How much ice at a temperature of -22.3 ∘C must be dropped into the water so that the final temperature of the system will be 30.0 ∘C ?How much ice at a temperature of -22.3 ∘C must be dropped into the water so that the final temperature of the system will be 30.0 ∘C ?

Answer:

(i) The period, T is 0.25 s

(ii) The angular velocity, ω is 25.136 rad/s

(iii) The linear velocity is, v is 50.27 m/s

Explanation:

Given;

number of revolution of the particle, N = 240 revolutions per minutes

radius of the circle, R = 2m

(i) The period is given by;

[tex]T = \frac{2\pi}{\omega}[/tex]

where;

ω is angular velocity  

[tex]\omega = (240 \frac{rev}{min}) *(\frac{1 \ min}{60 \ s} )*(2\pi \frac{rad}{rev})\\\\ \omega = 25.136 \ rad/s[/tex]

[tex]T = \frac{2\pi }{\omega}\\\\ T = \frac{2\pi}{25.136}\\\\T = 0.25 \ s[/tex]

(ii) angular velocity, ω  = 25.136 rad/s

(iii) linear velocity is given by;

v = ωR

v = (25.136 rad/s) x (2 m)

v = 50.27 m/s

Answer:

x₂ = 0.01336

Explanation:

In this exercise we use the translational equilibrium equation

          F₁₃ - F₂₃ = 0

          F₁₃ = F₂₃

at the point where charge 3 is placed the two electric forces have the same magnitude

let us use the expression of Coulomb's law for the electric force

         F₁₃ = ka q₁ q₃ / r₁₃²

         F₂₃ = ka q₂ q₃ / r₂₃²

we substitute

         k q₁ q₃ / r₁₃² = k q₂ q₃ / r₂₃²

now imprescriptibility suppose that particle 1 is at the origin of the coordinate system, particle 2 is at a distance d = 2.00cm = 2 10-2 m, therefore let's call the distance from particle 1 to particle 3 as x

            r₁₃ = x

            R₂₃ = d-x

In the exercise we are given the charges for the particle1 q1 = q, for the particle 2 the charge is q2 = 4q the distance between them is d = 2.00cm = 0.0200 m, the value of q = 2.00 nC = 2.00 10⁻⁹ C

let's substitute these values

              q₁ / x₂ = q₂ / (d-x)²

let's clear x

              (d-x)² = q₂ / q₁    x²

              d² - 2dx + x₂ = q₂ /q₁   x²

               x² (1-q₂ / q₁) - 2d x + d² = 0

let's substitute the values ​​and solve the quadratic equation

              x² (1 - 4q / q) - 2 0.02 x + 0.02² = 0

              -3 x² - 0.04 x + 0.0004 = 0

                x² + 0.0133 x - 0.0001333 = 0

               x = [-0.0133 ±√(0.0133² + 4 0.00013333)] / 2

                x = [-0.0133 + - 0.026664] /2

              x₁ = -0.01998 m

              x₂ = 0.01336 m

Since load 3 must be between charged 1 and 2 the correct answer is

x₂ = 0.01336

Answer:

   h ’= 0.6963 m

this is the height of the part of the post it supports.

Explanation:

For this exercise we can use the trigonometric relationships

Let's start by finding the vertical component of the height of the pole

             cos 30 = x / L

             x = L cos 30

             x = 6 cps 30

             x = 5.196 m

with this length, the post has the same height as the building, so there are

               Aₓ = L -x

               Δx = 6 -5.196

               Δx = 0.804 ft

the height of this section is

               cos 30 = h ’/ Dx

               h ’= 0.804 cos30

               h ’= 0.6963 m

this is the height of the part of the post it supports.

Answer:

9.2 s

Explanation:

Given:

v₀ = 46 m/s

v = 0 m/s

a = -5 m/s²

Find: t

v = at + v₀

0 m/s = (-5 m/s²) t + 46 m/s

t = 9.2 s

Answer: you would feel more gravity

Explanation: the more mass of an object the more pull

Answer:

A) 0.9844 s

B) x2 = 0.4587 m

C) v = 6.657 m/s

Explanation:

We are given;

Height of take off point above pool; x1 = 1.8 m

Initial take off velocity; u = 3 m/s

Final velocity at highest point before free fall; v = 0 m/s

B) To find the highest point above the board her feet reaches means the distance from take off to the top of the motion just before free fall.

