A speeder passes a parked police car with a speed of 65 km/h in a 50km/h zone. One second after the speeder has passed the police car, the police begin his pursuit. The police car accelerates with constant acceleration of 2m/s^2.Required:a. How long does it take for the police to catch the speeding car? b. How far did the police car travel before police caught up with the speeder? c. What is the speed of the police car when catches up with the speeder?

Answers

Answer 1

Answer:

a) The police will take 18.056 seconds to catch the speedy car, b) The police will travel 326.019 meters before catching the speedy car, c) The speed of the police car when catches up with the speeder is 36.112 meters per second.

Explanation:

Let suppose that speeder moves in a uniform motion, whereas police car has an uniformly accelerated motion.

a) How long does it take for the police to catch the speeding car:

Kinematic equation of each vehicle's position are described:

Speeder

[tex]s_{A} = s_{A,o}+v_{A}\cdot t[/tex]

Police Car

[tex]s_{B} = s_{B,o}+v_{B,o}\cdot t + \frac{1}{2}\cdot a_{B}\cdot t^{2}[/tex]

If [tex]s_{A} = s_{B}[/tex], [tex]s_{A,o} = s_{B,o}[/tex], [tex]v_{A} = 18.056\,\frac{m}{s}[/tex], [tex]v_{B,o} = 0\,\frac{m}{s}[/tex] and [tex]a_{B} = 2\,\frac{m}{s^{2}}[/tex], the resulting expression is done:

[tex]v_{A} \cdot t = v_{B,o}\cdot t +\frac{1}{2}\cdot a_{B}\cdot t^{2}[/tex]

[tex]\frac{1}{2}\cdot a_{B}\cdot t^{2}+(v_{B,o}-v_{A})\cdot t = 0[/tex]

[tex]t \cdot \left(\frac{1}{2}\cdot a_{B}\cdot t +v_{B,o}-v_{A} \right)= 0[/tex]

[tex]t = 0\,s\,\wedge\, t = \frac{2\cdot (v_{A}-v_{B,o})}{a_{B}}[/tex]

[tex]t = \frac{2\cdot \left(18.056\,\frac{m}{s}-0\,\frac{m}{s} \right)}{2\,\frac{m}{s^{2}} }[/tex]

[tex]t = 18.056\,s[/tex]

The police will take 18.056 seconds to catch the speedy car.

b) How far did the police car travel before police caught up with the speeder?

The distance travelled by the police is: ([tex]s_{B,o} = 0\,m[/tex],  [tex]a_{B} = 2\,\frac{m}{s^{2}}[/tex], [tex]v_{B,o} = 0\,\frac{m}{s}[/tex] and  [tex]t = 18.056\,s[/tex])

[tex]s_{B} = 0\,m + \left(0\,\frac{m}{s} \right)\cdot (18.056\,s)+\frac{1}{2}\cdot \left(2\,\frac{m}{s^{2}} \right) \cdot (18.056\,s)^{2}[/tex]

[tex]s_{B} = 326.019\,m[/tex]

The police will travel 326.019 meters before catching the speedy car.

c) What is the speed of the police car when catches up with the speeder?

The speed of the police car is represented by the following formula:

[tex]v_{B} = v_{B,o} + a_{B}\cdot t[/tex]

Where [tex]v_{B}[/tex] is the speed of the police car, measured in meters per second.

Given that [tex]v_{B,o} = 0\,\frac{m}{s}[/tex], [tex]t = 18.056\,s[/tex] and [tex]a_{B} = 2\,\frac{m}{s^{2}}[/tex], the final speed of the police car when catches up with the speeder is:

[tex]v_{B} = 0\,\frac{m}{s}+\left(2\,\frac{m}{s^{2}} \right)\cdot (18.056\,s)[/tex]

[tex]v_{B} = 36.112\,\frac{m}{s}[/tex]

The speed of the police car when catches up with the speeder is 36.112 meters per second.


