A circus van moves 50 km east in the first hour, 40 km west in the second hour, and 30 km north in the next
half hour. Calculate the average velocity of the travelling circus van.

The question is incomplete. The complete question is :

In your job as a mechanical engineer you are designing a flywheel and clutch-plate system. Disk A is made of a lighter material than disk B, and the moment of inertia of disk A about the shaft is one-third that of disk B. The moment of inertia of the shaft is negligible. With the clutch disconnected, A is brought up to an angular speed ?0; B is initially at rest. The accelerating torque is then removed from A, and A is coupled to B. (Ignore bearing friction.) The design specifications allow for a maximum of 2300 J of thermal energy to be developed when the connection is made. What can be the maximum value of the original kinetic energy of disk A so as not to exceed the maximum allowed value of the thermal energy?

Solution :

Let M.I. of disk A = [tex]$I_0$[/tex]

So, M.I. of disk B =  [tex]$3I_0$[/tex]

Angular velocity of A = [tex]$\omega_0$[/tex]

So the kinetic energy of the disk A = [tex]$\frac{1}{2}I_0\omega^2$[/tex]

After coupling, the angular velocity of both the disks will be equal to ω.

Angular momentum will be conserved.

So,

[tex]$I_0\omega_0 = I_0 \omega + 3I_0 \omega$[/tex]

[tex]$I_0\omega_0 = 4I_0 \omega$[/tex]

[tex]$\omega = \frac{\omega_0}{4}$[/tex]

Now,

[tex]$K.E. = \frac{1}{2}I_0\omega^2+ \frac{1}{2}3I_0\omega^2$[/tex]

[tex]$K.E. = \frac{1}{2}I_0\frac{\omega_0^2}{16}+ \frac{1}{2}3I_0\frac{\omega_0^2}{16}$[/tex]

[tex]$K.E. = \frac{1}{2}I_0\omega_0^2 \left(\frac{1}{16}+\frac{3}{16}\right)$[/tex]

[tex]$K.E. = \frac{1}{2}I_0\omega_0^2\times \frac{1}{4}$[/tex]

[tex]$\Delta K = \frac{1}{2}I_0\omega_0^2 - \frac{1}{2}I_0\omega_0^2 \times \frac{1}{4} $[/tex]

[tex]$2300=\frac{3}{4}\left(\frac{1}{2}I_0\omega_0^2\right)$[/tex]

[tex]$\frac{1}{2}I_0\omega_0^2=2300 \times \frac{4}{3 } \ J $[/tex]

Therefore, the maximum initial K.E. = 3066.67 J

Explanation:

this is the answer for your question if you have any doubt you can join

The acceleration is 1.2 m/s^2. The distance travelled is 1197 m. The time taken is 44.7 seconds.

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

Given that;

v = u - at

v = 0 m/s because the car comes to stop

u = at

a = u/t

a = 53.6 m/s/ 45 seconds

a = 1.2 m/s^2

The distance travelled is;

v^2 = u^2 - 2as

v = 0 m/s

u^2 = 2as

s = u^2/2a

a = ( 53.6 )^2/2 * 1.2

s = 1197 m

Time taken to stop is obtained from;

v = u - at

when v = 0 m/s

u = at

t = u/a

t = 53.6/1.2

t = 44.7 seconds

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So they give us this

V=IR

V= 1.8

I=0.4

R=?

So we insert the thing that we know.

1.8=0.4*R

We need to leave our unknown value alone. So if our value of 0.4 is multiplying the unknown value it passes to the other side dividing.

So we have this.

Lastly we solve.

R=4.5ohms

The formula to find R is V=IR

V/I=R

So the resistance will be the Voltage divided by the Current

Answer:

41.64 N

Explanation:

Applying,

v = √(T/m')................ Equation 1

Where v = velocity of the wave, T = Tension of the rope, m' = mass per unit length of the rope.

make T the subject of the equation,

T = v²m'................. Equation 2

But,

v = λf............... Equation 3

Where λ = wavelength, f = frequency

And

m' = m/L........... Equation 4

Where m = mass of the rope, L = length of the rope

Substitute equation 3 and equation 4 into equation 2

T = (λf)²(m/L).............. Equation 5

From the question,

Given: λ = 0.740 m, f = 42 Hz, m = 0.125 kg, L = 2.9 m

Substitute these values into equation 5

T = (42×0.74)²(0.125/2.9)

T = 41.64 N

Answer:

c

Explanation:

search it up

Answer:

Option A.