Thus, we will be using equation of motion and we have;

v² = u² + 2gs

Now, let s = x2 which will be the distance between take off and the top before free fall.

So;

v² = u² + 2g(x2)

Now,since the motion is against gravity, g will be negative.

Thus;

v² = u² + 2(-9.81)(x2)

Plugging in the relevant values to give;

0² = 3² - (19.62x2)

19.62(x2) = 9

x2 = 9/19.62

x2 = 0.4587 m

A) We want to find how long her feet is in air.. It means we want to find out the time to get to a distance of x1 and also the time to achieve the distance (x1 + x2) on free-fall.

Thus, using equation of motion;

v = u + gt

Again, g = -9.81

Thus;

0 = 3 - 9.81t1

9.81t1 = 3

t1 = 3/9.81

t1 = 0.3058 s

Now, for the time taken to achieve the distance (x1 + x2) on free-fall, we will use the formula;

s = ut + ½gt²

Where s = (x1 + x2) = 1.8 + 0.4587 = 2.2587 m

And now, u = 0 m/s because the start of the free fall is from maximum height with velocity of 0 m/s. Again, g = - 9.81 m/s²

Thus;

2.2587 = 0 - ½(-9.81)(t2)²

2.2587 = 4.905(t2)²

(t2)² = 2.2587/4.905

(t2)² = 0.4605

t2 = √0.4605

t2 = 0.6786 s

Thus, total time of feet in air = t1 + t2 = 0.3058 + 0.6786 = 0.9844 s

C) Velocity when feet hit the water would be given by;

v = u + gt

Where u = 0 m/s and t = t2 = 0.6786

Since it's in direction of gravity, g = 9.81 m/s

v = 0 + (0.6786 × 9.81)

v = 6.657 m/s

Answer:

The time taken to drive on this high way is 6.48 hours

Explanation:

Given;

average velocity, v = 24 m/s

distance of travel, d = 560 km = 560,000 m

The distance traveled is given by;

[tex]d = (\frac{v+u}{2} )t\\\\d =( 24 \ m/s)t\\\\t = \frac{d}{24} \\\\t = \frac{560,000}{24}\\\\t = 2333.33 \ s = 6.48 \ hrs[/tex]

Therefore, the time taken to drive on this high way is 6.48 hours

Answer:

7 × 60 (minute to seconds) = 420 seconds

1.38 × 420 = 579.6m

Answer:

16.80672269 m/s

Explanation:

100 divide by 5.95 seconds

Answer:

16.8m/s

Explanation:

S=D/T

=

[tex] \frac{100}{5.95} [/tex]

=16.80672269

=16.8

Answer:

c

Explanation:

energy isn't stored and it doesn't change to mass

The ball has some kinetic energy when it hits the floor, but some of it is changed, so it loses some of it each time it bounces. After a few bounces, the ball has so little kinetic energy remaining that it stops bouncing. Thus, option C is correct.

Since the gravitational potential energy, which can be transformed back into kinetic energy on the rebound, increases with the drop height, the consequent bounce height will likewise rise.

Because it has the most elasticity, the rubber ball will rebound the highest when all three balls are dropped from the same height. Because rubber is incredibly elastic, it squishes or compresses as it touches the ground but immediately snaps back to its original shape.

Therefore, Additionally, some energy is changed into other forms like heat and sound. The majority of these additional energy sources are lost and not recovered, causing the ball to bounce back to a lesser height.

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Distance = (speed) x (time)

Distance = (3.75 m/s) x (5.0 s)

Distance = 18.75 meters

Answer: 3.75 x 5.0= 18.75 meters

Explanation:

Answer:

6.416*10^ 8C

Explanation:

First we find mass of the water

Mass= density x volume

=1.2*1000= 1200g

To find number of moles we use

N= mass/ molar mass

= 1200/18.02= 66.6mol

So the total number of water molecules is

66.6* 6.022*10^23= 401.07x 10^23moles

But each h20 moles has 10 electrons so it will be 4.01*10^26electrons

Then finally total charge will be

4.01*10^26electrons x 1.6*10^-19

= 6.416*10^ 8C

The magnitude of total charge will be "6.416×10⁸ C".

