The sound of one student typing furiously on their History paper which is due at 8:00 am tomorrow morning is 60.0 dB. What will the decibel level be when three sleep deprived students are typing furiously in the same room?

Answer:

value of the acceleration of gravity on the planet is 5.00 m/s²

Explanation:

The problem is similar to a free fall exercise, with another gravity value, the expression they give us is the following:

       y-yo = ½ gₐ t²       (1)

They tell us that they make a squared time graph with the variation of the distance, it is appropriate to clarify this in a method to linearize a curve, which is plotted the nonlinear axis to the power that is raised, specifically, the linearization of a curve The square is plotted against the other variable.

  Let's continue our analysis, as we have a linear equation, write the equation of the line.

        y1 = m x1 + b       (2)

where  “y1” the dependent variable, “x1” the independent variable, “m” the slope and “b” the short point

In this case as the stone is released its initial velocity is zero which implies that b = 0,

We plot on the “y” axis the time squared “t²” and on the horizontal axis we place “y-yo”.  To better see the relationship we rewrite equation 1 with this form

        t² = 2 /gₐ  (y-yo)

With the two expressions written in the same way, let's relate the terms one by one

        y1 = t²

        x1 = (y-yo)

        m = 2/gap

        b= 0

We substitute and calculate

        m = 2/gp

        gₐ = 2/m

        gₐ = 2/ 0.400

        gₐ = 5.00 m / s²

This is the value of the acceleration of gravity on the planet, note that the decimals are to keep the figures significant

Answer:

after 6 second it will stop

he travel 36 m to stop

Explanation:

given data

speed = 12 m/s

distance = 100 m

decelerates rate = 2.00 m/s²

so acceleration a = - 2.00 m/s²

to find out

how long does it take to stop and how far does he travel

solution

we will apply here first equation of motion that is

v = u + at   ......1

here u is speed 12 and v is 0 because we stop finally

put here all value in equation 1

0 = 12 + (-2) t

t = 6 s

so after 6 second it will stop

and

for distance we apply equation of motion

v²-u² = 2×a×s  ..........2

here v is 0 u is 12 and a is -2 and find distance s

put all value in equation 2

0-12² = 2×(-2)×s

s = 36 m

so  he travel 36 m to stop

Answer:29.627 m

Explanation:

Given

Initial velocity of life preserver(u) is 1.6 m/s

it takes 2.3 s to reach the water

using equation of motion

v=u+at

[tex]v=1.6+9.81\times 2.3[/tex]

v=24.163 m/s

Let s be the height of life preserver

[tex]v^2-u^2=2gs[/tex]

[tex]24.163^2-1.6^2=2\times 9.81\times s[/tex]

[tex]s=\frac{581.29}{2\times 9.81}[/tex]

s=29.627 m

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.

Read more about Revolution here https://brainly.com/question/111640

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

Answer:

(a) Projectile B will travel 4 times as far as projectile A prior to landing

Explanation:

Initial velocity = v

Angle at which the projectile is shot at = θ

g = Acceleration due to gravity

Range of a projectile is given by

[tex]R=\frac {v^{2}\sin 2\theta}{g}[/tex]

When Initial velocity = v

[tex]R_A=\frac{v^{2}\sin 2\theta}{g}[/tex]

When Initial velocity = 2v

[tex]R_B=\frac{(2v)^{2}\sin 2\theta}{g}\\\Rightarrow R_B=\frac{4v^2\sin 2\theta}{g}[/tex]

Dividing the equtions, we get

[tex]\frac{R_A}{R_B}=\frac{\frac{v^{2}\sin 2\theta}{g}}{\frac{4v^2\sin 2\theta}{g}}[/tex]

Here, the angle at which the projectiles are fired at are equal.

[tex]\frac{R_A}{R_B}=\frac{1}{4}\\\Rightarrow R_B=4R_A[/tex]

Hence, projectile B will travel 4 times as far as projectile A prior to landing

To compare the ranges of the two projectiles, we can use the fact that the horizontal range of a projectile launched with initial speed [tex]\( v_0 \)[/tex]at an angle [tex]\( \theta \)[/tex] with respect to the horizontal is given by:

[tex]\[ R = \frac{v_0^2 \sin(2\theta)}{g} \][/tex]

Where:

-  R  is the range,

- [tex]\( v_0 \)[/tex] is the initial speed of the projectile,

- [tex]\( \theta \)[/tex] is the launch angle, and

- g is the acceleration due to gravity.