Related Questions

Public television station KQED in San Francisco broadcasts a sinusoidal radio signal at a power of 777 kW. Assume that the wave spreads out uniformly into a hemisphere above the ground. At a home 5.00 km away from the antenna,
(a) what average pressure does this wave exert on a totally reflecting surface,
(b) what are the amplitudes of the electric and magnetic fields of the wave, and
(c) what is the average density of the energy this wave carries?
(d) For the energy density in part (c), what percentage is due to the electric field and what percentage is due to the magnetic field?

Answers

Answer:

A) P = 3.3 × 10^(-11) Pa

B) Amplitude of electric field = 1.931 N/C

Amplitude of magnetic field = 6.44 × 10^(-9) T

C) μ_av = 1.65 × 10^(-11) J/m³

D) 50% each for the electric and magnetic field

Explanation:

A) First of all let's calculate intensity.

I = P_av/A

We are given;

P_av = 777 KW = 777,000 W

Distance = 5 km = 5000 m

Thus;

I = 777000/(2π × 5000²)

I = 0.00495 W/m²

Now, the average pressure would be given by the formula;

P = 2I/C

Where C is speed of light = 3 × 10^(8) m/s

P = (2 × 0.00495)/(3 × 10^(8))

P = 3.3 × 10^(-11) Pa

B) Formula for the amplitude of the electric field is gotten from;

E_max = √[2I/(εo•c)].

Where εo is the Permittivity of free space with a constant value of 8.85 × 10^(−12) c²/N.mm²

I and c remain as before.

Thus;

E_max = √[(2 × 0.00495)/(8.85 × 10^(−12) × 3 × 10^(8))]

E_max = √3.72881355932

E_max = 1.931 N/C

Formula for amplitude of magnetic field is gotten from;

B_max = E_max/c

B_max = 1.931/(3 × 10^(8))

B_max = 6.44 × 10^(-9) T

C) Formula for average density is;

μ_av = εo(E_rms)²

Now, E_rms = E_max/√2

Thus;

E_rms = 1.931/√2

μ_av = 8.85 × 10^(−12) × (1.931/√2)²

μ_av = 1.65 × 10^(-11) J/m³

D) The energy density for the electric and magnetic field is the same. So both of them will have 50% of the energy density.

if a cart goes around a turn at 20 km/h ,what remains constant

1.position
2.velocity
3.direction
4.speed

Answers

Answer: 4.speed

Explanation:

In this case, we know that the cart remains at a constant 20km/h.

Now, one could say that "the velocity remains constant, because it always is 20km/h"

But remember that velocity is a vector, so this has a direction, and if the cart is going around a turn, then the direction of motion is changing, which tell us that there is acceleration.

But the module of the velocity, the speed, remains constant at 20km/h.

Then the correct option is 4, speed.

A ball of mass m moving with speed V collides with another ball of mass 2m (e= 1/2) in a horizontal smooth fixed circular tube of radius R (R is sufficiently large R>>>d). The time after which next collision will take place is:________

Answers

Answer:

[tex]$ \frac{4\pi R}{V}$[/tex]

Explanation:

Given :

Mass of ball 1 = m

Mass of ball 2 = 2m

Since, R>>>d, the collision is head on.

Therefore, we get

[tex]$ \frac{v_1 -v_2}{V}=\frac{1}{2}$[/tex]

[tex]$ \therefore \frac{\text{velocity of seperation}}{\text{velocity of approach}}= v_1-v_2 = \frac{V}{2}$[/tex]

Relative velocity is given by V/2. So, we get the time when the masses will again collide as

[tex]$ t = \frac{2\pi R}{\frac{V}{2}}=\frac{4\pi R}{V} $[/tex]

2. If a cyclist in the Tour de France traveled southwest a distance of 12,250 meters in one hour, what would the velocity of the cyclist be?​

Answers

Answer:

12,25 km/h

≈ 3,4 m/s

Explanation:

v = d/t

= 12250m/h

= 12,25km/h

or

v = d/t

= 12250m/h

1h = 60m×60s = 3600s

= 12250m/3600s

≈ 3,4 m/s

(7) A 2500 lbm car moving at 25 mi/hr is accelerated at a constant rate of 15 ft/s2 up to a speed of 50 mi/hr. What is the force and total time required?

