Assuming two hypothetical maps that each cover a standard 8.5 by 11-inch sheet of paper, the larger-scale map would cover a larger geographic area than the smaller-scale map.
a) true
b) false

Answers

Answer 1

Answer:

b) false

Explanation:

The scale of a map is the ratio of a distance on the map to the corresponding distance on the ground. Scaling allow us to capture a large geographical area on a reduced platform while still retaining the relative sizes and positioning of places on the map to their real life sizes and positioning. If both maps cover a standard 8.5 by 11-inch sheet of paper, then the map with the smaller ratio will have the bigger geographical area.

To understand better, let us assume two geographical areas A and B. A is bigger than B. If we were to put them both on the same area of map paper, then we'll have to scale up the smaller geographical area B so as to fit into the map paper. This means that the geographical area with the smaller area B will have the larger scale on the map.


Related Questions

A uniform crate with a mass of 22 kg must be moved up along the 15° incline without tipping. The force P is horizontal. Determine the corresponding magnitude of force P.

Answers

Answer:

[tex]F_x=208.25\ N[/tex]

Explanation:

Given that,

Mass of a crate is 22 kg

It moved up along the 15 degrees incline without tipping.

We need to find the corresponding magnitude of force P. The force P is acting in horizontal direction.

It means that the horizontal component of force is given by :

[tex]F_x=F\cos\theta\\\\F_x=mg\cos\theta\\\\F_x=22\times 9.8\times \cos(15)\\\\F_x=208.25\ N[/tex]

So, the horizontal component of force is 208.25 N.

A uniform meter rule with a mass of 200g is suspended at zero mark pivotes at 22.0cm mark. calculate the mass of the rule.
pls answer quickly. Thanks​

Answers

Answer:

The mass of the rule is 56.41 g  

Explanation:

Given;

mass of the object suspended at zero mark, m₁ = 200 g

pivot of the uniform meter rule = 22 cm

Total length of meter rule = 100 cm

0                          22cm                                  100cm

-------------------------Δ------------------------------------

↓                                                                       ↓

200g                                                                 m₂  

Apply principle of moment

(200 g)(22 cm - 0)     = m₂(100 cm - 22 cm)

(200 g)(22 cm) = m₂(78 cm)

m₂ =  (200 g)(22 cm)  / (78 cm)

m₂ = 56.41 g  

Therefore,  the mass of the rule is 56.41 g                                            

Assume that helium behaves as an ideal monatomic gas. If 2 moles of helium undergo a temperature increase of 100 K at constant pressure, how much energy has been transferred to the helium as heat

Answers

Answer:

6235.5J

Explanation:

Using ( nစ)p= ncp x change in temp

But cp= ( 1+ f/2)R

So cp= ( 1+ 3/2R

Cp= 5R/2

So = n x 5R/2x 150k

= 2 x 5/2x 8.314 x150

= 6235.5J

Answer:

2500 J

Explanation:

Q=(3/2)nRΔT

Q=(3/2)*2 mol*(8.314 J/mol*k)*100 k

Q=2494 J

A physics professor wants to perform a lecture demonstration of Young's double-slit experiment for her class using the 633-nm light from a He-Ne laser. Because the lecture hall is very large, the interference pattern will be projected on a wall that is 5.0 m from the slits. For easy viewing by all students in the class, the professor wants the distance between the m=0 and m=1 maxima to be 35 cm. What slit separation is required in order to produce the desired interference pattern?

Answers

Answer:

The distance of separation is [tex]d = 9.04 *10^{-6 } \ m[/tex]

Explanation:

From the question we are told that

    The wavelength is [tex]\lambda = 633\ nm = 633 *10^{-9} \ m[/tex]

     The  distance of the screen is [tex]D = 5.0 \ m[/tex]

     The  distance between the fringes is  [tex]y = 35 \ cm = 0.35 \ m[/tex]

       

Generally the distance between the fringes is mathematically represented as

       [tex]y = \frac{ \lambda * D }{d }[/tex]

Here d is the distance of separation between the slit

=>    [tex]d = \frac{ \lambda * D }{y }[/tex]

=>    [tex]d = \frac{ 633 *10^{-9} * 5 }{ 0.35 }[/tex]

=>   [tex]d = 9.04 *10^{-6 } \ m[/tex]

Which is the property of mattter in which substance can transfer heat to electricity

Answers

Conductivity is the property of matter in which a substance can transfer heat or electricity.