Both projectiles are launched at the same angle with respect to the ground. Since the launch angle is the same for both projectiles, we can compare their ranges by comparing their initial speeds.

Let's denote the range of projectile A as [tex]\( R_A \)[/tex] and the range of projectile B as [tex]\( R_B \).[/tex]

For projectile A:

[tex]\[ R_A = \frac{v^2 \sin(2\theta)}{g} \][/tex]

For projectile B:

[tex]\[ R_B = \frac{(2v)^2 \sin(2\theta)}{g} = 4 \times \frac{v^2 \sin(2\theta)}{g} = 4R_A \][/tex]

So, the range of projectile B is four times the range of projectile A.

Therefore, the correct answer is:

(a) Projectile B will travel 4 times as far as projectile A prior to landing.

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

Explanation:

The maximum mass that can be hung from a string, m = 10 kg

Length of the string, l = 2 m

Mass of the object, m = 0.5 kg

Let v is the maximum speed that the mass can attain under these conditions without the string breaking. If T is the maximum tension in the string. So,

[tex]T_{max}=mg[/tex]

[tex]T_{max}=10\times 9.8=98\ N[/tex]

The centripetal force is provided by the tension in the string such that :

[tex]T_{max}=\dfrac{mv^2}{r}[/tex]

[tex]v=\sqrt{\dfrac{T_{max}r}{m}}[/tex]

[tex]v=\sqrt{\dfrac{98\times 2}{0.5}}[/tex]

v = 19.79 m/s

or

v = 20 m/s

So, the maximum speed that the mass can attain under these conditions without the string breaking is 20 m/s. Hence, this is the required solution.

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:

a)Q= + 0.71 nC , For the resultant electric field at the origin to be 45.0 N/C in the +x direction

b)Q= -2.83nC ,for the resultant electric field at the origin to be 45.0 N/C in the −x direction

Explanation:

Conceptual analysis

The electric field at a point P due to a point charge is calculated as follows:

E = k*q/d²

E: Electric field in N/C

q: charge in Newtons (N)

k: electric constant in N*m²/C²

d: distance from charge q to point P in meters (m)

The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.

Equivalences

1nC= 10⁻9 C

Data

k = 8.99*10⁹ N×m²/C²

q₁ =+5.45nC = 3*10⁻⁹C

d₁ =1.35 m

d₂ = 0.595m

a)Problem development : sign and magnitude of Q for the resultant electric field at the origin to be 45.0 N/C in the +x direction

We make the algebraic sum of fields at at the origin :

[tex]E_{o} =E_{q} +E_{Q}[/tex] Equation (1)

[tex]E_{q} =\frac{k*q_{1} }{d_{1}{2}  }[/tex]

Calculation of E(q)

[tex]E_{q} =\frac{8.99*10^{9} *5.45*10^{-9} }{1,35^{2} }[/tex]

[tex]E_{q} =26.88\frac{N}{C}[/tex] : in the +x direction .As the charge is negative, the field enters the charge

We replace [tex]E_{o}[/tex] and [tex]E_{q}[/tex] in the equation (1)

[tex]45=26.88+E_{Q}[/tex]

[tex]E_{Q} =45-26.88[/tex]

[tex]E_{Q} = 18.12 N/C[/tex] : in the +x direction .

Sign and magnitude of Q

Q must be positive for the field to abandon the load in the +x

[tex]E_{Q} =\frac{k*Q}{d_{2}^{2} }[/tex]

[tex]18.12=\frac{8.99*10^{9}*Q }{0.595^{2} }[/tex]

[tex]Q=\frac{18.12*0.595^{2} }{8.99*10^{9} }[/tex]

Q=0.71*10⁻⁹ C =0.71 nC

b)Sign and magnitude of Q for the resultant electric field at the origin to be 45.0 N/C in the −x direction

We make the algebraic sum of fields at at the origin :

[tex]E_{o} =E_{q} +E_{Q}[/tex]

[tex]-45=26.88+E_{Q}[/tex]

[tex]-71.88=E_{Q}[/tex]

[tex]71.88=\frac{8.99*10^{9} *Q}{0.595^{2} }[/tex]

Q= 2.83*10⁻⁹ C

Q= -2.83nC

Q must be negative for the field to enters the charge in the −x direction

The magnitude and sign of Q is given by the required magnitude and

sign of the charge at the origin due to the sum of the charges.

