A wheel is rotating freely at angular speed 530 rev/min on a shaft whose rotational inertia is negligible. A second wheel, initially at rest and with 9 times the rotational inertia of the first, is suddenly coupled to the same shaft. (a) What is the angular speed of the resultant combination of the shaft and two wheels

Answer:

The gauge pressure is equal to 147 kPa.

Explanation:

The pressure exerted by fluid is given by :

[tex]P=\rho gh[/tex]

Where

[tex]\rho[/tex] is density of water

h is height

So, put all the values,

[tex]P=1000\times 9.8\times 15\\\\P=147000\ Pa[/tex]

or

P = 147 kPa

So, the gauge pressure is equal to 147 kPa.

Answer:

The gauge pressure is 147000 Pa.

Explanation:

Height, h = 15 m

density of water, d= 1000 kg/m^3

gravity, g = 9.8 m/s^2

The gauge pressure is the pressure exerted by the fluid.

The pressure exerted by the fluid is given by

P  = h d g

P = 15 x 1000 x 9.8 =  147000 Pa

Answer:

d. straight line joining any two points on the mirror

Explanation:

Principal axis is the straight line that passes through the center of a mirror, which is also perpendicular to the surface of the mirror.

The principal axis connects the principal focus and the center of curvature of the mirror. In other words, it is a straight line or axis on which the center of curvature and principal focus can be found. The principal axis joins these two points; the center of curvature and principal focus.

Therefore, the correct option is "D"

d. straight line joining any two points on the mirror

Answer:

Fq = k q Q / R^2

Fp = k q Q / (6 R)^2

The force exerted between S and p is 1 / 36 of that between S and q

or the force between S and q is 36 times that between S and p.

Answer:

The magnitude of the electric field is [tex]8.99*10^{12}N/C[/tex] in the r direction.

Explanation:

The equation of the electric field is given by:

[tex]|\vec{E}|=k\frac{q}{r^{2}}[/tex]

Where:

[tex]|\vec{E}|=8.99*10^{9}\frac{12.5}{0.11^{2}}[/tex]    

[tex]\vec{E}=9.29*10^{12} \vac{r} \: N/C[/tex]

Therefore, the magnitude of the electric field is [tex]8.99*10^{12}N/C[/tex] in the r direction.

I hope it helps you!

Answer:

The velocity becomes [tex]v\sqrt 2[/tex].

Explanation:

The force acting on the bobber is centripetal  force.

The centripetal force is given by

[tex]F =\frac{mv^2}{r}[/tex]

when mass remains same, radius is doubled and the force is same, so the velocity is v'.

[tex]F =\frac{mv^2}{r}=\frac{mv'^2}{2r}\\\\v'=v\sqrt 2[/tex]

Answer:

a) Light that passes through the floor to reveal yourself (not shadow).

b) 2 rays of light that bounce between 2 transparent media.

c) I don't know what is Diffraction?

Answer:

220V a.c is more dangerous than 220V d.c because of the peak voltage of 220V a.c. which is much larger.

Explanation:

Given that,

Five ideal-gas molecules chosen at random are found to have speeds of 500, 600, 700, 800, and 900 m/s.

The rms speed for this collection is as follows :

[tex]v_{rms}=\sqrt{\dfrac{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2}{5}} \\\\v_{rms}=\sqrt{\dfrac{500^2+600^2+700^2+800^2+900^2}{5}} \\\\v_{rms}=714.14[/tex]

The average speed of these molecules is :

[tex]v_{a}=\dfrac{v_1+v_2+v_3+v_4+v_5}{5}} \\\\v_{a}={\dfrac{500+600+700+800+900}{5}} \\\\v_{a}=700\ m/s[/tex]

So, the rms speed is 714.14 m/s abd the average speed is 700 m/s.

Answer:

The speed of car A before collision is 3.5 km/h.

Explanation:

Mass of car A = 690 kg eastwards

Mass of car B = 520 kg at 74 km/h west wards

Distance, s = 6 m

coefficient of friction = 0.75

Let the speed after collision is v.

Use third equation of motion

[tex]v^2 = u^2 + 2 as \\\\0 =v^2- 2 \times 0.75\times9.8\times 6\\\\v = 9.4 m/s = 33.84 km/h[/tex]

Let the initial speed of car A is v'.