Answers

Answer:

The  force is  [tex]F =  1164.6\  lbf[/tex]

The time is   [tex]\Delta t =  2.44 \  s[/tex]

Explanation:

From the question we are told that

  The  mass of the car is  [tex]m  =  2500 \ lbm[/tex]

   The  initial velocity of the car is [tex]u  =  25 \ mi/hr[/tex]

   The final  velocity of the car is  [tex]v  =  50 \  mi/hr[/tex]

  The acceleration is  [tex]a =  15 ft/s^2 =  \frac{15 *  3600^2}{ 5280} =  36818.2 \  mi/h^2[/tex]

   

Generally the acceleration is mathematically represented as

      [tex]a =  \frac{v-u}{\Delta t}[/tex]

=>   [tex]36818.2 =  \frac{50 - 25 }{ \Delta t}[/tex]

=>   [tex]t = 0.000679 \  hr[/tex]

converting to seconds

       [tex]\Delta t =  0.0000679 *  3600[/tex]

=>     [tex]\Delta t =  2.44 \  s[/tex]

Generally the force is mathematically represented as

        [tex]F  =  m * a[/tex]

=>      [tex]F  =  2500 *  15[/tex]

=>      [tex]F  =  37500 \ \frac{lbm *  ft}{s^2}[/tex]

Now converting to foot-pound-second we have  

         [tex]F =  \frac{37500}{32.2}[/tex]

=>        [tex]F =  1164.6\  lbf[/tex]

Which scientist determined that electrons had predicted zones orbiting the nucleus?

Answers

Answer:

Rutherford

Explanation:

Because

Schrödinger I just took the unit review

Explain why a Merry-Go-Round and a Ferris Wheel have a constant acceleration when they are moving?

Answers

Answer:

merry go round and Ferris wheel have a constant acceleration due to the change in direction at every point.

Answer:

A merry-go-round is accelerating. Acceleration is a change in speed, direction, or both. Even though the speed of the merry-go-round does not change, its direction constantly changes as it spins.

Explanation:

Two pounds of water vapor at 30 psia fill the 4-ft3 left chamber of a partitioned system. The right chamber has twice the volume of the left and is initially evacuated. Determine the pressure of the water after the partition has been removed and enough heat has been transferred so that the temperature of the water is 40oF.

Answers

Answer:

3.38atm

Explanation:

Using data from the steam table we have that

Moles of water vapour = 907.19 / 18

= 50.4 moles

So

p1 = 30 psi = 30 x 0.68 = 2.04 atm

v1 = 4ft³= 113.2 L

Then from

PV= nRT

Then to find T we use

T1 = p1 V1 / n R

= 2.04 x 113.2 / 50.4 x 0.0821

= 55.8 K

Then to find volume two

v2 = 2v1 + v1

So

3 v1 = 339.6 K

The pressure two we use

P2 = n R T2 / V2

= 50.4 x 0.0821 x 277.6 / 339.6

So we have

= 3.38 atm =

A -5.40nC point charge is on the x axis at x = 1.25m . A second point charge Q is on the x axis at -0.625m.
A) What must be the charge Q for the resultant electric field at the origin to be 50.0N/C in the +x direction?
B) What must be the charge Q for the resultant electric field at the origin to be 50.0N/C in the -x direction?

Answers

Answe

a)  Q = 0.820 10⁻⁹ C ,   b)  Q = -3.52 10⁻⁹ C

Explanation:

The electric field is given by the formula

         E = k q / r²

where E is a vector quantity, so it must be added as a vector

          E_total = E₁ + E₂

let's look for the two electric fields

           E₁ = k q₁ / r₁²

           E₁ = 9 10⁹  5.4 10⁻⁹ / 1.25²

           E₁ = 31.10 N / C

           E2 = k Q / r₂²

           E2 = 9 10⁹ Q / 0.625²

           E2 = 23.04 10⁹ Q N / C           (1)

now we can solve the two cases presented

a) The total field is

            E_total = 50.0 N / C towards + x

since the test charge is positive the electric field E1 points to the right in the direction of the + x axis, so the equation is

