₂ = P₁V₁ / V₂

= 3866.45 X 10⁶ X 78 / 25.3

= 11920 X 10⁶ . Pa

Answer:

building height is 52.52 m

Explanation:

given data

initial speed v = 13 m/s

angle = 35°

time = 2.6 s

to find out

how tall is building

solution

we consider here h is height of building

so

initial velocity at angle 35 is express as

u = v sin(θ)     ...........1

u = 13 sin35 = 7.45 m/s

so

by distance formula

h = ut + 1/2 at²        ...........2

h = 7.45 ( 2.6) + 1/2 × (9.81) × (2.6)²

h = 52.52 m

so building height is 52.52 m

Final answer:

Using the formula for vertical displacement in projectile motion, with an initial velocity of 13 m/s downward and the ball being in the air for 2.6 seconds, the building's height is found to be approximately 52.5 meters.

Explanation:

To solve for the height of the building in the given scenario, we can use the vertical component of the projectile motion. We know that the child throws the tennis ball with an initial speed of 13 m/s at an angle of 35° below the horizontal, and it lands after 2.6 seconds. We can use the formula for the vertical motion under constant acceleration (gravity in this case) to find the height:

Equation for vertical displacement:

s = ut + ½at2

s is the vertical displacement (height of the building), u is the initial vertical velocity, a is the acceleration due to gravity (9.8 m/s2), and t is the time the ball is in the air.

First, we calculate the initial vertical velocity component (u):

u = vinitial × sin(θ)

u = 13 m/s × sin(35°) = 13 m/s × 0.5736 ≈ 7.457 m/s (downward)

Since the initial velocity is downward and we need upward to be positive, we will treat it as negative:

u = -7.457 m/s

Next, we can calculate the height of the building.

s = (-7.457 m/s)(2.6 s) + ½(-9.8 m/s2)(2.6 s)2

s = -19.3886 m - 33.118 m

s = -52.5066 m

Since the displacement is in the negative direction (downward), the height of the building is 52.5 meters (we must take the absolute value of the displacement to get the height).

Answer:

The distance covered by comet is [tex]200,000 km[/tex]

Explanation:

Speed is defined as the rate of change of distance with time. It is given by the equation speed= [tex]\bold{\frac{distance}{time}}[/tex]

Thus distance= [tex]speed*time[/tex]

In this problem it is given that speed of comet= [tex]\frac{50,000km}{hr}[/tex]

time travelled by the comet= 4 hours

Thus distance= [tex]speed*time[/tex]

                            = [tex]500000*4[/tex]

                            = [tex]\bold{200,000km}[/tex]

Answer:

[tex]1.189eV[/tex]

Explanation:

The electric potential energy is the potential energy that results from the Coulomb force and is associated with the configuration of two or more charges. For an electron in the presence of an electric field produced by a proton, the electric potential energy is defined as:

[tex]U=\frac{kq_{e}q_{p}}{r}[/tex]

where  [tex]q_{e}[/tex] is the electron charge, [tex]q_{p}[/tex] is the proton charge, r is the separation distance between the charges and k is the coulomb constant.

Knowing this, we can calculate how much electric potential energy was lost:

[tex]\Delta U=U_{f}-U{i}\\\Delta U=\frac{kq_{e}q_{p}}{r_{f}}-\frac{kq_{e}q_{p}}{r_{i}}\\\Delta U=kq_{e}q_{p}(\frac{1}{r_{f}}-\frac{1}{r_{i}})\\\Delta U=(8.99*10^9\frac{Nm^2}{C^2})(-1.60*10^{-19}C)(1.60*10^{-19}C)(\frac{1}{0.105*10^{-9}m}-\frac{1}{0.115*10^{-9}m})\\\Delta U=1.90*10^{-19}J*\frac{6.2415*10^{18}eV}{1J}=1.189eV[/tex]

The change in electric potential energy between an electron and a proton with changing separation can be calculated using the Coulomb potential energy equation and can be expressed in electron volts (eV).

The question involves calculating the change in electric potential energy of a system consisting of an electron and a proton as they change separation distance during a reaction. The energy of the system is described by the Coulomb electrostatic potential energy equation Epot = -e2 / (4πε₀r), where e is the charge of the electron, ε₀ is the vacuum permittivity, and r is the separation distance between the electron and proton.

To find the amount of potential energy lost, we calculate the potential energy at the initial separation (0.115 nm) and the final separation (0.105 nm), then take the difference between these values. The energy will be expressed in electron volts (eV), a common unit used to describe energy at subatomic scales.