According to the question,

Density, d = 1.2 L

Volume, V = 1000

The mass will be:

→ Mass = Density × Volume

By substituting the values,

            = [tex]1.2\times 1000[/tex]

            = [tex]1200[/tex] g

Now,

The number of moles will be:

→ N = [tex]\frac{Mass}{Molar \ mass}[/tex]

      = [tex]\frac{1200}{18.02}[/tex]

      = [tex]66.6[/tex] mol

The total no. of water molecules be:

= [tex]66.6\times 6.022\times 10^{23}[/tex]

= [tex]401.07\times 10^{23}[/tex] moles

hence,

The total charge be:

= [tex]4.01\times 10^{26}\times 1.6\times 10^{-19}[/tex]

= [tex]6.416\times 10^8[/tex] C

Thus the above response is correct.

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Answer:

tasteless

Explanation:

Answer:

Bolt

Explanation:

Answer:

57.42 N

Explanation:

We first find the acceleration knowing that it goes from rest to 8.3 m/s in 0.13 seconds,

Therefore, the acceleration is:

[tex]a=\frac{v_f-v_i}{time} =\frac{8.3-0}{0.13 } =63.8\,\,m/s^2[/tex]

Now we use this to calculate the average net force by multiplying mass times acceleration:

F = m * a = 0.9 (63.8) = 57.42 N

Explanation:

Some standard unit conversion are 1 mile = 1.609344 km, 1 ft.= 30.48 cm,

1 km= 1000 m or 1 m = 0.001 km and 1 m= 100 cm or 1 cm=0.01 m.

Now, use these values to convert the given lengths.

A. length of cave = 5 miles (given)

From standard value 1 mile = 1.609344 km

[tex]\Rightarrow[/tex] 5 miles = [tex]\times[/tex] 1.609344 km= 8.04672 km,

[tex]\Rightarrow[/tex] 5 miles =8.04672 [tex]\times[/tex] 1000 m= 8046.72 m [ as 1 km = 1000 m]

[tex]\Rightarrow[/tex] 5 miles = 8046.72 [tex]\times[/tex] 100 cm=804672 cm

B. Height of the waterfall = 1235 ft. (given)

1 ft.= 30.48 cm [ from standard value]

[tex]\Rightarrow[/tex] 1235.2 ft.= 1235.2[tex]\times[/tex] 30.48 cm=37648.896 cm,

[tex]\Rightarrow[/tex] 1235.2 ft.=37648.896 [tex]\times[/tex] 0.01 m =376.48896 m [ as 1 cm = 0.01 m]

[tex]\Rightarrow[/tex] 1235.2 ft.=376.48896 [tex]\times[/tex] 0.001 km =0.37648896 km [ as 1 m = 0.001 km]

C. Height of the mountain= 21320 ft. (given)

From standard value: 1 ft.= 30.48 cm

[tex]\Rightarrow[/tex] 21320 ft.= 21320 [tex]\times[/tex] 30.48 cm=649833.6 cm,

[tex]\Rightarrow[/tex] 21320 ft.=649833.6 [tex]\times[/tex] 0.01 m =6498.336 m [ as 1 cm = 0.01 m]

[tex]\Rightarrow[/tex] 21320 ft.=6498.336 [tex]\times[/tex] 0.001 km =6.498336 km [ as 1 m = 0.001 km].

D. Depth of canyon =6630 ft.

From standard value: 1 ft.= 30.48 cm

[tex]\Rightarrow[/tex] 6630 ft.= 6630 [tex]\times[/tex] 30.48 cm=202082.4 cm,

[tex]\Rightarrow[/tex] 6630 ft.=202082.4  [tex]\times[/tex] 0.01 m =2020.824  m [ as 1 cm = 0.01 m]

[tex]\Rightarrow[/tex] 6630 ft.=2020.824  [tex]\times[/tex] 0.001 km =2.020824  km [ as 1 m = 0.001 km].