Additional information:-

Matter : Anything which occupies space and has mass is called matter.

Chemical classifications

Pure Substances ( made of one kind of substance )

Impure Substances ( mixture )

Physical classifications

Solid

Liquid

Gas

Plasma ( made of ions and free electrons )

BEC ( Bose Einstein Condensate )

Fermionic Condensate ( It discovered in 2003 )

Turning the barrel of a 50-mm-focal-length lens on a manual-focus camera moves the lens closer to or farther from the sensor to focus on objects at different distances. The lens has a stated range of focus from 0.70 m infinity.
How far does the lens move between these two extremes?

Answers

Answer:

Explanation:

To focus object at .7m , the image distance can be measured as follows

object distance u = .7m

focal length f = .05 m

image distance v = ?

from lens formula

[tex]\frac{1}{v} -\frac{1}{u} = \frac{1}{f}[/tex]

[tex]\frac{1}{v} +\frac{1}{.7} = \frac{1}{.05}[/tex]

[tex]\frac{1}{v} =\frac{1}{.05} -\frac{1}{.7}[/tex]

v = .054 m

= 54 mm

when the object is at infinity , image is formed at focus ie at distance of

50 mm .

So lens position from sensor  where image is formed , varies from 54 mm to 50 mm .

Which statement about the ocean is true? A. No evaporation or precipitation in the water cycle occurs over the ocean. B. Most evaporation and precipitation in the water cycle occur over the ocean. C. All evaporation and precipitation in the water cycle occur over the ocean. D. Evaporation, but not precipitation, in the water cycle occurs over the ocean.

Answers

Answer:

A

Explanation:

Answer:

A

Explanation:

No evaporation or precipitation in the water cycle occurs over the ocean.

What is the heat-loss rate through the slab if the ground temperature is 5 ∘C while the interior of the house is 25 ∘C?

Answers

Complete question :

A 12 m x 15 m house is built on a 12-cm-thick concrete slab.

What is the heat-loss rate through the slab if the ground temperature is 5°C while the interior of the house is 25°C

Answer:

3kW

Explanation:

Given the following :

Dimension of house :

Length = 12m

Width = 15m

Thickness of concrete slab (t) = 12cm

t in metres :

100cm = 1m

12cm = (12/100)m

= 0.12m

Ground temperature (Tg) = 5°C

Interior temperature = (Th) = 25°C

Thermal conductivity of concrete (K) is approximately 1 Wm/k

Using the relation:

Q = KA * [ (Th - Tg) / d]

A = Length * width = (12 *15) = 180

Q = (1 * 180) * [(25°C - 5°C) / 0.12]

Q = 180 * (20/0.12)

Q = 180 * 16.6666

Q = 3,000W = 3kW

The heat-loss rate is 3kW

Given that,

Dimension of house :

Length = 12m

Width = 15m

Thickness of concrete slab (t) = 12cm

We know that

100cm = 1m

so,

12cm = (12/100)m

= 0.12m

And,

Ground temperature (Tg) = 5°C

Interior temperature = (Th) = 25°C

calculation of heat loss rate:

Q = KA * [ (Th - Tg) / d]

A = Length * width = (12 *15) = 180

Q = (1 * 180) * [(25°C - 5°C) / 0.12]

Q = 180 * (20/0.12)

Q = 180 * 16.6666

Q = 3,000W

= 3kW

learn more about the temperature here: https://brainly.com/question/16940730

6. Solve (5.87 x 10^7)(4.200 x 10^11). Be
sure your answer is in scientific notation.
Round to two decimal places.

Answers

Explanation:

We need to solve [tex](5.87\times 10^7)(4.2\times 10^{11})[/tex]

Firstly, multiplying 5.87 and 4.2 = 24.654

Now taking exponent of 10.

We know that : [tex]x^a{\cdot} x^b=x^{a+b}[/tex]

It means, [tex]10^7{\cdot} 10^{11}=10^{11+7}=10^{18}[/tex]

So,

[tex](5.87\times 10^7)(4.2\times 10^{11})=24.654\times 10^{18}[/tex]

In scientific notation,

[tex](5.87\times 10^7)(4.2\times 10^{11})=2.4654\times 10^{19}[/tex]

Hence, the value of [tex](5.87\times 10^7)(4.2\times 10^{11})[/tex] is [tex]2.4654\times 10^{19}[/tex]

Answer:

Explanation:

We need to solve

Firstly, multiplying 5.87 and 4.2 = 24.654

Now taking exponent of 10.