Answer:

t = 20.96 seconds

Explanation:

given,

lightning strikes at = 4.5 mi( 23,760 ft)

air temperature =74.0°F

74.0°F = 23.3 °C

speed of the sound at 23.3 °C

V = 331.5 + 0.6 × T

V = 331.5 + 0.6 × 23.3

   = 345.48 m/s

distance given = 4.5 mile

1 mile = 1609.4                

4.5 mile = 4.5 × 1609.4 = 7242.3 m            

time taken =

[tex]t =\dfrac{d}{v}[/tex]

[tex]t =\dfrac{7242.3}{345.48}[/tex]

t = 20.96 seconds

hence, time taken by the sound to reach by the observer is 20.96 sec.

Final answer:

To estimate the time elapsed between seeing lightning and hearing thunder, one must know the speed of sound, which varies with temperature, and the distance to the lightning strike. At a distance of 4.5 miles and an air temperature of 74.0°F, approximately 21 seconds will elapse.

Explanation:

Calculating the Time Difference Between Lightning and Thunder

The question involves the relationship between the speed of light and sound to determine how long it will take for the sound of thunder to reach an observer after the lightning is seen. Since light travels at approximately 300,000 kilometers per second (3 × 108 meters per second), lightning is seen almost instantaneously. However, the sound of thunder, which is caused by the rapid expansion of superheated air due to a lightning strike, travels much slower.

At a temperature of 74.0°F, we need to convert this to Celsius since the speed of sound in air is given by the formula v = 331.4 m/s + 0.6T, where T is in Celsius. The temperature in Celsius can be found through T(°C) = (T(°F) - 32) × 5/9, giving us T(°C) = (74 - 32) × 5/9 = 23.33°C.

The speed of sound at this temperature is v = 331.4 m/s + 0.6 × 23.33 = 345.4 m/s. To find the time it takes for the sound to travel 4.5 miles (or 23,760 feet), we need to convert the distance to meters since the speed of sound is given in meters per second. We have 1 mile = 1,609.34 meters, so 4.5 miles = 7,242 meters approximately. Finally, we calculate the time as time = distance/speed = 7,242 m / 345.4 m/s ≈ 20.97 s.

Therefore, approximately 21 seconds will elapse between seeing the lightning and hearing the thunder if the lightning strikes 4.5 miles away and the air temperature is 74.0°F.

Answer:

V=120m/s

Explanation:

A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.

Vf=Vo+a.t                                     (1)

{Vf^{2}-Vo^2}/{2.a} =X                (2)

X=Xo+ VoT+0.5at^{2}                   (3)

X=(Vf+Vo)T/2                                 (4)

Where

Vf = final speed

Vo = Initial speed

T = time

A = acceleration

X = displacement

In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve

for this problem you must divide the problem into two parts 1 and 2, when the rocket accelerates (1), and when the rocket decelerates (2).

Then you raise equations 3 and 1 in both parts.

finally you use algebraic methods to find the value of speed

I attach the complete procedure

Answer:

As from the given question,

The free body diagram of a ladder leans against a non vertical wall with friction that make a angle of 50 degree with ground is show below.

Friction force can be calculated by formula:

Friction force = μ N

As we know that

At equilibrium,

The sum of the total horizontal force is equal to '0'.

The sum of the total vertical force is equal to '0'.

Answer:

1. Running velocity (5 m/s)

2. Resting velocity (0 m/s)

3. Walking velocity (-1 m/s)

1. Running speed (5 m/s)

2. Walking speed (1 m/s)

3. Resting speed (0 m/s)

Explanation:

Attached you will find the plot of position vs time of Ellie´s movement.

The velocity is the displacement of the object over time relative to the system of reference. The speed, in change, is the traveled distance over time in disregard of the system of reference.

So, the velocity is calculated as follows:

v = Δx / Δt

where

Δx = final position - initial position

Δt = elapsed time

1) The average velocity of Ellie while running is:

v = 1000 m - 0 m / 200 s = 5 m/s

While resting:

v = 0 m - 0 m / 100 s = 0 m/s

And while walking back:

v = 0 m - 1000 m / 1000 s = - 1 m/s

Note that in this last case, the initial position is 1000 m because Ellie is 1000 m from the origin of the system of reference when she walks back. The final position will be the origin of the system of reference, 0 m.

Comparing with the graphic, the velocity is the slope of the function position(t).

