Assume that a projectile is fired horizontally from a "gun" located 1 meter above the ground. If the ball strikes the ground a distance of 2 meters from the end of the "gun" determine the muzzle velocity of the gun.

Answer:

5.05 m

Explanation:

Given:

According to the work-energy theorem, work done by the gravitational force will be equal to the kinetic energy change in the vaulter.

[tex]\therefore -mg(h)=\dfrac{1}{2}m(v^2-u^2)\\\Rightarrow -gh=\dfrac{1}{2}(v^2-u^2)\\\Rightarrow h=\dfrac{1}{-2g}(v^2-u^2)\\\Rightarrow h=\dfrac{1}{-2\times 9.8}((1)^2-(10)^2)\\\Rightarrow h=\dfrac{-99}{-19.6}\\\Rightarrow h=5.05[/tex]

Hence, the height of the vaulter is 5.05 m.

Final answer:

Maria Montessori is not typically associated with the creative curriculum model. Howard Gardner's theory does not include 'creative' as a standalone type of intelligence, although creativity can be an aspect of multiple intelligences.

Explanation:

The creative curriculum model is influenced by various educational researchers, but the one who is not typically associated with this model is A. Maria Montessori. The model is guided by the principles of Jean Piaget, Howard Gardner, and Erik Erikson, all of whom focused on developmental and educational psychology in different ways. For example, Jean Piaget is known for his theory of cognitive development, Erik Erikson for his theory of psychosocial development, and Howard Gardner for his theory of multiple intelligences.

Speaking of Howard Gardner, the correct response to the second question is A. creative. The types of intelligences Gardner identified in his theory are linguistic, logical-mathematical, musical, spatial, bodily-kinesthetic, interpersonal, intrapersonal, and naturalistic. While creativity is an aspect that can be present in multiple intelligences, Gardner did not identify 'creative' as a standalone type of intelligence in his original theory.

Answer:

The speed is 173 m/s.

Explanation:

Given that,

A = 47

B = 14

Length 1 urk = 58.0 m

An hour is divided into 125 time units named dorts.

3600 s = 125 dots

dorts = 28.8 s

Speed v= (25.0+A+B) urks/dort

We need to convert the speed into meters per second

Put the value of A and B into the speed

[tex]v=25.0+47+14[/tex]

[tex]v =86\ urk s/dort[/tex]

[tex]v=86\times\dfrac{58.0}{28.8}[/tex]

[tex]v=173.19\ m/s[/tex]

Hence, The speed is 173 m/s.

Final answer:

To convert the speed (25.0 + A + B) urks/dort to meters per second, add A and B to 25.0, convert urks/dort to urks per second, then to meters per second. Using sample values A=47, B=14, the final speed is approximately 1.582 m/s.

Explanation:

To convert a speed of (25.0 + A + B) urks/dort into meters per second (m/s), one must first understand the unit conversions involved: 1 urk equals 58.0 meters, and there are 125 dorts in an hour. Let's assume, for instance, A = 47 (the last two digits of the student ID) and B = 14 (the sum of the last three digits). Therefore, the speed in urks/dort is (25.0 + 47 + 14) urks/dort.

Performing the addition, we get 86 urks/dort. To convert this into meters per second, follow these steps:

Calculating this gives a speed of approximately 1.582 m/s, rounded to three significant figures.

Answer:

Normal Force is usually perpendicular to the movement and static friction usually means that there is no movement.

Explanation:

The work donde by any force on an object is equal to the displacement of the object multiplied by the component of the force that is in the direction of the displacement.

Normal force is usually perpendicular to the movement, so there is no component in the direction of the displacement. This is why it is zero in most circumstances.

Static friction on the other hand, usually means that there is no movement at all (it's static). It means that there is no displacement between the object and ground (in most cases). If there is no displacement, there is no work.

The normal force and the force of static friction do no work on an object for different reasons. The normal force acts at right angles to the displacement of the object, while the force of static friction only acts when there is no relative motion between the object and the surface.

The normal force and the force of static friction do no work on an object in most circumstances, but for different reasons.

The normal force is a force exerted by a surface that is perpendicular to the surface. It does no work because it acts at right angles to the displacement of the object.

The force of static friction is a force that opposes the motion of an object on a surface. It does no work because it only acts when there is no relative motion between the object and the surface.

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Answer:

1)The station is rotating at 2.11 revolutions per minute

2) The space station will have to rotate at as peed of 4.03 revolutions per minute, for the artificial gravity to equal that at the surface of earth (9.8 m/s²)

Explanation:

1)

[tex]a_{c}=[/tex]ω²r = ω²d/2

Here,

[tex]a_{c}=[/tex]2.7 m/s²

d = 110 m

therefore,

ω² = (2.7 m/s²)(2)/(110 m)

ω = √0.049 rad/s²

ω = 0.2216 rad/s

To convert into rev/min

ω = (0.2216 rad/s)(1 rev/2π rad)(60 s/ 1min)

ω = 2.11 rev/min

2)

Here,

[tex]a_{c}=g=[/tex]9.8 m/s²

d = 110 m

therefore,

ω² = (9.8 m/s²)(2)/(110 m)

ω = √0.178 rad/s²

ω = 0.4221 rad/s

To convert into rev/min

ω = (0.4221 rad/s)(1 rev/2π rad)(60 s/ 1min)

ω = 4.03 rev/min

This involves an object revolving around a center or axis.

We can use this formula for question 1

ω²r = ω²d/2

ω² = (2.7 m/s²)(2)/(110 m)

ω = √0.049 rad/s²

ω = 0.2216 rad/s

We then convert to rev/min

ω = (0.2216 rad/s)(1 rev/2π rad)(60 s/1min)

ω = 2.11 rev/min

ω²r = ω²d/2

ω² = (9.8 m/s²)(2)/(110 m)

ω = √0.178 rad/s²

ω = 0.4221 rad/s

We then convert into rev/min

ω = (0.4221 rad/s)(1 rev/2π rad)(60 s/ 1min)

ω = 4.03 rev/min.

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Answer:

a) During the reaction time, the car travels 21 m

b) After applying the brake, the car travels 48 m before coming to stop

Explanation:

The equation for the position of a straight movement with variable speed is as follows:

x = x0 + v0 t + 1/2 a t²

where

x: position at time t

v0: initial speed

a: acceleration

t: time

When the speed is constant (as before applying the brake), the equation would be:

x = x0 + v t

a)Before applying the brake, the car travels at constant speed. In 0.80 s the car will travel:

x = 0m + 26 m/s * 0.80 s = 21 m  

b) After applying the brake, the car has an acceleration of -7.0 m/s². Using the equation for velocity, we can calculate how much time it takes the car to stop (v = 0):

v = v0 + a* t

0 = 26 m/s + (-7.0 m/s²) * t

-26 m/s / - 7.0 m/s² = t

t = 3.7 s

With this time, we can calculate how far the car traveled during the deacceleration.

x = x0 +v0 t + 1/2 a t²

x = 0m + 26 m/s * 3.7 s - 1/2 * 7.0m/s² * (3.7 s)² = 48 m

a. The car travels 20.8 meters during the reaction time and 48.29 meters while decelerating, b. resulting in a total distance of 69.09 meters.

Part a: Distance Traveled During Reaction Time

The initial speed of the car is 26 m/s, and the driver has a reaction time of 0.80 seconds. During this reaction time, the car continues to move at the initial speed because the brakes haven't been applied yet.

Using the formula:

distance = speed × time

We get:

distance = 26 m/s × 0.80 s = 20.8 meters

Part b: Distance Traveled After Brakes Are Applied

The car decelerates at a rate of -7.0 m/s² until it comes to a stop. We need to find the distance traveled during this deceleration.

Using the kinematic equation:

v² = u² + 2as

where:

Rearranging the equation to solve for s:

0 = (26 m/s)² + 2(-7.0 m/s²) * s

Solving for s:

0 = 676 - 14s

14s = 676

s = 48.29 meters

The total distance the car travels includes both the distance during the reaction time and the distance while braking:

Total Distance = 20.8 meters + 48.29 meters = 69.09 meters

Answer:

(a) [tex]2.31\times10^{-8}\ N[/tex]

(b) [tex]1.44\times 10^{-19}\ eV[/tex]

Explanation:

Given:

*p = charge on proton = [tex]1.602\times 10^{-19}\ C[/tex]

*e = magnitude of charge on an electron = [tex]1.602\times 10^{-19}\ C[/tex]

*r = distance between the proton and the electron in the Rutherford's atom = [tex]1.0\times 10^{-10}\ m[/tex]

Part (a):

Since two unlike charges attract each other.

According to Coulomb's law:

[tex]F=\dfrac{kqe}{r^2}\\\Rightarrow F = \dfrac{9.0\times 10^{9}\times 1.602\times 10^{-19}\times 1.602\times 10^{-19}}{(1.0\times 10^{-10})^2}\\\Rightarrow F = 2.31\times 10^{-8}\ N[/tex]

Hence, the attractive electrostatic force of attraction acting between an electron and a proton of Rutherford's atom is  [tex]2.31\times 10^{-8}\ N[/tex].

Part (b):

Potential energy between two charges separated by a distance r is given by:

[tex]U= \dfrac{kqQ}{r}[/tex]

So, the potential energy between the electron and the proton of the Rutherford's atom is given by:

[tex]U = \dfrac{kqe}{r}\\\Rightarrow U = \dfrac{9\times 10^{9}\times1.602\times 10^{-19}\times e}{1.0\times 10^{-10}}\\\Rightarrow U = 1.44\times 10^{-19}\ eV[/tex]

Hence, the electrostatic potential energy of the atom is  [tex]1.44\times 10^{-19}\ eV[/tex].