We know that :

It means,

So,

In scientific notation,

Hence, the value of  is

Find the sum of the following vectors A=3i-12j and B=4i+7j

Answers

Answer:

(I). The sum of the vectors is (7i-5j).

(II). The sum of the vectors is (8i+7j).

Explanation:

Given that,

(I). Vector A [tex]A=3i-12j[/tex]

Vector B [tex]B=4i+7j[/tex]

Suppose, (II). Vector A [tex]A=6i+15j[/tex]

Vector B [tex]B=2i-8j[/tex]

(I). We need to calculate the sum of the vectors

Using formula of sum

[tex]\vec{C}=\vec{A}+\vec{B}[/tex]

Where,

[tex]\vec{A}= vector A[/tex]

[tex]\vec{B}= vector B[/tex]

[tex]\vec{C}= sum of the vector A and b

Put the value into the formula

[tex]\vec{C}=(3i-12j)+(4i+7j)[/tex]

[tex]\vec{C}=7i-5j[/tex]

(II). We need to calculate the sum of the vectors

Using formula of sum

[tex]\vec{C}=\vec{A}+\vec{B}[/tex]

Put the value into the formula

[tex]\vec{C}=(6i+15j)+(2i-8j)[/tex]

[tex]\vec{C}=8i+7j[/tex]

Hence, The sum of the vectors is (7i-5j).

The sum of the vectors is (8i+7j).

Select the correct answer. Physics is explicitly involved in studying which of these activities? A. the mixing of metals to form an alloy B. the metabolic functions of a living organism C. the motion of a spacecraft under gravitational influence D. the depletion of the atmospheric ozone layer due to pollutants E. the killing of cancerous cells by radiation therapy

Answers

Answer:

C. the motion of a spacecraft under gravitational influence

Two vehicles collide and stick together. After the collision, their combined y-momentum is 2.40 × 104 kilogram meters/second, and their x-momentum is 7.00 × 104 kilogram meters/second. What is the angle of the motion of the two vehicles, with respect to the x-axis?

Answers

Explanation:

It is given that,

Momentum in y direction is [tex]2.4\times 10^4\ kg-m/s[/tex]

Momentum in x direction is [tex]7\times 10^4\ kg-m/s[/tex]

We need to find the angle of the motion of the two vehicles, with respect to the x-axis. The angle between two vectors is given by :

[tex]\tan\theta=\dfrac{p_y}{p_x}\\\\\tan\theta=\dfrac{2.4\times 10^4}{7\times 10^{4}}\\\\\theta=\tan^{-1}\left(0.342\right)\\\\\theta=18.88^{\circ}[/tex]

So, the angle of the motion of the two vehicles is 18.88 degrees.

A large number of very industrious people make a very long pole. It is 10.0 light years long! ( As they measure it. ) Soon a spaceship flies along the length of the pole at 90% the speed of light. How much time passes on the spaceship from the moment the ship passes the first end of the pole to the moment the ship passes the second end of the pole

Answers

Answer:

L = L0 ( 1 - v^2/c^2))1/2     where L0 is the proper length

L = 10 L-y (1 - .9^2)^1/2 = 4.36 L-y   length of pole measured by ship

t = 4.36 L-y / .9 c = 4.84 y  since the ship travels at .9 c

What is the shortest possible time in which a bacterium to travel distance of 8.4cm across a Petri dish at a constant velocity of 1.2 cm/s

Answers

Answer:

[tex] \boxed{\sf Shortest \ possible \ time = 7 \ seconds} [/tex]

Given:

Distance travelled (s) = 8.4 cm

velocity (v) = 1.2 cm/s

To Find:

Shortest possible time (t) in which a bacterium travel a distance 8.4 cm across a Petri Dish

Explanation:

[tex] \boxed{ \bold{\sf Time \ (t) = \frac{Distance \ travelled \ (s)}{Velocity \ (v)}}}[/tex]

Substituting values of Distance travelled (s) & Velocity (v) in the equation:

[tex] \sf \implies t = \frac{8.4}{1.2} [/tex]

[tex] \sf \implies t = \frac{7 \times \cancel{1.2}}{ \cancel{1.2}} [/tex]

[tex] \sf \implies t = 7 \: s[/tex]

uncertainty propagation question #2

Hi all, I am trying to calculate the uncertainty and volume for a rectangular block with the measurements being 8.7cm, 5.2cm, 5.4cm. I am struggling with the uncertainty propagation, and I am unsure if I did this correctly. Heres what I've tried.