            E_total = E1 + E₂

             E₂ = E_toal - E₁

             E₂ = 50.0 -31.10

             E2 = 18.9 N /C

With the value of the electric field we can calculate the charge (Q) using equation 1

             E₂ = 23.04 10⁹ Q

              Q = E₂ / 23.04 10⁹

              Q = 18.9 / 23.04 10⁹

              Q = 0.820 10⁻⁹ C

the charge on Q is positive

b) E_total = -50.0 N / C

              E_total = E₁ + E₂

              E₂ = E_total - E₁

              E2 = -50.0 - 31.10

               E2 = -81.10 N /C

we calculate the charge

             Q = E2 / 23.04 10⁹

             Q = -81.1 / 23.04 10⁹

              Q = -3.52 10⁻⁹ C

for this case the charge is negative

If one object (a) is moving at 60m/s^2, and the other object (b) is moving at 65m/s^2, at what time will the faster moving object be 10m ahead of the other object?

Answers

Answer:

a is moving at 60m and the other object

A commuter backs her car out of her garage with an acceleration of 1.40 m/s2.A) How long does it take her to reach a speed of 2.00 m/s?B) If she then breaks to a stop in 0.800 s, what is her deceleration?

Answers

Answer:

(A) 1.43secs

(B) -2.50m/s^2

Explanation:

A commuter backs her car out of her garage with an acceleration of 1.40m/s^2

(A) When the speed is 2.00m/s then, the time can be calculated as follows

t= Vf-Vo/a

The values given are a= 1.40m/s^2 , Vf= 2.00m/s, Vo= 0

= 2.00-0/1.40

= 2.00/1.40

= 1.43secs

(B) The deceleration when the time is 0.800secs can be calculated as follows

a= Vf-Vo/t

= 0-2.00/0.800

= -2.00/0.800

= -2.50m/s^2

A satellite dish has the shape of a parabola when viewed from the side. The dish is inches wide and inches deep. How far is the receiver from the bottom of the dish if the receiver is placed at the focus

Answers

Complete question is;

A satellite dish has the shape of a parabola when viewed from the side. The dish is 60 inches wide and 45 inches deep. How far is the receiver from the bottom of the dish if the receiver is placed at the focus?

Answer:

the receiver should be put 40 inches from the bottom of the dish on the concave side of the dish

Explanation:

The base of the dish would simply be the vertex of parabola.

Since we want to find how far the receiver is from the bottom, the place where we'll place the receiver is simply the focus of the parabola.

Now, for example, if this is a parabola that opens upward and has it's vertex at the origin, then half of the diameter at a height of 45 inches gives the two points (60, 22.5) and (-60, 22.5)

Standard form equation of parabola with vertex at origin and pointing upwards is given by;

x² = 4ay

Plugging in the values of x and y gives;

60² = 4a(22.5)

3600/90 = a

a = 40 inches

Thus, the receiver should be put 40 inches from the vertex on the concave side of the dish

Finally, consider the expression (6.67 x 10^-11)(5.97 x 10^24)/(6.38 x 10^6)^2 Determine the values of a and k when the value of this expression is written in scientific rotation. Enter a and k, separated by commas.

Answers

Explanation:

We need to find the value of following expression :

[tex]\dfrac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{(6.38 \times 10^6)^2}[/tex]

Firstly, solving the numerator of the above expression :

[tex]=\dfrac{39.8199\times 10^{-11+24}}{40.7044\times 10^{12}}\\\\=\dfrac{39.8199\times 10^{13}}{40.7044\times 10^{12}}\\\\=9.7827[/tex]

Rounding off the result = 9.78

In scientific notation : [tex]9.78\times 10^0[/tex]

The value of a = 9.78 and k = 0.

What happens to the temperature as altitude increases in the exosphere? Does it increase or decrease the higher it goes?