Explanation:

Because this is a lunar eclipse it normally happens 2 times a year.  This is a decently rare phenomenon due to the positions they have to be in to make a lunar eclipse.  Therefore, it is option A.

Answer:

[tex]175\ \text{N/C}[/tex]

Explanation:

[tex]E_1[/tex] = Initial electric field = 3.5 N/C

[tex]E_2[/tex] = Final electric field

[tex]r_1[/tex] = Initial distance = 10 m

[tex]r_2[/tex] = Final distance = 20 cm

Electric field is given by

[tex]E=\sqrt{\dfrac{2P}{\pi r^2c\varepsilon_0}}[/tex]

So,

[tex]E\propto \dfrac{1}{r}[/tex]

[tex]\dfrac{E_2}{E_1}=\dfrac{r_1}{r_2}\\\Rightarrow E_2=E_1\dfrac{r_1}{r_2}\\\Rightarrow E_2=3.5\dfrac{10}{0.2}\\\Rightarrow E_2=175\ \text{N/C}[/tex]

The electric field amplitude at the required point is [tex]175\ \text{N/C}[/tex].

Answer:

It is example of Closed system

Answer:

If we use the equation for the transformation of velocities for moving frames:

v' = (v - u) / (1 - u * v / c^2) where we measure the speed of v' approaching from the left where v is in a frame moving at -u towards v'

v' = (.6 c - (-.6 c)) / (1 - (-.6 c) * .6 c / c^2) = 1.2 c / (1 + .6 * .6)

or v' = 1.2 c / (1 + .36) = .88 c

v is approaching from the left at .6 c in the reference frame and the other frame approaches from the right at -.6 c with speed u  (-.6 c) and we measure the speed of v as seen in the frame moving to the left

Answer:

The name of law is mass engery law

Answer:

Properties of matter

Explanation:

All properties of matter are either extensive or intensive and either physical or chemical. Extensive properties, such as mass and volume, depend on the amount of matter that is being measured. Intensive properties, such as density and color, do not depend on the amount of matter.

Answer:

The pythagoream theorem

Explanation:

Answer:

d = 1.19 x 10⁻⁴ m = 0.119 mm

Explanation:

This problem can be solved by using Young's double-slit experiment formula:

[tex]Y = \frac{\lambda L}{d}[/tex]

where,

Y = fringe spacing = 32 mm = 0.032 m

L = slit to screen distance = 6 m

λ = wavelength of light = 633 nm = 6.33 x 10⁻⁷ m

d = slit width = ?

Therefore,

[tex]0.032\ m = \frac{(6.33\ x\ 10^{-7}\ m)(6\ m)}{d}\\\\d = \frac{(6.33\ x\ 10^{-7}\ m)(6\ m)}{0.032\ m}[/tex]

d = 1.19 x 10⁻⁴ m = 0.119 mm

Answer:

Eubacteria Kingdom :)

Explanation:

Answer: eubacteria

Explanation:

Answer:

Explanation:

We shall divide the stretch in two stage .

from .2 m to .25 m ( extension is .05 )

from .25 m to .3 m

force when length  .2 m = 0

force when length .25 m = 17 x .05 = .85 N .

average force = (0 + .85) / 2

= .425 N

work done = average force x extension

= .425 x .05

= .02125 J .

from .25 m to .3 m

force when length  .25 m = .85 N

force when length .3 m = 17 x .1 = 1.7 N .

average force = (1.7 + .85) / 2

= 1.275 N

work done = average force x extension

= 1.275 x .05

= .06375 J .

Answer:

Q = 8368 Joules.

Explanation:

Given the following data;

Mass = 50 g

Initial temperature = 30°C

Final temperature = 70°C

Specific heat capacity = 4.184 J/g °C

To find the amount of heat absorbed by the water;

Heat capacity is given by the formula;

[tex] Q = mcdt[/tex]

Where;

Q represents the heat capacity or quantity of heat.

m represents the mass of an object.

c represents the specific heat capacity of water.

dt represents the change in temperature.

dt = T2 - T1

dt = 70 - 30

dt = 40°C

Substituting the values into the equation, we have;

[tex] Q = 50*4.184*40[/tex]

Q = 8368 Joules.