I found the uncertainty for each individual measurement to be .1 because they all have 1 decimal place. I Added this to the formula with the measurements, took the square root of the sum of the squares with the uncertainty for each individual measurement being in the numerator and the measurement in the denominator, as follows: √(.1/8.7)\^2 + (.1/5.2)\^2 + (.1/5.4)\^2. My final answer was volume= 244.296 +/- .029 cm\^3. I rounded the uncertainty that I got from the equation to 2 significant figures because that’s what the smallest measurement has. Did I do this correctly?

Answers

Answer:

The correct treatment of uncertainties for the volume is shown below

Explanation:

In order to estimate the uncertainty in the volume which is derived via the formula:

[tex]V = w*l*h[/tex]

you normally start with the relative errors  [tex](\frac{\delta Q}{Q})[/tex]  of each quantity (Q) measured, since they are so easy to handle, stating that the relative error in the Volume is the addition of the relative errors in each quantity:

[tex]\frac{\delta V}{V} =\frac{\delta w}{w} +\frac{\delta l}{l} +\frac{\delta h}{h}[/tex]

and finally solve for [tex]\delta V[/tex] by multiplying both sides by the volume you calculated.

In your case, this becomes:

[tex]\delta V =V \left \{\frac{\delta w}{w} +\frac{\delta l}{l} +\frac{\delta h}{h}\right \} \\\delta V = 244.296 \left \{\frac{0.1}{5.4} +\frac{0.1}{8.7} +\frac{0.1}{5.2}\right \}\\\delta V = 244.296 \, (0.04924354)\\\delta V = 12.03 \,\,cm^3[/tex]

Then, since the standard practice is to write the uncertainty with ONLY ONE significant figure, the rounding of your uncertainty becomes:

[tex]\delta V=10\,\,cm^3[/tex]

Giving this, you need to express the final measurement as:

[tex]V=240\,\,cm^3\,+/- 10 \,\,cm^3[/tex]

making sure that the expression for the volume doesn't have significant figures passed the limitation imposed by its uncertainty (in this case the tenths).

Please notice as well that in the treatment you did, you:

1) ended up with an uncertainty even smaller than the relative uncertainty of each measurement (which cannot be possible since relative uncertainties add-up)

2) are not rounding your uncertainty to ONE SIG FIG.

If you are driving 95 km????h along a straight road and you look to the side for 2.0 s, how far do you travel during this inattentive period?

Answers

Answer:

52.7 m

Explanation:

Given that

speed of the vehicle, v = 95 km/h

time of inattentiveness, t = 2 s

distance travelled, s = ?

Since we have the speed in km/h and the time in s, it would be best if we converted one of them to make sure we have all units in the same rank.

95 km/h = 95 * 1000/3600 m/s

95 km/h = 95000/3600 m/s

95 km/h = 26.38 m/s

Now, we use our derived speed in m/s

Speed of a moving vehicle is given by,

v = s/t, where

v = speed in m/s

s = distance travelled, in m

t = time spent, in s

if we make d the subject of formula by rearranging the equation, we have

s = v * t

distance travelled, s = 26.38 * 2

distance travelled, s = 52.7 m

therefore, during this inattentive period, 52.7 m was travelled.

A parachuter, falling with a constant speed, drops 456m in 14.4s.
Determine all unknowns.
S =
51
m/s
d =
456
m
t =
14.4
S

Answers

Answer:

The parachuter is falling at a speed of 31.667 meters per second.

Explanation:

Given that parachuter falls at constant speed and travelled distance and time are known, the unknown is speed, measured in meters per second, which is obtained by the following kinematic expression:

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

Where:

[tex]d[/tex] - Travelled distance, measured in meters.

[tex]t[/tex] - Time, measured in seconds.