The attractive electrostatic force at the atomic size in Rutherford's model is about 2.3 x 10^-9 N. The electrostatic potential energy at this distance is approximately 14.4 eV.

The question asks for the attractive electrostatic force between an electron and a proton at a distance of 1.0 x 10-10 m (Rutherford's atomic size) and the electrostatic potential energy in eV. We use Coulomb's Law to find the force and potential energy equations in physics.

(a) The attractive electrostatic force (F) is given by the equation: F = (k * e^2) / r^2, where k is Coulomb's constant (8.99 x 10^9 N m^2/C^2), e is the charge of an electron/proton (1.6 x 10^-19 C), and r is the distance (1.0 x 10^-10 m). Plugging in these values, we get F = (8.99 x 10^9 N m^2/C^2 * (1.6 x 10^-19 C)^2) / (1.0 x 10^-10 m)^2 ≈ 2.3 x 10^-9 N.

(b) The electrostatic potential energy (U) is given by the equation: U = (k * e^2) / r, which results in U = (8.99 x 10^9 J m/C^2 * (1.6 x 10^-19 C)^2) / (1.0 x 10^-10 m) ≈ 2.3 x 10^-18 J. Because 1 J = 6.242 x 10^18 eV, we convert this to eV to get approximately U ≈ 14.4 eV.

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Answer:

a) 27.2 rad/min

b) 260 rev/h

Explanation:

The passenger is traveling at 9 mph, this is the tangential speed.

The relation between tangential speed and angular speed is:

v = r * w

Where

v: tangential speed

r: radius

w: angular speed

Also, the radius is

r = d/2

d is the diameter

Therefore:

v = (d * w)/2

Rearranging:

w = 2*v/d

w = (2*9 mile/h)/(58 feet)

We need to convert the feet to miles

w = (2*9 mile/h)/(0.011 miles) = 1636 rad/h

We divide this by 60 to get it in radians per minute

w = 1636/60 = 27.2 rad/min

Now the angular speed is in radians, to get revolutions we have to divide by 2π

n = v/(π*d)

n = (9 mile/h)/(π*0.011 mile) = 260 rev/h

The angular speed of a Ferris wheel with a 58-foot diameter, while carrying a passenger traveling at a speed of 9 miles per hour, is approximately 27.31 radians per minute. This Ferris wheel makes approximately 259 revolutions per hour.

To solve this problem, we first need to convert the linear speed from miles per hour to feet per minute, as the unit of the Ferris wheel’s diameter is in feet. One mile is equivalent to 5280 feet, and one hour is 60 minutes. Therefore, the passenger's speed in feet per minute (ft/min) is 9 miles/hour x 5280 feet/mile ÷ 60 minutes/hour = 792 ft/min.

(a) The angular speed in radians per minute can be found by dividing the linear speed by the radius of the wheel (which is half of the diameter). So, the wheel’s radius is 58 feet ÷ 2 = 29 feet, and thus, the angular speed is 792 ft/min ÷ 29 feet = 27.31 rad/min.

(b) The number of revolutions per hour is found by dividing the linear speed by the circumference of the wheel (which is the diameter × π). Therefore, the wheel's circumference is 58 feet x π. Consequently, the number of revolutions per hour is 792 ft/min x 60 min/hour ÷ (58 feet x π) ≈ 259 revolutions per hour, when rounded to the nearest whole number.

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Answer:

shearing force is [tex]3.40\times 10^{-4} lb[/tex]

Explanation:

we know that

force can be determined [tex]F = \tau \times A[/tex]

Area can be determine as

[tex]A  = \frac{\pi}{4} d^2 = \frac{\pi}{4} [\frac{0.2}{12}]^2 =   2.18\times 10^{-4} ft^2[/tex]

linear velocity can be determines as

[tex]\tau = \mu_{air} \frac{U}{b}[/tex]

dynamic viscosity of air [tex]\mu_{air} = 3.74\times 10^{-7} lb-s/ft^2[/tex]

veolcity of disc

[tex]U =\omega R[/tex]

[tex]U = \frac{2\pi N}{60} \times R = \frac{2\pi 10,000}{60} \times \frac{2}{12}[/tex]

U = 174.5 ft/s

so

[tex]\tau = 3.74\times 10^{-7} \times \frac{174.5}{\frac{0.0005}{12}}[/tex]

[tex]\tau = 1.56 lb/ft^2[/tex]

[tex]F = 1.56\times 2.18\times 10^{-4} = 3.40\times 10^{-4} lb[/tex]

shearing force is [tex]3.40\times 10^{-4} lb[/tex]

Answer:

(a) A = [tex][LT^{- 2}][/tex]

B = [tex][LT^{- 3}][/tex]

[tex]C = [LT^{- 5}][/tex]

(b) A = [tex]ms^{- 2}[/tex]

B = [tex]ms^{- 3}[/tex]

C = [tex]ms^{- 5}[/tex]

Solution:

The acceleration of a body is the rate at which the velocity of the body changes.