Answers

Answer:

it gets colder the higher you go

Gwen releases a rock at rest from the top of a 40-m tower. If g = 9.8 m/s2 and air resistance is negligible, what is the speed of the rock as it hits the ground?

Answers

Answer:

[tex]28\; \rm m \cdot s^{-1}[/tex].

Explanation:

Short Explanation

Apply the SUVAT equation [tex]\left(v^2 - u^2) = 2\, a \, x[/tex], where:

[tex]v[/tex] is the final velocity of the object,[tex]u[/tex] is the initial velocity of the object, [tex]a[/tex] is the acceleration (should be constant,) and[tex]x[/tex] is the displacement of the object while its velocity changed from [tex]v[/tex] to [tex]u[/tex].

Assume that going downwards corresponds to a positive displacement. For this question:

[tex]v[/tex] needs to be found.[tex]u = 0[/tex] because the rock is released from rest.[tex]a = g = 9.8 \; \rm m\cdot s^{-2}[/tex].[tex]x = 40\; \rm m[/tex].

Solve this equation for [tex]v[/tex]:

[tex]\displaystyle v = \sqrt{2\, a\, x + u^2} = \sqrt{2\times 9.8 \times 40} = 28\; \rm m \cdot s^{-1}[/tex].

In other words, the rock reached a velocity of [tex]28\; \rm m\cdot s^{-1}[/tex] (downwards) right before it hits the ground.

Explanation

Let [tex]v[/tex] be the velocity (in [tex]\rm m \cdot s^{-1}[/tex]) of this rock right before it hits the ground. Under the assumptions of this question, it would take a time of [tex]t = (v / 9.8)[/tex] seconds for this rock to reach that velocity if it started from rest and accelerated at [tex]9.8\; \rm m \cdot s^{-2}[/tex].

Note that under these assumptions, the acceleration of this rock is constant. Therefore, the average velocity of this rock would be exactly one-half the sum of the initial and final velocity. In other words, if [tex]u[/tex] denotes the initial velocity of this rock, the average velocity of this rock during the fall would be:

[tex]\displaystyle \frac{u + v}{2}[/tex].

On the other hand, [tex]u = 0[/tex] because this stone is released from rest. Therefore, the average velocity of this rock during the fall would be exactly [tex](v / 2)[/tex].

The displacement of an object over a period of time is equal to the length of that period times the average velocity over that period. For this rock, the length of this fall would be [tex]t = (v / 9.8)[/tex], while the average velocity over that period would be [tex](v / 2)[/tex]. Therefore, the displacement (in meters) of the rock during the entire fall would be:

[tex]\displaystyle \left(\frac{v}{2}\right) \cdot \left(\frac{v}{9.8}\right) = \frac{v^2}{19.6}[/tex].

That displacement should be equal to the change in the height of the rock, [tex]40\; \rm m[/tex]:

[tex]\displaystyle \frac{v^2}{19.6} = 40[/tex].

Solve for [tex]v[/tex]:

[tex]v = 28\; \rm m \cdot s^{-1}[/tex].

Once again, the speed of the rock would be [tex]28\;\rm m \cdot s^{-1}[/tex] right before it hits the ground.

A Lotus will travel 275 meters in 4.71 seconds. What is this car's average speed?

Answers

390 Because it is what it is

The speed of a bus increases uniformly from
15 ms- to 60 ms in 20 seconds. Calculate
a. the average speed,
b. the acceleration,
C. the distance travelled during the entire
period The speed of a bus increases uniformly from
15 ms- to 60 ms in 20 seconds. Calculate
a. the average speed,
b. the acceleration,
C. the distance travelled during the entire
period​

Answers

Explanation:

a. For constant acceleration:

v_avg = ½ (v + v₀)

v_avg = ½ (60 m/s + 15 m/s)

v_avg = 37.5 m/s

b. a = (v − v₀) / t

a = (60 m/s − 15 m/s) / 20 s

a = 2.25 m/s²

c. x = v_avg t

x = (37.5 m/s) (20 s)

x = 750 m

which water molecules have the greatest kinetic energy

Answers

The higher the temperature the substance is, the more energy in it because the particles are moving a lot more around in it. So therefore, steam which has the highest temperature, has the greatest kinetic energy.