If [tex]d = 456\,m[/tex] and [tex]t = 14.4\,s[/tex], the speed of the parachuter is:

[tex]s = \frac{456\,m}{14.4\,s}[/tex]

[tex]s = 31.667\,\frac{m}{s}[/tex]

The parachuter is falling at a speed of 31.667 meters per second.

Each wheel of a 320 kg motorcycle is 52 cm in diameter and has rotational inertia 2.1 kg m2 . The cycle and its 75 kg rider are coasting at 85 km/h on a flat road when they encounter a hill. If the cycle rolls up the hill with no applied power and no significant internal friction, what vertical height will it reach

Answers

Answer:

The value is  [tex]h = 32.91 \ m[/tex]

Explanation:

From the question we are told that

    The diameter of each wheel is  [tex]d = 52 \ cm = 0.52 \ m[/tex]

    The mass of the motorcycle is  [tex]m = 320 \ kg[/tex]

    The rotational kinetic inertia is  [tex]I = 2.1 \ kg \ m^2[/tex]

    The  mass of the  rider is  [tex]m_r = 75 \ kg[/tex]

     The  velocity is  [tex]v = 85 \ km/hr = 23.61 \ m/s[/tex]

      Generally the radius of the wheel is mathematically represented as

      [tex]r = \frac{d}{2}[/tex]

=>     [tex]r = \frac{0.52}{2}[/tex]

=>    [tex]r = 0.26 \ m[/tex]

Generally from the law of energy conservation

     Potential energy  attained  by  system(motorcycle and rider )  =  Kinetic  energy of the system  +  rotational kinetic energy of  both wheels of the motorcycle

=>  [tex]Mgh = \frac{1}{2} Mv^2 + \frac{1}{2} Iw^2 + \frac{1}{2} Iw^2[/tex]

=>    [tex]Mgh = \frac{1}{2} * Mv^2 + Iw^2[/tex]

Here  [tex]w[/tex] is the angular velocity which is mathematically represented as

     [tex]w = \frac{v }{r }[/tex]

So

    [tex]Mgh = \frac{1}{2} * Mv^2 + I \frac{v}{r} ^2[/tex]

Here [tex]M = m_r + m[/tex]

         [tex]M = 320 + 75[/tex]

          [tex]M = 395 \ kg[/tex]

[tex]395 * 9.8 * h = 0.5 * 395 * (23.61)^2 + 2.1 *[\frac{ 23.61}{ 0.26} ] ^2[/tex]

=>   [tex]h = 32.91 \ m[/tex]

   

The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.420 with the floor. If the train is initially moving at a speed of 57.0 km/h, in how short a distance can the train be stopped at constant acceleration without causing the crates to slide over the floor

Answers

Answer:

The distance is  [tex]s= 30.3 \ m[/tex]

Explanation:

From the question we are told that

   The  coefficient of static friction is  [tex]\mu_s = 0.42[/tex]

    The  initial speed of the train is  [tex]u = 57 \ km /hr = 15.8 \ m/s[/tex]

   

For the crate not to slide the friction force must be equal to the force acting on the train i.e

       [tex]-F_f = F[/tex]

The negative sign shows that the two forces are acting in opposite direction

=>   [tex]mg * \mu_s = ma[/tex]

=>   [tex]-g * \mu_s = a[/tex]

=>   [tex]a = -9.8 * 0.420[/tex]

=>   [tex]a = -4.116 m/s^2[/tex]

From equation of motion

  [tex]v^2 = u^2 + 2as[/tex]

Here  v =  0 m/s since it came to a stop

=>   [tex]s= \frac{v^2 - u^2 }{ 2 a}[/tex]

 =>   [tex]s= \frac{0 -(15.8)^2 }{ - 2 * 4.116}[/tex]

=>     [tex]s= 30.3 \ m[/tex]

Experts in model airplanes develop a supersonic plane to scale, it moves horizontally in the air while it is conducting a flight test. The development team defines that the space that the airplane travels as a function of time is given by the function: e (t) = 9t 2 - 6t + 3 Determine what acceleration the scale airplane has (Second derivative).

Answers

Explanation:

e(t) = 9t² − 6t + 3

The velocity is the first derivative:

e'(t) = 18t − 6

The acceleration is the second derivative:

e"(t) = 18

#1 A boy pushes forward a cart of groceries with a total mass of 40 kg. What is
the acceleration of the cart if the net force on the cart is 60 N?