Responses:

The charge at the origin is given as follows;

When the charge at the origin is 45.0 N/C, we have;

[tex]45 = \mathbf{\dfrac{8.99 \times 10^{9} \times -5.45 \times ^{-9} }{1.35^2} + \dfrac{8.99 \times 10^{9} \times Q} {(-0.595)^2}}[/tex]

Which gives;

[tex]Q = \dfrac{\left(45 - \dfrac{8.99 \times 10^{9} \times -5.45 \times ^{-9} }{1.35^2} \right) \times (-0.595)^2}{8.99 \times 10^{9} } \approx \mathbf{2.83 \times 10^{-9}}[/tex]

When the charge at the origin is [tex]E_0[/tex] = 45 N/C, we have;

When the charge at the origin is [tex]E_0[/tex] = 45 N/C in the -x direction, we have;

[tex]Q = \dfrac{\left(-45 - \dfrac{8.99 \times 10^{9} \times -5.45 \times ^{-9} }{1.35^2} \right) \times (-0.595)^2}{8.99 \times 10^{9} } \approx -7.134 \times 10^{-10}[/tex]

Therefore;

The charge at the origin is [tex]E_0[/tex] = 45 N/C in the -x direction, we have;

Learn more about electric field strength here:

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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:

[tex]\Delta p=(0,3.10)kg*m/s\\[/tex]  

Explanation:

Momentum change:

[tex]\Delta p=p_{f}-p_{o}\\[/tex]   :  vector

p=mv

[tex]p_{o}=(p_{ox, p_{oy}}}=(m*v*sin(\theta),-m*v*cos(\theta) )\\[/tex]   : the ball move downward with an angle theta to the vertical

[tex]p_{f}=(p_{fx, p_{fy}}}=(m*v*sin(\theta),+m*v*cos(\theta) )\\[/tex]     :the ball move upward with the same angle theta to the vertical, with same speed

So:

[tex]\Delta p=p_{f}-p_{o}=(0,2m*v*cos(\theta))=(0,2*0.609*3*cos(32))=(0,3.10)kg*m/s\\[/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

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:

93.29 gallons

Explanation:

Given:

Number of solar panels = 3

Area of each solar panel = 1 m × 2 m = 2 m²

Total area of solar panels = 3 × 2 = 6 m²

Time = 4 hrs = 4 × 60 × 60 = 14400 seconds

Change in temperature, ΔT = 120° C - 40° C = 80° C

Now,

the solar power received on the Earth = 1368 W/m²

Thus,

The Heat energy received = Power × Area × Time

or

The Heat energy received = 1368 × 6 × 14400 =  118195200 J

Also,

Heat = mCΔT

where, C is the specific heat of the water

m is the mass of the water = 4.184 J/g.C

thus,

118195200 J = m × 4.184 × 80

or

mass of water that can be heated, m = 353116.63 grams = 353.116 kg

Also,

1 gallon of water = 3.785 kg

thus,

1 kg of water = 0.2642 gallons

Hence,

353.116 kg of water = 93.29 gallons

i.e 93.29 gallons of water can be heated

To determine how many gallons of water can be heated from 40°C to 120°C by three solar panels, we must calculate the energy absorbed by the panels using insolation data and then apply the specific heat capacity of water. The actual insolation value is necessary for the precise calculation, which was not provided in the question.

The amount of water that can be heated from 40°C to 120°C each day by solar panels can be calculated using principles from physics, specifically thermodynamics and energy transfer. First, we need to determine the energy incident on the solar panels. Assuming normally incident sunlight for 4 hours and a total solar panel area of 6 m² (as there are three 1 m x 2 m panels), we can calculate the energy absorbed.

Next, we use the specific heat capacity equation to find how much water this energy can heat from 40°C to 120°C. The specific heat capacity of water is approximately 4.18 J/g°C. Afterwards, we'll convert this amount of water from liters (or kilograms, as 1 L of water is approximately 1 kg) into gallons for the answer.

To complete this calculation, we would need to know the actual insolation in the specific location in terms of energy per unit area per unit time (e.g., kW/m²), which is not provided in the question or the reference material. But, if we assume insolation similar to the examples given in the reference material, we could use that to approximate the answer.