An 85 kg skydiver is falling through the air at a constant speed of 195 km h-1. At what rate does air resistance remove energy from the skydiver?

Answers

Answer:

46041J

Explanation:

Using Energy lost= mgh

Changing to standard its we have

= 195*1000/3600=54.2m/s

So = 85*54.2*10= 46041J

Answer:

45167.15 J/s

Explanation:

mass of the man = 85 kg

The man's speed = 195 km/h = 195 x 1000/3600 = 54.167 m/s

The man's weight = mg

where

m is the mass

g is acceleration due to gravity = 9.81 m/s^2

weight = 85 x 9.81 = 833.85 N

The rate at which energy is removed from the man = speed x weight

==> 54.167 x 833.85 = 45167.15 J/s

A paper airplane is thrown horizontally with a velocity of 20 mph. The plane is in the air for 7.63 s before coming to a standstill on the ground. What is the acceleration of the plane?

Answers

Answer:

-1.17 m/s²

Explanation:

Given:

v₀ = 20 mph = 8.94 m/s

v = 0 m/s

t = 7.63 s

Find: a

v = at + v₀

0 m/s = a (7.63 s) + 8.94 m/s

a = -1.17 m/s²

The acceleration of the plane will be:

"-1.17 m/s²".

Acceleration and Velocity

According to the question,

Velocity, v₀ = 20 mph or,

                   = 8.94 m/s

and,

                v = 0 m/s

Time, t = 7.63 s

We know the relation,

→ v = at + v₀

By substituting the values,

  0 = a × 7.63 + 8.94

7.63a = - 8.94

      a = -[tex]\frac{8.94}{7.63}[/tex]

         = - 1.17 m/s²  

Thus the response above is correct.

Find out more information about velocity here:

https://brainly.com/question/742413

A ball with a mass of 3.7 kg is thrown downward with an initial velocity of 8 m/s from a high building. How fast will it be moving after 3 seconds?

Answers

Answer:

v=37.4 m/s

Explanation:

It is given that,

Mass of a ball, m = 3.7 kg

Initial velocity of the ball is u = 8 m/s

We need to find its velocity after 3 seconds. It is moving downwards. The equation of motion is this case is

v=u+gt

[tex]v=8+9.8\times 3\\\\v=37.4\ m/s[/tex]

So, the velocity of the ball after 3 seconds is 37.4 m/s.

A bungee jumper with mass 65.0 kg jumps from a high bridge. After reaching his lowest point, he oscillates up and down, hitting a low point eight more times in 38.0 s. After many oscillations, he finally comes to rest 25.0 m below the level of the bridge. Calculate the spring stiffness constant and the unstretched length of the bungee cord.

Answers

Explanation:

It is given that,

Mass of a bungee jumper is 65 kg

The time period of the oscillation is 38 s, hitting a low point eight more times.It means its time period is

[tex]T=\dfrac{38}{8}\\\\T=4.75\ s[/tex]

After many oscillations, he finally comes to rest 25.0 m below the level of the bridge.

For an oscillating object, the time period is given by :

[tex]T=2\pi \sqrt{\dfrac{m}{k}}[/tex]

k = spring stiffness constant

So,

[tex]k=\dfrac{4\pi ^2m}{T^2}\\\\k=\dfrac{4\pi ^2\times 65}{(4.75)^2}\\\\k=113.43\ N/m[/tex]

When the cord is in air,

mg=kx

x = the extension in the cord

[tex]x=\dfrac{mg}{k}\\\\x=\dfrac{65\times 9.8}{113.6}\\\\x=5.6\ m[/tex]

So, the unstretched length of the bungee cord is equal to 25 m - 5.6 m = 19.4 m

The spring stiffness constant is 116.7 N/m and the the unstretched length of the bungee cord is 19.54 m.