Thus

[tex]a = \frac{\Delta v_{o}}{\Delta t}[/tex]

The SI unit of velocity of an object is [tex]ms^{- 1}[/tex] and its dimension is [LT^{- 1}] and for time, T the SI unit is second, s and dimension is [T] and hence

The SI unit and dimension for the acceleration of an object is [tex]ms^{- 2}[/tex] and [LT^{- 2}] respectively.

Now, as per the question:

acceleration, a = [tex]A + Bt + Ct^{3}[/tex]

(a) Now, according to the homogeneity principle in dimension, the dimensions on both the sides of the eqn must be equal,

For the above eqn:

[tex]LT^{- 2} = A + Bt + Ct^{3}[/tex]

Thus the dimensions of :

A = [tex][LT^{- 2}][/tex]

BT =  [tex][LT^{- 2}][/tex]

Thus for B

B = [tex][LT^{- 3}][/tex]

[tex]CT^{3} = LT^{- 2}[/tex]

[tex]C = [LT^{- 5}][/tex]

(b) For the units of A, B and C, we will make use of their respective dimensional formula from part (a)

where

L corresponds to length in meter(m)

T corresponds to time in seconds(s)

Now, for:

A = [tex][LT^{- 2}] = ms^{- 2}[/tex]

B = [tex][LT^{- 3}] = ms^{- 3}[/tex]

C = [tex][LT^{- 5}] = ms^{- 5}[/tex]

Answer:

Surely Achilles will catch the Tortoise, in 400 seconds

Explanation:

The problem itself reduces the interval of time many times, almost reaching zero. However, if we assume the interval constant, then it is clear that in two hours Achilles already has surpassed the Tortoise (20 miles while the Tortoise only 3).

To calculate the time, we use kinematic expression for constant speed:

[tex]x_{final}=x_{initial}+t_{tor}v_{tor}=1+t_{tor}\\x_{final}=x_{initial}+t_{ach}v_{ach}=10t_{ach}[/tex]

The moment that Achilles catch the tortoise is found by setting the same final position for both (and same time as well, since both start at the same time):

[tex]1+t=10t\\t=1/9 hour=0.11 hours[/tex]

Answer:

a)

3.43*10^{14} Hz

8.75*10^{-7} m

b)

2.70*10^{14} Hz

1.10*10^{-6} m

Explanation:

GIVEN DATA:

a)

i)we know that

E = h\nu

where E is energy

h = plank's constant = 6.625* 10^{-34} j-s

[tex]\nu = frequency[/tex]

[tex]\nu = \frac{E}{h} = \frac{1.42*1.6*10^{-19} v}{6.625*10^{-34}}[/tex]

[tex]\nu = 3.43*10^{14} Hz[/tex]

ii)wavelength  is given as

[tex]\lambda = \frac{c}{\nu}[/tex]

            [tex]= \frac{3*10^8}{3.43*10^{14}} = 8.75*10^{-7} m[/tex]

b) i) i)we know that

[tex]E = h\nu[/tex]

where E is energy

h = plank's constant = 6.625* 10^{-34} j-s

[tex]\nu = frequency[/tex]

[tex]\nu = \frac{E}{h} = \frac{1.12*1.6*10^{-19} v}{6.625*10^{-34}}[/tex]

[tex]\nu =2.70*10^{14} Hz[/tex]

ii)wavelength  is given as

[tex]\lambda = \frac{c}{\nu}[/tex]

[tex]= \frac{3*10^8}{2.70*10^{14}} = 1.10*10^{-6} m[/tex]

Answer:

Rt≅6377Km

Explanation:

Take a look at the image. The horizontal line is the horizon, and the angle α corresponds to the earth rotation during the 77 seconds.

With this information, we can know the value of α:

α = [tex]\alpha= \frac{77s}{1day}*\frac{1day}{24H}*\frac{1H}{60min}*\frac{1min}{60s}*2*\pi  =0.0056rad[/tex]

Since we have formed a rectangle triangle:

[tex]cos\alpha =\frac{Rt}{Rt+100m}[/tex]   Solving for Rt:

Rt≅6377467m=6377Km

Answer:

(a) 16.777mi

(b)Yes, he was speeding

Explanation:

(a)

Let's do the proper operations in order to convert km to mi:

[tex]27km*\frac{1000m}{1km} *\frac{1mi}{1609.34m} =16.77706389mi[/tex]

We can conclude that the trip length in miles was:

[tex]d=16.77706389mi[/tex]

(b)

Let's calculate the speed of the man during the trip:

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

But first, let's do the proper operations in order to convert min to h:

[tex]16min*\frac{1h}{60min} =2.666666667h[/tex]

Now, the speed is:

[tex]v=\frac{16.77706389mi}{2.666666667h} =62.91398959\frac{mi}{h}[/tex]

As we can see:

[tex]62.91398959\frac{mi}{h}>55\frac{mi}{h}[/tex]

So, we can conclude that the driver was speeding

Final answer:

The trip was 16.77 miles long, and the driver was speeding by traveling at an approximate speed of 62.78 miles per hour, surpassing the 55 mph speed limit.