Use conservation of momentum

690 x v' - 520 x 74 = (690 + 520) x 33.8

690 v' + 38480 = 40898

v' = 3.5 km/h

Answer:

 I = - m 16  the two impulses are the same,

Explanation:

The impulse is given by the relationship

         I = Δp

         I = p_f - p₀

in this case the final velocity is zero therefore p_f = 0

        I = -p₀

For driver A the steering wheel impulse is

        I = - m v₀

        I = - m 16

For driver B, the airbag gives an impulse

        I = - m 16

We can see that the two impulses are the same, the difference is that in the air bag more time is used to give this impulse therefore the force on the driver is less

Answer: I do

Explanation:

Resistance opposes current thereby reducing the amount of current that flows through a circuit. In other words, it leads to a loss of electrical energy.

Ideally speaking, a good circuit should have no internal resistance as this would lead to more energy having to be supplied to overcome that resistance. External resistance however, is not a bad thing. For instance, oxygen being removed from lightbulbs.

Explanation:

Given that,

Bill is riding his bicycle at 5 m/s eastward: and Carlos is driving his car at 15 m/s westward.

Taking eastward as positive direction, we have:

[tex]v_B=+5\ m/s[/tex]is the velocity of Bill with respect to Amy (which is stationary)

[tex]v_c=15\ m/s[/tex] is the velocity of Carlos with respect to Amy.

Bill is moving 5 m/s eastward compared to Amy at rest, so the velocity of Bill's reference frame is

[tex]v_B=+5\ m/s[/tex]

Therefore, Carlos velocity in Bill's reference frame will be

[tex]v_c'=-15\ m/s-(+5\ m/s)\\\\=-20\ m/s[/tex]

So, the magnitude is 20 m/s and the direction is westward (negative sign).

Answer:

the instantaneous velocity is 51 m/s

Explanation:

Given;

acceleration, a = 2 + 5t²

Acceleration is the change in velocity with time.

[tex]a = \frac{dv}{dt} \\\\a = 2 + 5t^2\\\\The \ acceleration \ (a) \ is \ given \ so \ we \ have \ to \ find \ the \ velocity \ (v)\\\\To \ find \ the \ velocity, \ integrate\ both \ sides \ of \ the \ equation\\\\2 + 5t^2 = \frac{dv}{dt} \\\\\int\limits^3_0 {(2 + 5t^2)} \, dt = dv\\\\v = [2t + \frac{5t^3}{3} ]^3_0\\\\v = 2(3) + \frac{5(3)^3}{3} \\\\v = 6 + 5(3)^2\\\\v = 6 + 45\\\\v = 51 \ m/s[/tex]

Therefore, the instantaneous velocity is 51 m/s

Answer:

19,224 N

Explanation:

The given parameters are;

The mass limit of the elevator = 2,400 kg

The maximum ascent speed = 18.0 m/s

The maximum descent speed = 10.0 m/s

The maximum acceleration = 1.80 m/s²

Given that the acceleration due to gravity, g ≈ 9.81 m/s²

The minimum upward force that the elevator cable exert on the elevator car, [tex]F_{min}[/tex] , is given in the downward motion as follows;

[tex]F_{min}[/tex] = m·g - m·a

∴ [tex]F_{min}[/tex] = 2,400 kg × 9.81 m/s² - 2,400 kg × 1.80 m/s² = 19,224 N

The minimum upward force that the elevator cable exert on the elevator car, [tex]F_{min}[/tex] = 19,224 N

Answer:

C. the product of the power rating of the light bulb and the time that it remains lit.

Explanation:

The power rating of the light is bulb is defined as the energy supplied to the light bulb divided by the time the bulb is lit up. Therefore,

[tex]P = \frac{E}{t}\\\\E = Pt[/tex]

where,

E = Energy Supplied to the bulb = Energy stored in capacitor = ?

P = Power rating of the bulb

t = time the bulb is lit up

Hence, the correct option is:

C. the product of the power rating of the light bulb and the time that it remains lit.

Answer:

Sorry I dont know this answer sorry

Answer:

b. parallel

Explanation:

Answer:

  f = 276.6 Hz

Explanation:

This musical instrument can be approximated to a tube system where each tube has one end open and the other closed.