Then:

1. Running velocity (5 m/s)

2. Resting velocity (0 m/s)

3. Walking velocity (-1 m/s)

2) The speed is the distance traveled over time:

Running speed = 1000 m / 200 s = 5m /s

Resting speed = 0 m / 100 s = 0 m/s

Walking speed = 1000 m/ 1000 s = 1 m/s

Then:

1. Running speed (5 m/s)

2. Walking speed (1 m/s)

3. Resting speed (0 m/s)  

Answer:

a) v = 15.6 m/s

b) 0.65 s are needed to reach a height of 12.3 m

Explanation:

The equations that describe the height and velocity of the stone are the following:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where

y = height of the stone at time t

y0 = initial height

v0 = initial speed

t = time

g = acceleration due to gravity

b) First, let´s find the time at which the stone reaches a height of 12.3 m:

y = y0 + v0 · t + 1/2 · g · t²

12.3 m = 0 m + 22.0 m/s · t + 1/2 · (-9.8 m/s²) · t²    (y0 = 0 placing the center of the frame of reference at the point at which the stone is thrown.)

-4.9 m/s² · t² + 22.0 m/s · t - 12.3 m = 0

t = 0.65 s (when the stone goes upward) and t = 3.84 s ( when the stone returns downward) .

So, 0.65 s are needed to reach a height of 12.3 m

a) The velocity at that time will be:

v = v0 + g · t

v = 22.0 m/s - 9.8 m/s² · 0.65 s = 15.6 m/s

The stone moves at approximately 15.62 m/s when it reaches a height of 12.3 meters. The time required to reach this height is approximately 0.652 seconds.

Initial velocity (u) = 22.0 m/s

Acceleration (a) = -9.8 m/s² (due to gravity)

Height (h) = 12.3 m

We can use the equation:

v² = u² + 2a(s), where v is the final velocity, u is the initial velocity, a is acceleration, and s is the displacement.

Substituting the given values:

v² = (22.0 m/s)² + 2(-9.8 m/s²)(12.3 m)

v² = 484 - 240.12

v² = 243.88

v = √243.88 ≈ 15.62 m/s

Thus, the stone is moving at approximately 15.62 m/s when it reaches a height of 12.3 m.

We can use the equation:

s = ut + 1/2 at², where s is the displacement, u is the initial velocity, t is time, and a is acceleration.

12.3 m = (22.0 m/s)t + 1/2(-9.8 m/s²)t²

12.3 = 22.0t - 4.9t²

Rearrange to form a quadratic equation:

-4.9t² + 22.0t - 12.3 = 0

Using the quadratic formula:

t = [ -b ± √(b² - 4ac) ] / 2a

t = [ 22.0 ± √(484 - 4(-4.9)(-12.3)) ] / 2(-4.9)

t = ( 22.0 ± √(484 - 240.12) ) / 9.8

t = ( 22.0 ± √243.88 ) / 9.8

t = 22.0 ± 15.62 / 9.8

We get two possible values for t:

t₁ = (22.0 - 15.62) / 9.8 ≈ 0.652 s

t₂ = (22.0 + 15.62) / 9.8 ≈ 3.83 s

The correct time to reach 12.3 m as the stone ascends is approximately 0.652 seconds.

Answer:

[tex]W=2eV[/tex]

Explanation:

Energy from a light source (photons):

E=h*c/λ

Photoelectric effect and Work function (W):

E=W+Ek

Ek: maximum kinetic energy from photoelectrons

then:

h*c/λ=W+Ek

Case 1:

[tex]h*c/lambda_{1}=W+Ek_{1}[/tex]  (1)

Case 2:

[tex]h*c/lambda_{2}=W+Ek_{2}[/tex]

but  [tex]lambda_{2}=lambda_{1}/2[/tex]

[tex]2h*c/lambda_{1}=W+Ek_{2}[/tex]  (2)

If we divide (2) by (1):

[tex]2=\frac{W+Ek_{2}}{W+Ek_{1}}[/tex]

[tex]W=Ek_{2}-2Ek{1}=2eV[/tex]

Answer:1.1 eV

Explanation:

Given

Energy(E)=2.9 eV

Work function of the target=1.8 eV

We know that

[tex]Energy(E)=W_0+KE_{max}[/tex]

Where [tex]W_0=work\ function[/tex]

[tex]2.9=1.8+KE_{max}[/tex]

[tex]KE_{max}=2.9-1.8=1.1 eV[/tex]

Answer:

1) 1001

=1 × 2³ + 0 × 2² + 0 × 2¹ + 1 × 2⁰

= 8 + 1 = (9)₁₀

2) 10101

=1 × 2⁴ + 0 × 2³ + 1 × 2² + 0 × 2¹ + 1 × 2⁰

= 16 + 4 + 1 = (21)₁₀

3) 1010001

=1 × 2⁶ + 0 × 2⁵ + 1 × 2⁴ + 0 × 2³ + 0 × 2² + 0 × 2¹ + 1 × 2⁰

= 64 + 16 + 1 = (81)₁₀

4) 1010.1010

= 1 × 2³ + 0 × 2² + 1 × 2¹ + 0 × 2⁰ + 1 × 2⁻¹ + 0 × 2⁻² + 1 × 2⁻³ + 0 × 2⁻⁴

= 8 + 2 + 0.5 + 0.125 = (10.625)₁₀

Answer:

1) 1001

=1 × 2³ + 0 × 2² + 0 × 2¹ + 1 × 2⁰

= 8 + 1 = (9)₁₀

2) 10101

=1 × 2⁴ + 0 × 2³ + 1 × 2² + 0 × 2¹ + 1 × 2⁰

= 16 + 4 + 1 = (21)₁₀

3) 1010001

=1 × 2⁶ + 0 × 2⁵ + 1 × 2⁴ + 0 × 2³ + 0 × 2² + 0 × 2¹ + 1 × 2⁰

= 64 + 16 + 1 = (81)₁₀

4) 1010.1010

= 1 × 2³ + 0 × 2² + 1 × 2¹ + 0 × 2⁰ + 1 × 2⁻¹ + 0 × 2⁻² + 1 × 2⁻³ + 0 × 2⁻⁴

= 8 + 2 + 0.5 + 0.125 = (10.625)₁₀

Explanation:

Answer:

Explanation:

The spacetime interval [tex]\Delta s^2[/tex] is given by

[tex]\Delta s ^2 = \Delta (c t) ^ 2 - \Delta \vec{x}^2[/tex]

please, be aware this is the definition for the signature ( + - - - ), for the signature (- + + + ) the spacetime interval is given by:

[tex]\Delta s ^2 = - \Delta (c t) ^ 2 + \Delta \vec{x}^2[/tex].

Lets work with the signature ( + - - - ), and, if needed in the other signature, we can multiply our interval by -1.

[tex]\Delta \vec{x}^2 = (7.5 \ 10 \ m)^2[/tex]

[tex]\Delta \vec{x}^2 = 5,625 m^2[/tex]

[tex]\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 3 \ s)^2[/tex]

[tex]\Delta (c t) ^ 2 = (899,377,374 \ m)^2[/tex]

[tex]\Delta (c t) ^ 2 = 8.0888 \ 10^{17} m^2[/tex]

so

[tex]\Delta s ^2 = 8.0888 \ 10^{17} m^2 - 5,625 m^2[/tex]

[tex]\Delta s ^2 = 8.0888 \ 10^{17} m^2[/tex]

[tex]\Delta \vec{x}^2 = (5 \ 10 \ m)^2[/tex]

[tex]\Delta \vec{x}^2 = 2,500 m^2[/tex]

[tex]\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 0.58 \ s)^2[/tex]

[tex]\Delta (c t) ^ 2 = (173,879,625.6 \ m)^2[/tex]

[tex]\Delta (c t) ^ 2 = 3.0234 \ 10^{16} m^2[/tex]

so

[tex]\Delta s ^2 = 3.0234 \ 10^{16} m^2 - 2,500 m^2[/tex]

[tex]\Delta s ^2 = 3.0234 \ 10^{16} m^2[/tex]

[tex]\Delta \vec{x}^2 = (5 \ 10 \ m)^2[/tex]

[tex]\Delta \vec{x}^2 = 2,500 m^2[/tex]

[tex]\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 58 \ s)^2[/tex]

[tex]\Delta (c t) ^ 2 = (1.73879 \ 10^{10} \ m)^2[/tex]

[tex]\Delta (c t) ^ 2 = 3.0234 \ 10^{20} m^2[/tex]

so

[tex]\Delta s ^2 = 3.0234 \ 10^{20} m^2 - 2,500 m^2[/tex]

[tex]\Delta s ^2 = 3.0234 \ 10^{20} m^2[/tex]

Answer:

Ax  =7 cm

Explanation:

Analysis:

Let A be the vector of rectangular components Ax and Ay:

Where:

Ay=7 cm  : The The y-component of  vector A-component of  vector A

α =45       : Angle of A with the positive x-axis

Ax           : The x-component of  vector A

Because Ax and Ay are the components of a right triangle we apply the following formula:

[tex]tan\alpha =\frac{A_{y} }{A_{x} }[/tex]

[tex]tan45=\frac{7}{A_{x} }[/tex]

[tex]A_{x} =\frac{7}{tan45}[/tex]

Ax  =7 cm

Answer:

Explanation:

At the topmost position,  the car does not have zero velocity but it has velocity of v so that

v² /r = g or centripetal acceleration should be equal to g ( 9.8 )

Considering that,  the car must fall from a height of 2r + h where

mgh = 1/2 mv²

= 1/2 m gr

So h = r/2

Hence the ball must fall from a height of

2r + r /2

= 2.5 r . So that it can provide velocity of v  at the top where

v² / r = g .