Answers

Explanation:

∑F = ma

60 N = (40 kg) a

a = 1.5 m/s²

What is force? What creates it?

Answers

Answer:

its an interaction that can move an object; push or pull makes it or gravity, magnetism

Explanation:

its all in the answer

Answer:

In physics, a force is any interaction that, when unopposed, will change the motion of an object. A force can cause an object withmass to change its velocity (which includes to begin moving from a state of rest), i.e., toaccelerate. 

A lightning bolt with 13 kA strikes an object for 14 μ s. How much charge is deposited on the object?

Answers

Answer:

0.182C

Explanation:

Using Q= It

= 13x10^3 . 14x10^-6

= 0.182C

Two parallel very long straight wires carrying current of 5A each are kept at a separation of 1m. If the currents are in the same direction, the force per unit length between them is __________

Answers

Answer:

The force per unit between the two parallel wires with same current flowing in the same direction is 5 x 10⁻N/m repulsive force.

Explanation:

Given;

current though the two parallel wires, I₁ and I₂ = 5A

distance between the two wires, R = 1 m

The force per unit of the wires is calculated as;

[tex]\frac{F}{L} = \frac{\mu_o I_1I_2}{2\pi R}[/tex]

Where;

μ₀ is permeability of free space = 4π x 10⁻⁷ m/A

Substitute in the given values into the equation and determine the force per unit length (F/L).

[tex]\frac{F}{L} = \frac{\mu_o I_1I_2}{2\pi R} \\\\ \frac{F}{L} = \frac{4\pi *10^{-7}*5*5}{2\pi *1}\\\\ \frac{F}{L} = 5*10^{-6} \ N/m \ (repulsive)[/tex]

Therefore , the force per unit between the two parallel wires with same current flowing in the same direction is 5 x 10⁻ N/m repulsive force.

If you unbend a paper clip made from 1.5 millimeter diameter wire and push one end against the wall, what force must you apply to give a pressure of 120 atmospheres

Answers

Answer:

The force is  [tex]F = 21.48 \ N[/tex]

Explanation:

From the question we are told that

   The diameter of the wire is  [tex]d = 1.5 \ mm = 1.5 *10^{-3} \ m[/tex]

     The  pressure is  [tex]P = 120 \ a.t.m = 120 * 101.3 *10^{3} = 12156000 Pa[/tex]

Generally the radius of the of the wire is  

     [tex]r = \frac{d}{2}[/tex]

=>    [tex]r = \frac{ 1.5 *10^{-3}}{2}[/tex]  

=>    [tex]r = 7.5 *10^{-4} \ m[/tex]

The Area is evaluated as

     [tex]A = \pi r^2[/tex]

=>    [tex]A = 3.142 * 7.5 *10^{-4}[/tex]

=>    [tex]A = 1.7673*10^{-6} \ m^2[/tex]

Generally pressure is mathematically represented as

     [tex]P = \frac{F}{A }[/tex]

=>   [tex]F = P* A[/tex]

=>    [tex]F = 12156000 * 1.767*10^{-6}[/tex]

=>    [tex]F = 21.48 \ N[/tex]

The tires of a car make 77 revolutions as the car reduces its speed uniformly from 92.0 km/h to 60.0 km/h. The tires have a diameter of 0.84 m.
1. What was the angular acceleration of the tires?
2. If the car continues to decelerate at this rate, how much more time is required for it to stop?
3. If the car continues to decelerate at this rate, how far does it go? Find the total distance.

Answers

The angular acceleration of the tires is -2.2 rad/s².

If the car continues to decelerate at this rate, the time required to stop is 27.66 s.

The total distance traveled by the car before stopping is 210.96 revolutions.