In the closed part there is a node and in the open part a belly or antinode. Therefore the wavelength is

          L = λ/ 4

speed is related to wavelength and frequency

          v = λ f

          λ = v / f

we substitute

          L = v / 4f

          f = v / 4L

the speed of sound at 20ºC is

          v = 343 m / s

let's calculate

          f = [tex]\frac{343 }{4 \ 0.31}[/tex]

          f = 276.6 Hz

Answer:

 A_resulting = 0.2 m

Explanation:

Let's analyze the impact of the pulse with the pole, this is a fixed obstacle that does not move therefore by the law of action and reluctant, the force that the pole applies on the rope is of equal magnitude to the force of the rope on the pole (pulse), but opposite directional, so the reflected pulse reverses its direction and sense.

With this information we analyze a point on the string where the incident pulse is and each reflected with an amplitude A = 0.1 m, the resulting is

           A_res = 2A

           A_resultant = 2 .01

           A_resulting = 0.2 m

In vacuum, going at 2.99×10^8 m/s.

Answer:

0.83 J of work

Explanation:

2 J of work is required to stretch a spring from 34cm to 46cm

So that is 12cm stretched with 2 J of work

We can make that 6cm for 1 J of work

So, we need the find the work for stretching 36cm to 41cm

Which is 5cm

So, What is the work required to stretch 5cm?

1 J of work for 6cm

x work for 5cm

So, by proportion method

1 : 6 :: x : 5

6 * x = 1 * 5

6x = 5

x = 5/6

= 0.83

So to stretch 36cm to 41cm we need 0.83 J of work

Answer:

the magnetic field must double

Explanation:

For this exercise we use Ampere's law

         ∫ B . ds = μ₀ I

Where the bold indicate vectors

With this expression we can see that if we double the current, keeping the same trajectory, the magnetic field must double

Answer:

The final speed of the bullet is 480.4 m/s.

Explanation:

mass of bullet, m = 1.9 g

initial speed, u = 500 m/s

thickness, d = 0.75 inch = 0.01905 m

Force, F = 960 N

Let the final speed is v.

According to the work energy theorem,

Work = change in kinetic energy

[tex]W = F d = 0.5 m{\left (v^2 - u^2 \right )}[/tex]

-960 x 0.01905 = 0.5 x 0.0019 x (v^2 - 500 x 500)

-18.288 = 0.00095 (v^2 - 250000)

v = 480.4 m/s

Answer:

f = 106.3 N

Explanation:

The force applied on the sled must be equal to the static frictional force to move the sled:

Tension Force Horizontal Component = Static Frictional Force

[tex]TCos\theta = \mu W\\TCos\theta = \mu mg[/tex]

where,

T = Tension = 120 N

θ = angle of rope = 37°

μ = coefficient of static friction = ?

m = mass of children plus sled = 55 kg

g = acceleration due to gravity = 9.81 m/s²

Therefore,

[tex](120\ N)Cos\ 37^o = \mu (55\ kg)(9.81\ m/s^2)\\\\\mu = \frac{95.84\ N}{(55\ kg)(9.81\ m/s^2)}\\\\\mu = 0.18[/tex]

Now, the static friction acting on the mother will be:

[tex]f = \mu mg = (0.18)(61\ kg)(9.81\ m/s^2)\\[/tex]

f = 106.3 N

Answer:

i can help you i know this answer

Answer: the side two are 50 then the other two are 140  i thank

Explanation:

Answer:

1. True

2. True

3. False

4. True

5. True

Answer:

the correct answer is a

Explanation:

The torque is

        τ = F x r

where the bold letters indicate vectors, in this case the vector of the center of mass is perpendicular to the weight of the body

          τ = mg r

in body weight it is applied at the point of the center of mass, therefore as the distance of the force from the axis of rotation (center of amas) is zero, the die is zero

the correct answer is a

Answer:

[tex]T=8.1N[/tex]

Explanation:

From the question we are told that:

Mass m=0.40

Radius r=1.8m

Angle Beneath the Horizontal \theta =40 \textdegree

Speed v=5.0m/s

The Tension Angle

 [tex]\alpha=90-\theta\\\\\alpha=90-40[/tex]

 [tex]\alpha=50 \textdegree[/tex]

Generally the equation for Tension is is mathematically given by

 [tex]T=\frac{mv^2}{r}+mgcos \alpha[/tex]

 [tex]T=\frac{0.40*5^2}{1.8}+0.40*5cos50[/tex]

 [tex]T=8.1N[/tex]