The given parameters;

mass of the bungee jumper, m = 65 kgtime of motion, t = 38 sdistance to come to rest, d = 25 m

The period of oscillation of the bungee jumper is calculated as follows;

[tex]T = \frac{t}{n} \\\\T = \frac{38}{8} \\\\T = 4.75 \ s[/tex]

The spring stiffness constant is calculated as follows;

[tex]T = 2\pi \sqrt{\frac{m}{k} } \\\\\sqrt{\frac{m}{k} } = \frac{T}{2\pi} \\\\k = m \times \frac{T^2}{4\pi^2} \\\\k = 65 \times \frac{(4.75)^2}{4\pi ^2} \\\\k = 116.7 \ N/m[/tex]

The extension of the cord is calculated as follows;

[tex]F = kx\\\\mg = kx\\\\x = \frac{mg}{k} \\\\x = \frac{65 \times 9.8}{116.7} \\\\x = 5.46 \ m[/tex]

The unstretched length of the bungee cord is calculated as;

[tex]\Delta x = l_2-l_1\\\\l_1 = l_2 - \Delta x\\\\l_1 = 25 - 5.46\\\\l_1 = 19.54 \ m[/tex]

Learn more here:https://brainly.com/question/18648366

Assuming it is a van der Waals gas, calculate the critical temperature, pressure and volume for CO2. (a = 3.610 atm L2 mol-2, b = 0.0429 L mol-1)
pc = ___ atm
Tc = ___ K
Vc = ___ L/mol

Answers

Answer

To get critical pressure

We use

Pc = a/(27b²)

So

= 3.610/(27 X 0.0429²)

We have

= 72.7 atm

Critical temperaturewe

We use

Tc = 8a/27Rb

= 8 x 3.610/(27 x 0.0812 x 0.0429)

= 307 K

Critical volume

We use

Vc =3b =

3 x 0.0429

= 0.129L/mol

4. Lead has a density of 11.5g/cmº. A rectangular block of lead measures 7cm x5cmx2cm.
a) Find the volume of the block of lead.
b) Find the mass of the block of lead

Answers

Answer:

(a) 70cm³

(b) 805 grams

Explanation:

(a) V = L×B×H

= 7cm×5cm×2cm

= 35cm×2cm

= 70cm³

(b) Mass = Volume × Density

= 70cm³ × 11.5g/cm³

= 805 grams

A car stops in 120 m. If it has an acceleration of –5m/s 2 , how long did it take to stop

Answers

Answer:

t=240s

Explanation:

Distance=120m

Acceleration=-5m/s^2

v=0

Let u=x m/s

Using equation v^2-u^2=2as:-

0-x=2(-5)(120)

-x=-1200

x=1200m/s

Using now equation v=u+at:-

0=1200+-5t

5t=1200

t=240s

If a car stops at 120 meters. If it has an acceleration of –5 meters/second², then it would take  6.928 seconds to stop.

What is acceleration?

The rate of change of the velocity with respect to time is known as the acceleration of the object.

As given in the problem a car stops at 120 meters. If it has an acceleration of –5 meters/second², then we have to find out how long it would take seconds to stop.

By using the second equation of motion,

s = ut + 1/2at²

The distance traveled by car before stopping = 120 meters

acceleration =  –5 meters/second²

-120 = 0 + 0.5×( –5)t²

t² = 120/2.5

t² =48

t = 6.928 seconds

Thus, the time taken by the car before stopping would be 6.928 seconds.

To learn more about acceleration from here, refer to the link;

brainly.com/question/2303856

#SPJ2

A stone is thrown vertically upward with a speed of 28.0 m/s how much time is required to reach this height

Answers

A stone is thrown vertically upward with a speed of 17.0 m/s. How fast is it moving when it reaches a height of 11.0 m? How long is required to reach this height?

Let’s review the 4 basic kinematic equations of motion for constant acceleration (this is a lesson – suggest you commit these to memory):

s = ut + ½ at^2 …. (1)

v^2 = u^2 + 2as …. (2)

v = u + at …. (3)

s = (u + v)t/2 …. (4)

where s is distance, u is initial velocity, v is final velocity, a is acceleration and t is time.