Explanation:

To solve the question:

Thus, the trip was 16.77 miles long, and the driver was speeding, going approximately 62.78 miles per hour when the speed limit was 55 miles per hour.

Answer:

distance r from the uranium atom is 18.27 nm

Explanation:

given data

uranium and iron atom distance R = 44.10 nm

uranium atom = singly ionized

iron atom = doubly ionized

to find out

distance r from the uranium atom

solution

we consider here that uranium electron at distance = r

and electron between uranium and iron so here

so we can say electron and iron  distance = ( 44.10 - r ) nm

and we know single ionized uranium charge q2= 1.602 × [tex]10^{-19}[/tex] C

and charge on iron will be q3 = 2 × 1.602 × [tex]10^{-19}[/tex] C

so charge on electron is q1 =  - 1.602 × [tex]10^{-19}[/tex] C

and we know F = [tex]k\frac{q*q}{r^{2} }[/tex]  

so now by equilibrium

Fu = Fi

[tex]k\frac{q*q}{r^{2} }[/tex]  =  [tex]k\frac{q*q}{r^{2} }[/tex]

put here k = [tex]9*10^{9}[/tex] and find r

[tex]9*10^{9}\frac{1.602 *10^{-19}*1.602 *10^{-19}}{r^{2} }[/tex]  =  [tex]9*10^{9}\frac{1.602 *10^{-19}*1.602 *10^{-19}}{(44.10-r)^{2} }[/tex]

[tex]\frac{1}{r^{2} } = \frac{2}{(44.10 -r)^2}[/tex]

r = 18.27 nm

distance r from the uranium atom is 18.27 nm

Explanation:

The nucleus of an atom can be modeled as several protons and neutrons closely packed together.

Mass of the particle, [tex]m=1.67\times 10^{-27}\ kg[/tex]

Radius of the particle, [tex]R=10^{-15}\ m[/tex]

(a) The density of the nucleus of an atom is given by mass per unit area of the particle. Mathematically, it is given by :

[tex]d=\dfrac{m}{V}[/tex], V is the volume of the particle

[tex]d=\dfrac{m}{(4/3)\pi r^3}[/tex]

[tex]d=\dfrac{1.67\times 10^{-27}}{(4/3)\pi (10^{-15})^3}[/tex]

[tex]d=3.98\times 10^{17}\ kg/m^3[/tex]

So, the density of the nucleus of an atom is [tex]3.98\times 10^{17}\ kg/m^3[/tex].

(b) Density of iron, [tex]d'=7874\ kg/m^3[/tex]

Taking ratio of the density of nucleus of an atom and the density of iron as :

[tex]\dfrac{d}{d'}=\dfrac{3.98\times 10^{17}}{7874}[/tex]

[tex]\dfrac{d}{d'}=5.05\times 10^{13}[/tex]

[tex]d=5.05\times 10^{13}\ d'[/tex]

So, the density of the nucleus of an atom is [tex]5.05\times 10^{13}[/tex] times greater than the density of iron. Hence, this is the required solution.

Answer:2.83 m/s

Explanation:

Given

Object starts at x=-13 m at t=0 s

object takes 18 s to travel 51 m with constant velocity

i.e. there is no acceleration

and [tex]distance =speed\times times[/tex]

[tex]51=v\times 18[/tex]

v=2.83 m/s

Answer:

[tex]v_{ft} = 21.11 \frac{m}{s}[/tex]   :  Speed of the truck immediately after the collision , to the east.

Explanation:

Theory of collisions

Linear momentum is a vector magnitude (same direction and direction as velocity) and its magnitude is calculated like this:

P=m*v

where

P:Linear momentum

m: mass

v:velocity

There are 3 cases of collisions : elastic, inelastic and plastic.

For the three cases the total linear momentum quantity is conserved:

P₀=Pf   Formula (1)

P₀ :Initial  linear momentum quantity

Pf : nitial  linear momentum quantity

Nomenclature and data

mc: car mass= 1.43*10³ kg  = 1430kg

V₀c: initial car speed,  = 25.0 m/s

Vfc: final car speed = 18.0 m/s

mt: truck mass =  9000 kg  

V₀t: initial truck speed, = 20.0 m/s

Vft: final truck speed

Problem development

For this problem the collision is inelastic because after the collision the objects are deformed .