Answer:

Image is virtual and formed on the same side as the object, 19.29 cm from the lens.

The height of the image is 0.40509 cm

Image is upright as the magnification is positive and smaller than the object.

Explanation:

u = Object distance =  100 cm

v = Image distance

f = Focal length = -23.9 cm (concave lens)

[tex]h_u[/tex]= Object height = 2.1 cm

Lens Equation

[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{-23.9}-\frac{1}{100}\\\Rightarrow \frac{1}{v}=\frac{-1239}{23900} \\\Rightarrow v=\frac{-23900}{1239}=-19.29\ cm[/tex]

Image is virtual and formed on the same side as the object, 19.29 cm from the lens.

Magnification

[tex]m=-\frac{v}{u}\\\Rightarrow m=-\frac{-19.29}{100}\\\Rightarrow m=0.1929[/tex]

[tex]m=\frac{h_v}{h_u}\\\Rightarrow 0.1929=\frac{h_v}{2.1}\\\Rightarrow h_v=0.1929\times 2.1=0.40509\ cm[/tex]

The height of the image is 0.40509 cm

Image is upright as the magnification is positive and smaller than the object.

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]

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: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:

(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]

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 distance is [tex]9.57\times10^{13}\ m[/tex]

Explanation:

Given that,

Diameter of telescope d= 5.08 m

Diameter of sun spot y= 10000 mi

[tex]y =1609.3\times10^{4}\ m[/tex]

We need to calculate the distance

Using formula of distance

[tex]y =\dfrac{1.22\lambda D}{d}[/tex]

[tex]D=\dfrac{d y}{1.22\times\lambda}[/tex]

Put the value into the formula

[tex]D=\dfrac{5.08\times1609.3\times10^{4}}{1.22\times700\times10^{-9}}[/tex]

[tex]D=9.57\times10^{13}\ m[/tex]

Hence, The distance is [tex]9.57\times10^{13}\ m[/tex]

Final answer:

The most distant star on which the Hale Telescope could resolve a sunspot of this size is approximately 73 trillion km away.

Explanation:

To determine the most distant star that this telescope could resolve a sunspot on, we need to calculate the angular resolution of the telescope. The angular resolution of a telescope is given by the formula θ = 1.22 * (λ / D), where θ is the angular resolution, λ is the wavelength of light, and D is the diameter of the telescope's mirror.

To resolve the diameter of a sunspot, we can use the approximate diameter of a large sunspot of 10,000 miles. Converting this to kilometers, we get 16,093.44 km. Assuming an average wavelength of visible light of 550 nm, we can calculate the angular resolution as follows:

θ = 1.22 × (550 nm / 5.08 meters) = 0.013 arc seconds

Now, we need to determine the distance at which this angular resolution corresponds to a sunspot diameter of 16,093.44 km. We can use the small angle formula to calculate the distance:

D = diameter / tan(θ)

D = 16,093.44 km / tan(0.013 arc seconds) = 73,259,925,487,382 km

Therefore, the most distant star on which the Hale Telescope could resolve a sunspot of this size is approximately 73 trillion km away.

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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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.

A molecule of hydrogen moving at a speed of 115 cm/s would take approximately 79.5 seconds to travel the length of a football field (100 yd).

To compute the time it would take a molecule of hydrogen to move the length of a football field, you would use the formula time = distance/speed. However, we must first convert the length of the football field from yards to centimeters for consistency.

One yard is approximately 91.44 cm, so a football field which is 100 yards is about 9144 cm. Using the given speed of a hydrogen molecule which is 115 cm/s, the time it would take can be computed as follows: time = 9144 cm / 115 cm/s = approx. 79.5 seconds.

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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:

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:

The magnitude of force is [tex]4.26\times 10^{- 6} N[/tex]

Solution:

As per the question:

The strength of Electric field due west at a certain point, [tex]\vec{E_{w}} = 710,000 N/C[/tex]

Charge, Q = - 6 C

Now, the force acting on the charge Q in the electric field is given by:

[tex]\vec{F} = Q\vec{E_{w}}[/tex]

[tex]\vec{F} = -6\times 710,000 = - 4.26\times 10^{- 6} N[/tex]

Here, the negative sign indicates that the force acting is opposite in direction.

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:

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.