Answer:

112.5 lb/ft³

405.93 kg/m³

0.406

Explanation:

given that

Weight of the container, w(c) = 450 lb

Volume of fluid, v(f) = 4 ft³

Weight of empty container, w(e) = 50 lb

Specific weight is said to be the weight per unit volume. Thus, the specific weight of this fluid is

w(s) = 450 / 4

w(s) = 112.5 lb/ft³

Mass density is the ratio of the mass of the fluid, with respect to the volume.

ρ = m/v

Since the question gives us the mass in weight, we'd have to convert it to mass from weight.

mass = weight / acceleration due to gravity

mass = 450 / 9.81

mass = 45.87 kg

Also, since the mass is going to be in kg, the volume also has to be in m³, thus needing a conversion.

4 ft³ = 0.113 m³

Mass density = 45.87 / 0.113

Mass density = 405.93 kg/m³

Specific gravity of the fluid is simply given as mass density / 1000. Therefore,

g(s) = 405.93 / 1000

g(s) = 0.406

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Answer:

3.68×10¯⁶ m

Explanation:

The following data were obtained from the question:

Mass (m) = 5 g

Tension (T) = 180 N

Frequency (f) = 260 Hz

Wavelength (λ) = 0.60 m.

Length (L) =?

Next, we shall determine the period of the wave. This can be obtained as follow:

Period is simply defined as the time taken to complete 1 oscillation. Mathematically, it is represented as:

Period (T) = 1/frequency (f)

T = 1/f

Thus, with the above formula, the period of the wave can be obtained as shown below:

Frequency (f) = 260 Hz

Period (T) =?

T = 1/f

T = 1/260

T = 3.85×10¯³ s

Finally, we shall determine the length of the string as follow:

Period (T) = 3.85×10¯³ s

Acceleration due to gravity (g) = 9.8 m/s²

Pi (π) = 3.14

Length (L) =?

T = 2π√(L/g)

3.85×10¯³ = 2 × 3.14 × √(L/9.8)

3.85×10¯³ = 6.28 × √(L/9.8)

Divide both side by 6.28

3.85×10¯³ / 6.28 = √(L/9.8)

Take the square of both side

(3.85×10¯³ / 6.28)² = L/9.8

Cross multiply

L = (3.85×10¯³ / 6.28)² × 9.8

L = 3.68×10¯⁶ m

Therefore, the length of the string is 3.68×10¯⁶ m

Answer:

Trevor's taxi would cause higher levels of air pollution

Explanation:

Trevor's taxi use diesel oil.

Diesel is less cleaner than LPG.

Compared to automotive pollution from petrol and diesel, pollutants from LPG-driven cars include lower amounts of petroleum hydrocarbons, nitrogen oxides , sulphur oxides, ozone contamination and particulate matter.

Answer:

4

Explanation:

friction

weight

normal reaction

force to overcome inertia

Answer:

1

Explanation:

--->

Density = (mass) / (volume)  (definition)

(Density) x (volume)  =  (mass)

(volume)  =  (mass) / (density)

Volume = (57.3 g) / (1.38 g/mL)

Volume = 41.5 mL

Answer:

41.5

Explanation:

Got answer correct on EDG

Answer:

Explanation:

A. Q= It

30E-3A=q

Ne = q

30E-3/1.6*10-19= N

N= 1.8*10^16 electrons

B. Power= I x v

= 30*10^-3A x 100*10^3v

= 3000watts

Answer:

c)Average acceleration is always equal to the change in velocity of an object divided by the time interval.

Explanation:

Because average acceleration= a2+a1/2

Which is equal to change in velocity which is a time rate of change of velocity v-u/t which actually explains average acceleration

Answer:

The value is [tex]v  =  2.999 *10^{8} \  m/s[/tex]

Explanation:

From the question we are told that

   The time taken to travel to the planet from earth is [tex]t = 189 \ light-years[/tex]

    The  time to be spent on the ship is  [tex]t_{s} =  12 \  years[/tex]

Generally speed can be obtained using the mathematical relation represented below

       [tex]t_s  =  2 * t *  \sqrt{1 -  \frac{v^2}{c^2 } }[/tex]

The 2 in the equation show that the trip is a round trip i.e going and coming back

=>    [tex]12 =  2 * 189 *  \sqrt{1 -  \frac{v^2}{(3.0*10^{8})^2 } }[/tex]

=>     [tex]v  =  2.999 *10^{8} \  m/s[/tex]

Answer:

A diet high in fat, calories, and cholesterol increases your risk of diabetes. A poor diet can lead to obesity (another risk factor for diabetes) and other health problems. A healthy diet is high in fiber and low in fat, cholesterol, salt, and sugar. Also, remember to watch your portion size.