Because the known speeds go east they are positive, we assume that the truck continues moving east after the collision and its speed will also be positive:

We apply formula (1)

P₀=Pf

mc*V₀c+mt*V₀t=mc*Vfc+mt*Vft

1430*25+9000*20=1430*18+9000*Vft

215750=25740+9000*Vft

[tex]v_{ft} =\frac{215750-25740}{9000} = 21.11 \frac{m}{s}[/tex]

[tex]v_{ft} = 21.11 \frac{m}{s}[/tex]

Because the response was positive the truck moves east after the collision

Rate of change of mass is given by,

[tex]\frac{dm}{dt}=18.816\,g/s[/tex]

In the question,

We have the Base Area of the vertical container = 14 cm x 16 cm

Now,

Let us take the height of the container = h

Rate of change of height with time, dh/dt = 0.21 cm/s = 2.1 mm/s

So,

Volume of the container = Base Area x Height

So,

V = 14 x 16 x h

V = 140 x 160 x h (because, 1 cm = 10 mm)

V = (22400)h

Now,

Volume of one of the candy = 50 mm³

Mass of the candy = 0.0200 g

So,

Density of the candy = Mass/ Volume

So,

[tex]Density=\frac{0.0200}{50}\\Density=0.0004[/tex]

Now,

V = (22400)h

On differentiating with respect to time, t, we get,

[tex]\frac{dV}{dt}=\frac{22400h}{dt}\\\frac{d}{dt}(\frac{mass}{density})=22400.\frac{dh}{dt}\\\frac{1}{density}.\frac{dm}{dt}=22400.\frac{dh}{dt}\\[/tex]

Therefore, on putting the value of density in the equation and also the value of rate of change of height with time, we get,

[tex]\frac{1}{density}.\frac{dm}{dt}=22400.\frac{dh}{dt}\\\frac{1}{0.0004}.\frac{dm}{dt}=22400(2.1)\\\frac{dm}{dt}=18.816\,g/s[/tex]

Therefore, the rate of change of mass with respect to time is given by,

[tex]\frac{dm}{dt}=18.816\,g/s[/tex]

Answer:

168 m^2, 380 m^2

Explanation:

length of the room, l = 9.72 m

width of the room, b = 17.30 m

Area of teh rectangle is given by

A = length x width

So, A = 9.72 x 17.30 = 168.156 m^2

the significant digits should be 3 in the final answer

So, A = 168 m^2

Now length = 72 m

width = 17.39 feet = 5.3 m

Area, A = 72 x 5.3 = 381.6 m^2

There should be two significant digits in the answer so, by rounding off

A = 380 m^2

Answer:

Option (D)

Explanation:

The formula for the potential energy between the two charges is given by

[tex]U=\frac{KQq}{r}[/tex]

where, r is the distance between the two charges.

In first case the distance between the two charges is r1.

The potential energy is

[tex]U_{1}=\frac{KQq}{r_{1}}[/tex]

In first case the distance between the two charges is r2.

The potential energy is

[tex]U_{2}=\frac{KQq}{r_{2}}[/tex]

The change in potential energy is

[tex]\Delta U = U_{2}-U_{1}[/tex]

[tex]\Delta U=\frac{KQq}{r_{2}}-\frac{KQq}{r_{1}[/tex]

[tex]\Delta U=KQq \times \left ( \frac{1}{r_{2}} -\frac{1}{r_{2}} \right )[/tex]

The change in the potential energy of the charge +q during this process is [tex]\mathbf{\Delta U = KQq \Big (\dfrac{1}{r_2}- \dfrac{1}{r_1} \Big) }[/tex]

Option D is correct.

In an electrical circuit, the electric potential energy is the entire potential energy of a unit charge will have if it is positioned at any point in space.

The electric potential generated by a charge at any point in space is directly proportional to its magnitude and also varies inversely proportional to the distance out from the point charge.

Mathematically, it can be expressed as:

[tex]\mathbf{U = \dfrac{KQq}{r}}[/tex]

here;

So,

The potential energy for the first scenario can be expressed as:

[tex]\mathbf{U_1 = \dfrac{KQq}{r_1}}[/tex]

The potential energy for the second scenario can be expressed as:

[tex]\mathbf{U_2= \dfrac{KQq}{r_2}}[/tex]

Therefore, the change in the potential energy is:

[tex]\mathbf{\Delta U =U_2 -U1}[/tex]

[tex]\mathbf{\Delta U = \dfrac{KQq}{r_2}- \dfrac{KQq}{r_1}}[/tex]

[tex]\mathbf{\Delta U = KQq \Big (\dfrac{1}{r_2}- \dfrac{1}{r_1} \Big) }[/tex]

Learn more about electric potential energy here:
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Answer:

for -12db

 [tex]\frac{V1}{V2}=0.251\\\\\frac{P1}{P2}=0.0631[/tex]

for 3db

[tex]\frac{V1}{V2}=\sqrt{2}\\\\\frac{P1}{P2}=2[/tex]

for 10db

[tex]\frac{V1}{V2}=3.16\\\\\frac{P1}{P2}=10[/tex]

for 0db

[tex]\frac{V1}{V2}=1\\\\\frac{P1}{P2}=1[/tex]

Explanation:

The decibel is a logaritmic value given by:

[tex]db=10*log(\frac{P1}{P2})=20*log(\frac{V1}{V2})[/tex]

we use 10 for power values and 20 for other values such voltages or currents.

[tex]\frac{V1}{V2}=10^{\frac{db}{10}}\\\\\frac{P1}{P2}=10^{\frac{db}{20}}[/tex]

for -12db

[tex]\frac{V1}{V2}=10^{\frac{-12}{10}}=0.251\\\\\frac{P1}{P2}=10^{\frac{-12}{20}}=0.0631[/tex]

for 3db

[tex]\frac{V1}{V2}=10^{\frac{3}{10}}=\sqrt{2}\\\\\frac{P1}{P2}=10^{\frac{3}{20}}=2[/tex]

for 10db

[tex]\frac{V1}{V2}=10^{\frac{10}{10}}=3.16\\\\\frac{P1}{P2}=10^{\frac{10}{20}}=10[/tex]

for 0db

[tex]\frac{V1}{V2}=1\\\\\frac{P1}{P2}=1[/tex]

Answer:

Part a)

[tex]y_m = 0.157 mm[/tex]

part b)

[tex]k = 101.8 rad/m[/tex]

Part c)

[tex]\omega = 579.3 rad/s[/tex]

Part d)

here since wave is moving in negative direction so the sign of [tex]\omega[/tex] must be positive

Explanation:

As we know that the speed of wave in string is given by

[tex]v = \sqrt{\frac{T}{m/L}}[/tex]

so we have

[tex]T = 17.5 N[/tex]

[tex]m/L = 5.4 g/cm = 0.54 kg/m[/tex]

now we have

[tex]v = \sqrt{\frac{17.5}{0.54}}[/tex]

[tex]v = 5.69 m/s[/tex]

now we have

Part a)

[tex]y_m [/tex] = amplitude of wave

[tex]y_m = 0.157 mm[/tex]

part b)

[tex]k = \frac{\omega}{v}[/tex]

here we know that

[tex]\omega = 2\pi f[/tex]

[tex]\omega = 2\pi(92.2) = 579.3 rad/s[/tex]

so we  have

[tex]k = \frac{579.3}{5.69}[/tex]

[tex]k = 101.8 rad/m[/tex]

Part c)

[tex]\omega = 579.3 rad/s[/tex]

Part d)

here since wave is moving in negative direction so the sign of [tex]\omega[/tex] must be positive

Answer:

a) 567.6Hz

b) 516.8Hz

Explanation:

Using the formula for doppler effect:

[tex]f=\frac{C - V_{m} }{C - V_{o}} *f_{o}[/tex]   where:

[tex]V_{m}=15m/s; V_{o} = 30m/s; f_{o}=540Hz; C=343m/s[/tex]

Replacing the values we get:

f=567.6Hz

After the owl passes the mockingbind, the direction of sound relative to the owl and mockingbind changes direction, so the equation will be:

[tex]f=\frac{C + V_{m} }{C + V_{o}} *f_{o}[/tex]

Replacing the values we get:

f=516.8Hz

The Doppler Effect explains how the frequency changes due to the relative motion of the sound source and observer. When the mockingbird approaches the owl, it hears a frequency of 574 Hz. After the owl passes, the frequency it hears reduces to 510 Hz.

This question relates to the Doppler Effect, which is the change in frequency of a sound due to the relative motion between the source of the sound and the observer. The formula to calculate the frequency heard by an observer moving towards a source is given by: f' = f * (v + vo) / v, and the frequency heard by an observer moving away from a source is given by: f' = f * v / (v + vs).

(a) When the mockingbird is approaching the owl, the frequency it hears is calculated using the formula for the observer moving towards the source: f' = 540 * (343 + 14) / 343 = 574 Hz.

(b) When the mockingbird is moving in the same direction as the owl (after the owl has passed), the frequency it hears is lower. This can be calculated using the formula for the observer moving away from the source: f' = 540 * 343 / (343 + 30) = 510 Hz.

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Answer:

2.568 × 10⁻³ kg

Explanation:

The amount of sodium to be taken by an adult is measured in terms of a limit, as given here.

That limit has been set as 2568 mg , when measured in milligrams.

The basic conversion from milligrams to g is done by dividing with 1000. Then 2568 milligrams will be 2.568 grams.

Now 100 grams are present in 1 kilogram.

So 2.568 grams are divided with 1000 to get the specified mass in kg.

The gives it as 2.568 × 10⁻³.

Answer:

The correct answer is option 'c': Smaller stone rebounds while as larger stone remains stationary.

Explanation:

Let the velocity and the mass of the smaller stone be 'm' and 'v' respectively

and the mass of big rock be 'M'

Initial momentum of the system equals

[tex]p_i=mv+0=mv[/tex]

Now let after the collision the small stone move with a velocity v' and the big roch move with a velocity V'

Thus the final momentum of the system is

[tex]p_f=mv'+MV'[/tex]

Equating initial and the final momenta we get

[tex]mv=mv'+MV'\\\\m(v-v')=MV'.....i[/tex]

Now since the surface is frictionless thus the energy is also conserved thus

[tex]E_i=\frac{1}{2}mv^2[/tex]

Similarly the final energy becomes

[tex]E_f=\frac{1}{2}mv'^2+\frac{1}{2}MV'^2[/tex]\

Equating initial and final energies we get

[tex]\frac{1}{2}mv^2=\frac{1}{2}mv'^2+\frac{1}{2}MV'^2\\\\mv^2=mv'^2+MV'^2\\\\m(v^2-v'^2)=MV'^2\\\\m(v-v')(v+v')=MV'^2......(ii)[/tex]

Solving i and ii we get

[tex]v+v'=V'[/tex]

Using this in equation i we get

[tex]v'=\frac{v(m-M)}{(M-m)}=-v[/tex]

Thus putting v = -v' in equation i  we get V' = 0

This implies Smaller stone rebounds while as larger stone remains stationary.

Answer:

125 N-m

Explanation:

We have given force F= 25 N

Rupel pushes the box by 5 meter

So Distance S = 5 meter

Distance S = 5 meter

Work done in displacing a body is given by

Work done = force ×distance

So [tex]w=25\times 5=125N-m[/tex]

So work done by rupel pushes the box by 5 meter is 125 N-m

Now we know that 1 j = 1 N-m

So work done = 125 j

Rupel does 125.0 Joules of work when she pushes the box.

The work done by Rupel can be calculated using the equation:

Work = Force x Distance

Work = 25.0 N x 5.00 m = 125.0 Joules

Therefore, Rupel does 125.0 Joules of work when she pushes the box.

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To calculate average speed, divide the total distance by the total time taken. In this case, the average speed is 30.0 km/h.

To calculate average speed, we divide the total distance traveled by the total time taken. In this case, you drove 10.0 km in 20.0 minutes. To convert minutes to hours, we divide by 60. The average speed in kilometers per hour is calculated as:

Average speed = Total distance / Total time = 10.0 km / (20.0 min / 60 min/h) = 30.0 km/h

Answer:

option B

Explanation:

given,

seed of airplane = 70 m/s

height = 300 m

we know,

[tex]s = ut + \dfrac{1}{2}at^2[/tex]

[tex]300 = 0 + \dfrac{1}{2} \times 9.81\times t^2[/tex]

t = 7.82 s                                  

now, the range of the crate

 R = V × t

     = 70 × 7.82                      

     = 547.44 ≅ 548 m                          

hence, the correct answer is option B

Part A: The charge of sphere A after adding the electrons is [tex]\( q_A = -1.6 \times 10^{-7} \, \text{C} \).[/tex]

Part B: The charge of sphere B after separation is [tex]\( q_B = -8 \times 10^{-8} \, \text{C} \).[/tex]

Let's solve the problem step by step.

Part A: Charge of Sphere A

Initially, both spheres A and B are neutral, meaning they have no net charge. When [tex]\(1.0 \times 10^{12}\)[/tex] electrons are added to sphere A, these electrons carry a negative charge. The charge of one electron is approximately [tex]\( -1.6 \times 10^{-19} \)[/tex] coulombs.

The total charge added to sphere A can be calculated as:

[tex]\[ q_A = n \times e \][/tex]

where [tex]\( n = 1.0 \times 10^{12} \)[/tex] is the number of electrons and [tex]\( e = -1.6 \times 10^{-19} \)[/tex] C is the charge of one electron.

[tex]\[ q_A = 1.0 \times 10^{12} \times (-1.6 \times 10^{-19}) \][/tex]

[tex]\[ q_A = -1.6 \times 10^{-7} \, \text{C} \][/tex]

So, the charge of sphere A after adding the electrons is:

[tex]\[ q_A = -1.6 \times 10^{-7} \, \text{C} \][/tex]

Part B: Charge of Sphere B

When the two spheres are in contact, they form a system that will share the total charge equally due to their identical nature and the principle of electrostatic equilibrium. Since they are initially neutral, the total charge is just the charge added to sphere A, which is [tex]\( -1.6 \times 10^{-7} \, \text{C} \).[/tex]

When the charge is shared equally between the two identical spheres, each sphere will have:

[tex]\[ q_{\text{shared}} = \frac{q_A + q_B}{2} \][/tex]

Since [tex]\( q_B \)[/tex] is initially 0 (as it starts neutral), we have:

[tex]\[ q_{\text{shared}} = \frac{-1.6 \times 10^{-7} \, \text{C}}{2} \][/tex]

[tex]\[ q_{\text{shared}} = -0.8 \times 10^{-7} \, \text{C} \][/tex]

[tex]\[ q_{\text{shared}} = -8 \times 10^{-8} \, \text{C} \][/tex]

Therefore, after the separation, the charge on sphere B will be:

[tex]\[ q_B = -8 \times 10^{-8} \, \text{C} \][/tex]