In this case, we know u = 17.0m/s, a = -g = -9.81m/s^2, s = 11.0m and we want to know v and t, so from equation (2):

v^2 = u^2 + 2as

v^2 = 17.0^2 -2(9.81)(11.0)

v = √73.18 = 8.55m/s

now from equation (3):

v = u + at

8.55 = 17.0 – 9.81t

t = (8.55 – 17.0)/(-9.81) = 0.86s

When charges qa, qb, and qc are placed respectively at the corners a, b, and c of a right triangle, the potential at the midpoint of the hypotenuse is 20 V. When the charge qa is removed, the potential at the midpoint becomes 15 V. When, instead, the charge qb is removed (qa and qc both in place), the potential at the midpoint becomes 12 V. What is the potential at the midpoint if only the charge qc is removed from the array of charges?

Answers

Answer:

8v

Explanation:

First we apply super position principle

Vt= v1 + v2+ v3

Remove qa

But vt= 20v

So V = v2+v3

V1= 20-15

= 5v

Remove qb

V= v1+v3

V=8v

So the potential when qa and qc are remove is the potential due to qb

Which is 8v

A ski lift has a one-way length of 1 km and a vertical rise of 200 m. The chairs are spaced 20 m apart, and each chair can seat three people. The lift is operating at a steady speed of 10 km/h. Neglecting friction and air drag, and assuming that the average mass of each loaded chair is 250 kg, determine the power required to operate this ski lift. Also, estimate the power required to accelerate this ski lift in 17 s to its operating speed when it is first turned on.

Answers

Answer:

Explanation:

The question states that the chairs are spaced 20 m apart through a length of 1 km, or say, 1000 m.

It also does say that each chair weighs 250 kg, and as such the load is

M = 50 * 250

M = 12500.

Taking into consideration, the initial and final heights, we have

h1 = 0, h2 = 200 m

The work needed to raise the chairs,

W = mgh, where h = h2 - h1

W = 12500 * 9.81 * (200 - 0)

W = 2.54*10^7 J

The work is done at a rate of 10 km/h, and at a distance of 1 km, time taken would be

t = 1/10 = 0.1 h or say, 360 s

The power needed thus, is

P = W/t

P = 2.54*10^7 / 360

P = 68125 W, or 68 kW

Initial velocity, u = 0 m/s

Final velocity, v = 10 km/h = 2.78 m/s

Startup time, t is 17 s

Acceleration during the startup then is

a = (v - u)/t

a = 2.78/17

a = 0.163 m/s²

The power needed for the acceleration is

P = ½m [(v² - u²)/t]

P = ½ * 12500 * [2.78²/17]

P = 6250 * 0.455

P = 2844 W

Two 100kg bumper cars are moving towards eachother in oppisite directions. Car A is moving at 8 m/s and Car B at -10 m/s when they collide head on. If the resulting velocity of Car B after the collision is 8 m/s, what is the velocity of Car A after the collision

Answers

Answer:

[tex]-10 m/s[/tex]

Explanation:

When two cars collide then the momentum of two cars will remains conserved

Mass of two cars = 100 kg Speed of car A = 8 m/s Speed of car B = - 10 m/s After collision the speed of car B = +8 m/s

By momentum conservation equation

               [tex]m1v1i+m2v2i=m1v1f + m2v2f[/tex]

               [tex](100)(8)+(100)(-10)=(100v)+(100)(8)\\ v=-10 m/s[/tex]

How does sleep affect your ability to handle stress?

Answers

Answer: Stress can adversely affect sleep quality and duration, while insufficient sleep can increase stress levels. Both stress and a lack of sleep can lead to lasting physical and mental health problems.

Explanation:

Many report that there stress increases when the length and quality of their sleep decreases. When you do not get enough sleep, 21 percent of adults report feeling more stressed.

Sleep affects your ability to handle stress because when you have more sleep you are able to think more clearly and you’re more energized and happy throughout the day.
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