The given parameters;

number of revolutions of the tire, N = 77 revinitial linear speed of the car, u  = 92 km/h = 25.56 m/sfinal linear speed of the tire, v = 60 km/h = 16.67 m/sdiameter of the tire, d = 0.84 mradius of the tire, r = 0.42 m

The angular acceleration of the tire is calculated as follows;

[tex]\omega _f^2 = \omega _i ^2 + 2\alpha \theta\\\\(\frac{16.67}{0.42} )^2 = (\frac{25.56}{0.42} )^2 + 2( 77 \ rev \times \frac{2 \pi \ rad}{1 \ rev} ) \alpha \\\\1575.33 = 3703.59 \ + \ 967.736 \alpha \\\\-2128.26 = 967.736 \alpha\\\\\alpha = \frac{-2128.26}{967.736} \\\\\alpha = - 2.2 \ rad/s^2[/tex]

When the car stops, the final angular speed = 0. The time for the motion is calculated as;

[tex]\omega _f = \omega _i + \alpha t\\\\0 = \omega _i + \alpha t\\\\0 = 60.86 + (-2.2)t\\\\0 = 60.86 - 2.2t\\\\2.2t = 60.86\\\\t = \frac{60.86}{2.2} \\\\t = 27.66 \ s[/tex]

The total distance traveled by the car before stopping;

[tex]\theta = \omega_i t + \frac{1}{2} \alpha t^2\\\\\theta = (60.86 \times 27.66) \ + \ (0.5 \times -2.2\times 27.66^2)\\\\\theta = 841.8 \ rad\\\\\theta = 841.8 \ rad \times\frac{1 \ rev}{2\pi \ rad} = 133.96 \ rev[/tex]

total distance = 133.96 + 77 = 210.96 revolutions.

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What is the Malebioncy of a Capacitor?

Answers

Answer:

The switching rate between the steady state and the normal state of a capacitor

Explanation:

This was a hard one! Could only find it in my textbook. Anyways this basically is the rate which the capacitors switches back from steady state and normal state from when it charges and discharges over time. This has many purposes as a special type of diode or any other transistor type device etc etc.

Monochromatic light of wavelength, lambda, is traveling in air. The light then strikes a thin film having an index of refraction, n1 that is coating a material having an index of refraction n2. If n1 is larger than n2, what minumim film thickness will result in minimum reflection of this light?A. lambda/(4*n2)B. lambda/n2C. lambda/4D. lambda(2*n1)E. lambdaF. lambda/(2*n2)G. lambda/n1H. lambda/(4n1)I. lambda/2

Answers

Answer:

The  correct option is  H

Explanation:

From the question we are told that

    The index of refraction of  coating is  [tex]n_1[/tex]

       The  index of refraction of material  is  [tex]n_2[/tex]

   

Generally the condition for constructive for a thin film interference is mathematically represented

            [tex]2 * t = [ m + \frac{1}{2}] \frac{\lambda}{n_1 }[/tex]

Here  t represents the thickness

For minimum thickness  m =  0

So

           [tex]2 * t =0 + \frac{1}{2}\frac{\lambda}{n_1 }[/tex]

=>        [tex]t =\frac{\lambda}{4n_1 }[/tex]

4. How does the type of medium affect a sound wave?

Answers

Answer:

The type of medium affects a sound wave as sound travels with the help of the vibration in particles.

Explanation:

As different mediums have different amount and size of particles, for example, the speed of sound is faster through solid than liquid as solids have closely packed particles whereas liquids are loosely packed.

The speed of sound in a given medium is determined by its density and stiffness (or compressibility in the case of gases).The speed of sound increases with the rigidity (or lack of compressibility) of the medium. The speed of sound decreases with increasing medium density.

What type of medium affect a sound wave?

Any material or area through which a wave is transmitted is referred to as a medium. Four variables impact a wave's speed: wavelength, frequency, medium, and temperature. The wavelength and frequency are multiplied to determine the wave speed (speed = l × f).

Therefore, The rate at which energy is transferred through a medium depends on the amplitude of the vibrations of its constituent particles; the higher this rate, the more powerful the sound wave.

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You drive 6.00 km at 50.0 km/h and then another 6.00kmat 900 km/h Your average speed over
the 12.0 km drive will be

Answers

Explanation:

average speed = total distance travelled / total time travelled

time to travel the first 6km: 6 / 50 = 3/25 (h)

time to travel the next 6km: 6 / 90 = 1/15 (h)

[I think there's problem in the question 'cause 900km/h sounds impossible for normal person to travel in normal condition]

The total time: 3/25 + 1/15 = 14/75 (h)

Average speed over the 12 km drive will be:

[tex] \frac{12}{ \frac{14}{75} } = \frac{450}{7} = 64.3 \: km{h}^{ - 1} [/tex]

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