Answer:

The ping pong ball is hollow.

Explanation:

Since a ping pong ball, unlike the golf ball, is only filled with air, it is less dense. Air is notably LESS dense than solids, which is what comprises the inside of a golf ball.

Answer:

x = 78.9 cm

Explanation:

The instantaneous velocity v of the marble as it slides past the putty, as a function of time t, is given by

[tex]v(t)=u-bt+ct^3[/tex]

u = 2.50 m/s, b = 3.55 m/s², and c = 6.70 m/s⁴

As the marble exits the region covered in putty, the marble\'s instantaneous acceleration goes to zero.

[tex]a=\dfrac{dv}{dt}\\\\a=\dfrac{d(u-bt+ct^3)}{dt}\\\\a=-b+3ct^2[/tex]

or

[tex]0-3.55+3\times 6.7t^2\\\\0=-3.55+20.1t^2\\\\t=\sqrt{\dfrac{3.55}{20.1}} \\\\t=0.42\ s[/tex]

Let x is the length of the track.

[tex]v=\dfrac{dx}{dt}\\\\\text{or}\\\\\int\limits_0^x {dx} =\int\limits^{4.2}_0 {vdt} \\\\x=\int\limits^{4.2}_0 {u-bt+ct^3} \, dt\\\\x=[ut-\dfrac{bt^2}{2}+\dfrac{ct^4}{4}]_0^ {0.42}\\\\\text{Putting limits}\\\\x=0.42u-\dfrac{b(0.42)^2}{2}+\dfrac{c(0.42)^4}{4}\\\\\text{Now, put values of u,b and c}\\\\x=0.42(2.5)-\dfrac{3.55\times (0.42)^2}{2}+\dfrac{6.7\times (0.42)^4}{4}\\\\x=0.789\ m\\\\x=78.9\ cm[/tex]

So, the length of the track that is covered with putty is 78.9 cm.

The length of the track is covered with putty is 0.789 m.

The instantaneous velocity v of the marble as a function of time t is given below.

[tex]v(t)=u-bt+ct^3[/tex]

Where t = 0 at the moment when the marble first comes into contact with the putty. Also given that, u = 2.50 m/s, b = 3.55 m/s², and c = 6.70 m/s⁴.

Given that the acceleration = 0 as the marble exits the region covered in putty. The acceleration is the rate change in the velocity with respect to time.

[tex]a = \dfrac {dv}{dt}[/tex]

[tex]a = \dfrac {d(u-bt+ct^3)}{dt}[/tex]

[tex]a = -b + 3ct^2[/tex]

Substituting the values in the above equation, we get the time.

[tex]0 = -3.55 + 3 \times 6.70 t^2[/tex]

[tex]t^2 = 0.1766[/tex]

[tex]t = 0.42 \;\rm s[/tex]

Hence the time interval is t =0 to t = 0.42 seconds.

Let us consider that the length of the track is l. Then the velocity can be written as the function of length is given below.

[tex]v = \dfrac {dl}{dt}[/tex]

[tex]dl = vdt[/tex]

[tex]\int_{0}^{t}dl = \int_{0}^{0.42} vdt[/tex]

[tex]l = \int_{0}^{0.42} (u-bt+ct^3) dt[/tex]

[tex]l =[ ut-\dfrac {bt^2}{2} + \dfrac {ct^4}{4} ]_0^{0.42}[/tex]

Putting the value t = 0.42,

[tex]l = 2.50 \times 0.42 - \dfrac {3.55 \times (0.42)^2}{2} + \dfrac { 6.70 \times (0.42)^4}{4}[/tex]

[tex]l = 0.789 \;\rm m[/tex]

Hence the length of the track is covered with putty is 0.789 m.

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