Answer:

F = 668.9N

Explanation:

Using the formula;

m1d1 = m2d2

m1 and m2 are the masses

d1 and d2 are the distances

Substitute;

15 * 2.5 = x * (55/100) [55cm was converted to metres]

37.5 = 0.55x

Swap

0.55x = 37.5

x = 37.5/0.55

x = 68.18kg

Since F = mg

F = 68.18 * 9.81

F = 668.9N

This gives the required force

Answer:

speed = 0.8c

Explanation:

Given :

Distance from earth to the distant planet = 10 ly

Time taken by the astronauts for the entire journey = 25 years

The time taken to reach the planet is [tex]$t_1=\frac{25}{2}$[/tex]

                                                                  = 12.5 years

Therefore, speed of the starship can be calculated by :

[tex]$\text{Speed} = \frac{\text{distance}}{\text{time}}$[/tex]

[tex]$v=\frac{10 \times c \times 3.15 \times 10^7}{12.5 \times 3.15 \times 10^7}$[/tex]

  [tex]$=0.8c$[/tex]

Therefore the speed of the starship is 0.8c

Answer:

B - As one variable increases the other decreases. (-1 to +1)     coefficient

Complete question:

A light bulb emits light that travels uniformly in all directions. Detailed measurements show that at a distance of 56 m from the bulb, the amplitude of the electric field is 3.78 V/m. What is the average intensity of the light?

Answer:

The average intensity of the light is 0.02 W/m²

Explanation:

Given;

Amplitude of the electric field, E₀ = 3.78 V/m

The average intensity of the light is calculated as follows;

[tex]I_{avg} = \frac{c\epsilon_0 E_0^2}{2}[/tex]

where;

[tex]I_{avg}[/tex] is the average intensity of the light

c is speed of light = 3 x 10⁸ m/s

[tex]I_{avg} = \frac{(3\times 10^8)(8.85 \times 10^{-12}) (3.78)^2}{2} \\\\I_{avg} = 0.01897 \ W/m^2\\\\I_{avg} = 0.02 \ W/m^2[/tex]

Therefore, the average intensity of the light is 0.02 W/m²

Answer:

it would be a

Explanation:

If kinetic energy is highest at the bottom of the pendulum, then that is where potential energy is the lowest. So potential energy would be highest when the pendulum is at its highest point on either side of its motion where it is stationary for an instant.

The ball will have the most potential energy at the extreme position 1. So, the correct option is A.

Potential energy is a form of stored energy that is dependent on the arrangement of system components.

Here,

The potential energy of a pendulum executing simple harmonic oscillation is given by,

PE = 1/2 mω²y²

where m is mass of the bob, ω is the angular frequency and y is the displacement of the pendulum.

So, the potential energy will be maximum, when the displacement of the pendulum becomes maximum. The amplitude is the maximum displacement in an SHM.

So, y = A

Therefore, maximum potential energy,

PE(max) = 1/2mω²A²

Hence,

The ball will have the most potential energy at the extreme position 1.

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Answer:

static friction=0.126

Answer:

Y = 3.87 x 10⁻³ m = 3.87 mm

Explanation:

This problem can be solved by using Young's double-slit experiment formula:

[tex]Y = \frac{\lambda L}{d}[/tex]

where,

Y = fringe spacing = ?

L = slit to screen distance = 2 m

λ = wavelength of light = 580 nm = 5.8 x 10⁻⁷ m

d = slit width = 0.3 mm = 3 x 10⁻⁴ m

Therefore,

[tex]Y = \frac{(5.8\ x\ 10^{-7}\ m)(2\ m)}{3\ x\ 10^{-4}\ m}[/tex]

Y = 3.87 x 10⁻³ m = 3.87 mm

Answer:

5*10^2

Explanation:

A p e x

Answer:

Title) 50 (Bii) 45

Explanation:reflection, same angle as incidence

Answer:45

Explanation:

Answer:

= 1200m/s or 1.2 x [tex]10^{3}[/tex] m/s

Explanation: