on a day when the speed of sound in air is 340 m/s a bat emits a shriek whose echo returns to it 0.0250 seconds later. How far away is the mosquito that reflects back the shriek?

Answer:

a) 7250.5 N

b) 4.6 m/s²

Explanation:

a)

F = applied force = 8000 N

θ = angle with the horizontal = 65 deg

Consider the motion along the vertical direction :

[tex]F_{y}[/tex] = Applied force in vertical direction in upward direction = F Sinθ = 8000 Sin65 = 7250.5 N

[tex]F_{g}[/tex] = weight of the plane in vertical direction in downward direction = ?

[tex]a_{y}[/tex] = Acceleration in vertical direction = 0 m/s²

Taking the force in upward direction as positive and in downward direction as negative, the force equation along the vertical direction can be written as

[tex]F_{y}-F_{g} = m a_{y}[/tex]

[tex]7250.5 -F_{g} = m (0)[/tex]

[tex]F_{g}[/tex] = 7250.5 N

b)

m = mass of the plane

force of gravity is given as

[tex]F_{g} = mg [/tex]

[tex]7250.5 = m(9.8) [/tex]

m = 739.85 kg

Consider the motion along the horizontal direction

[tex]F_{x}[/tex] = Applied force in horizontal direction = F Cosθ = 8000 Cos65 = 3381 N

[tex]a_{x}[/tex] = Acceleration in horizontal direction

Acceleration in horizontal direction is given as

[tex]a_{x}=\frac{F_{x}}{m}[/tex]

[tex]a_{x}=\frac{3381}{739.85}[/tex]

[tex]a_{x}[/tex] = 4.6 m/s²

Answer:

Speed of bullet = 389.61 m/s.

Explanation:

Considering the vertical motion of bullet

Initial vertical speed = 0 m/s

Vertical displacement = 2.9 cm = 0.029 m

Vertical acceleration = 9.81 m/s²

Substituting in s = ut + 0.5at²

    0.029 = 0 x t + 0.5 x 9.81 x t²

    t = 0.077 s

So ball hits the target after 0.077 s.

Now considering the vertical motion of bullet

Time = 0.077 s

Horizontal displacement = 30 m

Horizontal acceleration = 0 m/s²

Substituting in s = ut + 0.5at²

    30 = u  x 0.077 + 0.5 x 0 x 0.077²

     u = 389.61 m/s

Speed of bullet = 389.61 m/s.

Answer:

390 m/s

Explanation:

Let the horizontal speed be u.

Horizontal distance , x = 30 m

Vertical distance, y = 2.9 cm

Let time taken be t.

Use second equation of motion in vertical direction

H = uy × t + 1/2 gt^2

0.029 = 0.5 × 9.8 × t^2

t = 0.077 s

Horizontal distance = horizontal velocity × time

30 = u × 0.077

u = 390 m/s

Answer:

explained

Explanation:

When the intensity of light is increased on a piece of metal only the number of electron ejected will increase because all other things independent of intensity of light.

Light below certain frequency will not cause any electron emission no matter how intense.

The intensity produces more electron but does not change the maximum kinetic energy of electrons.

Work function is independent of the intensity of light, because it is an intrinsic property of a material.

Explanation:

The kinetic energy of an object is associated with its motion. Mathematically, it is given by :

[tex]E_k=\dfrac{1}{2}mv^2[/tex]........(1)

Where

m is the mass of the car

v is the velocity of the car

And the momentum of an object is equal to the product of mass and velocity i.e.

[tex]momentum(p)=mass(m)\times velocity(v)[/tex]..........(2)

From equation (1) and (2) it is clear that the kinetic energy and the momentum is directly proportional to the velocity of the object. As a car increases velocity, its kinetic energy or momentum increase faster.

Answer:

Part a)

P = 8.23 kg m/s

Part b)

v = 3.05 m/s

Explanation:

Part a)

momentum of cart 1 is given as

[tex]P_1 = m_1v_1[/tex]

[tex]P_1 = (2.7)(3.7) = 9.99 kg m/s[/tex]

Momentum of cart 2 is given as

[tex]P_2 = m_2v_2[/tex]

[tex]P_2 = (1.1)(-1.6) = -1.76 kg m/s[/tex]

Now total momentum of both carts is given as

[tex]P = P_1 + P_2[/tex]

[tex]P = 8.23 kg m/s[/tex]

Part b)

Since two carts are moving towards each other due to mutual attraction force and there is no external force on two carts so here momentum is always conserved

so here we will have

[tex]P_i = P_f[/tex]

[tex](2.7 kg)v = 8.23[/tex]

[tex]v = 3.05 m/s[/tex]

Answer:

a)

365.3 N/C

b)

3.85 x 10⁴ m/s

Explanation:

q = magnitude of charge on proton = 1.6 x 10⁻¹⁹ C

m = mass of the proton = 1.67 x 10⁻²⁷ kg

d = distance between the plates = 2.10 cm = 0.021 m

t = time interval = 1.10 x 10⁻⁶ sec

E = magnitude of electric field

v₀ = initial velocity = 0 m/s

a = acceleration of the proton

Using the equation

d = v₀ t + (0.5) a t²

0.021 = (0) (1.10 x 10⁻⁶) + (0.5) a (1.10 x 10⁻⁶)²

a = 3.5 x 10¹⁰ m/s²

acceleration of the proton inside electric field is given as

[tex]a =\frac{qE}{m}[/tex]

[tex]3.5\times 10^{10} =\frac{(1.6\times 10^{-19})E}{1.67\times 10^{-27}}[/tex]

E = 365.3 N/C

b)

v = final speed of the proton

using the equation

v = v₀ + a t

v = 0 + (3.5 x 10¹⁰) (1.10 x 10⁻⁶)

v = 3.85 x 10⁴ m/s

Answer:

C

Explanation:

Givens

m = 2 kg

F = 2 * 9.81

F =  19.62 N

x = 0.15 m

Formula

F = k*x

Solution

19.62 = k*0.15

k = 19.62/0.15

k = 130.8 which rounded to the nearest given answer is C

The magnetic force on a wire carrying current towards the south under a magnetic field directed vertically upwards will point towards the East. In order to determine this, use the right-hand rule.

The direction of the magnetic force on a current-carrying wire under a magnetic field can be deduced using the right-hand rule. In this case, with the current flowing towards the south and the magnetic field directed vertically upward, you would point your right thumb in the direction of the current (southwards) and curl your fingers in the direction of the magnetic field (upwards). The palm of your hand will then face toward the direction of the force. In this case, the force would be pointing toward the East.

The right-hand rule is a vital principle in the study of electromagnetism as it aids in identifying the direction of various quantities in magnetic fields.  The magnetic force on a current-carrying wire represents the phenomenon underlying the working of many electric motors.

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Answer:

The distance to the first anti-nodal line is [tex]2.11\times10^{-3}\ m[/tex].

Explanation:

Given that,

Wavelength = 488 nm

Width [tex]d=6.23\times10^{-4}\ m[/tex]

Distance D =2.7 m

We need to calculate the distance to the first anti-nodal line

Using formula of the distance for first anti-nodal line

[tex]y_{n}=\dfrac{n\lambda D}{d}[/tex].....(I)

Where, n = number of fringe

d = width

D = distance from the screen

[tex]\lambda[/tex]=wavelength of light

Put the all value in the equation (I)

[tex]y_{n}=\dfrac{488\times10^{-9}\times2.7}{6.23\times10^{-4}}[/tex]

[tex]y_{n}=2.11\times10^{-3}\ m[/tex]

Hence, The distance to the first anti-nodal line is [tex]2.11\times10^{-3}\ m[/tex].

Explanation:

0.05

momentum =Mv

then,

Mv=mv

v2=0.12*6/14

v2=0.05kgm/s

The change in the magnitude of the momentum of the ball is 2.4 kg·m/s.

To calculate the change in the magnitude of the momentum of the ball, we need to find the initial momentum and the final momentum of the ball. Momentum is calculated by multiplying the mass of an object by its velocity. Let's start by calculating the initial momentum:

Initial momentum = mass × initial velocity

Final momentum = mass × final velocity

Change in momentum = final momentum - initial momentum

Plugging in the given values:

Initial momentum = 0.12 kg × 6 m/s

Final momentum = 0.12 kg × (-14 m/s) (since the ball changes direction)

Change in momentum = (-1.68 kg·m/s) - (0.72 kg·m/s) = -2.4 kg·m/s.

The change in the magnitude of the momentum of the ball is 2.4 kg·m/s. However, since the question is asking for the magnitude, we take the absolute value of the change in momentum, which is 2.4 kg·m/s.

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Answer:

anterior

anterior

Explanation:

In the given question is asked that

If a person is standing erect and flexes the trunk on the hip, the center of mass will move ___________ and the line of gravity moves___________ within the base of support.

The current answer to the blanks will be

anterior

anterior

hope this helps any further query can be asked in comment section.

The average current passing through a device is given by:

I = Q/Δt

I is the average current

Q is the amount of charge that has passed through the device

Δt is the amount of elapsed time

Given values:

Q = 600C

Δt = 0.500hr = 1800s

Plug in the values and solve for I:

I = 600/1800

I = 0.333A

The average current  flowing in the flashlight bulb is 0.333 A

The given parameters:

quantity of the charge, Q = 600 C

time of the current flow, t = 0.5 hour

To find:

The average current is calculated from quantity of charge that flows in the given time period:

Q = It

where:

I is the average current

Q is the quantity of the charge

t is the time period

From the equation above, make the average current the subject of the formula:

[tex]I = \frac{Q}{t} \\\\I = \frac{600 \ C}{0.5 \ hr \times 3600 \ s} \\\\I = \frac{600 \ C}{1800 \ s} \\\\I = 0.333 \ A[/tex]

Thus, the average current is 0.333 A

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Answer:

275.4 kg

Explanation:

specific gravity = 0.918

Density = 0.918 g/cm^3 = 918 kg/m^3

Volume = 100 gallons = 300 litres = 300 x 10^-3 m^3 = 0.3 m^3

Mass = volume x density = 0.3 x 918 = 275.4 kg

Answer:

The emf is 0.574 volt.

Explanation:

Given that,

Magnetic field [tex]B =3.00\times10^{-5}\ T[/tex]

Length = 75.0 m

Velocity = 255 m/s

We need to calculate the emf

Using formula of emf

[tex]\epsilon=Blv[/tex]

Where, v = velocity

l = length

B = magnetic field

Put the value into the formula

[tex]\epsilon= 3.00\times10^{-5}\times75.0\times255[/tex]

[tex]\epsilon=0.574\ volt[/tex]

Hence, The emf is 0.574 volt.

The emf induced between the wingtips of a jet airplane flying at 255 m/s with a 75.0 m wingspan in Earth's magnetic field of strength 3.00 x 10^-5 T is 0.57 V. This is calculated using the formula for motional emf: E = Blv.

To calculate the emf induced between the airplane wings due to Earth's magnetic field, we use the formula for motional emf: E = Blv, where B is the magnetic field strength, l is the length of the conductor moving through the field (in this case, the wingspan of the airplane), and v is the velocity of the conductor.

Substituting the given values into the formula we get:
E = (3.00 x 10^-5 T) * (75.0 m) * (255 m/s) = 0.57 V

This means that a 0.57 Volt emf is induced between the wingtips of the airplane due to the Earth's magnetic field. This is a direct application of Faraday's law of electromagnetic induction, which states that a changing magnetic field will induce an emf in a conductor.

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Answer:

A) 854.46 kPa

Explanation:

P₁ = initial pressure of the gas = 400 kPa

P₂ = final pressure of the gas = ?

T₁ = initial temperature of the gas = 110 K

T₂ = final temperature of the gas = 235 K

Using the equation

[tex]\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}[/tex]

Inserting the values

[tex]\frac{400}{110}=\frac{P_{2}}{235}[/tex]

P₂ = 854.46 kPa

Answer:It is rising at a rate of [tex]7.5cm/min[/tex]

Explanation:

We have volume of trapezoid equals

[tex]V=Area\times Length\\\\V=\frac{1}{2}(a+b)h\times L[/tex]

Thus at any time 't' we have

[tex]V(t)=\frac{1}{2}(a(t)+b(t))h(t)\times L\\\\\therefore V(t)=\frac{1}{2}(20+b(t))\times h(t)\times L[/tex]

Differentiating both sides with respect to time we get

[tex]\frac{dV(t)}{dt}=\frac{1}{2}b'(t)h(t)L+\frac{1}{2}(20+b(t))\times h'(t)L[/tex]

Applying values we have

[tex]b(t)=20+h(t)\\b'(t)=h'(t)[/tex]

Thus we have

[tex]\frac{dV(t)}{dt}=\frac{1}{2}h'(t)h(t)L+\frac{1}{2}(20+20+h(t))\times h'(t)L\\\\2V'(t)=h'(t)L[h(t)+(40+h(t))]\\\\\therefore h'(t)=\frac{2V'(t)}{L(h(t)+(40+h(t)))}[/tex]

Applying values we get

[tex]h'(t)=0.075m/min=7.5cm/min[/tex]

Answer:

0.71 kg

Explanation:

L = length of the steel wire = 3.0 m

d = diameter of steel wire = 0.32 mm = 0.32 x 10⁻³ m

Area of cross-section of the steel wire is given as

A = (0.25) πd²

A = (0.25) (3.14) (0.32 x 10⁻³)²

A = 8.04 x 10⁻⁸ m²

ΔL = change in length of the wire = 1.3 mm = 1.3 x 10⁻³ m

Y = Young's modulus of steel = 20 x 10¹⁰ Nm⁻²

m = mass hanging

F = weight of the mass hanging

Young's modulus of steel is given as

[tex]Y = \frac{FL}{A\Delta L}[/tex]

[tex]20\times 10^{10} = \frac{F(3)}{(8.04\times 10^{-8})(1.3\times 10^{-3})}[/tex]

F = 6.968 N

Weight of the hanging mass is given as

F = mg

6.968 = m (9.8)

m = 0.71 kg

From conservation of linear momentum, the magnitude of the velocity of the wreckage after collision is 15.6 m/s while its direction is 53 degrees.

There are four types of collision

In elastic collision, both momentum and energy are conserved. While in inelastic collision, only momentum is conserved.

From the given question, the following parameters are given.

Since the collision is inelastic, they will both move with a common velocity after collision.

Horizontal component

[tex]m_{1}[/tex][tex]v_{1}[/tex] = ([tex]m_{1}[/tex] + [tex]m_{2}[/tex] ) V

1500 x 25 = (1500 + 2500) V

37500 = 4000V

V = 37500 / 4000

V = 9.375 m/s

Vertical component

[tex]m_{2}[/tex][tex]v_{2}[/tex] = ([tex]m_{1}[/tex] + [tex]m_{2}[/tex])V

2500 x 20 = (1500 + 2500)V

50000 = 4000V

V = 50000 / 4000

V = 12.5 m/s

The net velocity will be the magnitude of the velocity of the wreckage after collision

V = [tex]\sqrt{9.4^{2} + 12.5^{2} }[/tex]

V = [tex]\sqrt{244.61}[/tex]

V = 15.6 m/s

The direction will be

Tan Ф = 12.5 / 9.4

Ф = [tex]Tan^{-1}[/tex](1.329)

Ф = 53 degrees.

Therefore,  the magnitude of the velocity of the wreckage after collision is 15.6 m/s while its direction is 53 degrees.

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The velocity of the wreckage after the collision is 32.015 m/s in the direction 38.66° north of east.

This problem is related to the conservation of linear momentum which states the total momentum of an isolated system remains constant if no external forces act on it. Momentum is a vector quantity having both magnitude and direction. The total initial and final momentum in both horizontal (x) and vertical (y) directions must be equal.

Initial momentum of the car is (1500 kg)*(25.0 m/s) = 37500 kg*m/s in the east direction whereas initial momentum of the van is (2500 kg)*(20.0 m/s) = 50000 kg*m/s in the north direction.

Since the collision is perfectly inelastic, the two vehicles stick together after the collision and move as one. Therefore, the final momentum is the vector sum of the individual momenta. We therefore calculate the magnitude of the resultant velocity using Pythagoras theorem, √[(25.0 m/s)² + (20.0 m/s)²] = 32.015 m/s.

To find the direction of the final velocity, we use the tangent of the angle which is equal to the vertical component divided by the horizontal component, which gives us an angle of 38.66° north of east.

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Answer:

1.18 x 10^-3 m^3/s

Explanation:

η = 0.25 N s/m^2, radius, r = 5 mm = 0.005 m, l = 25 cm = 0.25 m

P = 300 kPa = 300,000 Pa, Volume of flow = ?

By use of Poiseuillie's equation

Volume of flow = π P r^4 / (8 η l)

Volume of flow = (3.14 x 300000 x 0.005^4) / (8 x 0.25 x 0.25)

Volume of flow = 1.18 x 10^-3 m^3/s

Using Poiseuille's Law for viscous flow, the volume of oil flowing through the tube per second is calculated to be 1.8 liters/sec.

The volume of oil flowing through the tube per second can be calculated using Poiseuille's Law, which is an equation for viscous flow. This Law states that the flow rate of fluid through a pipe (Q) is directly proportional to the fourth power of the radius (r) of the tube, the pressure difference (ΔP) and inversely proportional to the viscosity (η) of the fluid, and the length (L) of the tube.

Using Poiseuille's Law, we have:

Q = (π/8) * (ΔP/r^4L) * η

By substituting the provided values:

Q = (π/8) * (300,000 Pa/(0.005 m)^4 * 0.25 m) * 0.25 N ∙ s/m^2

This yields: Q = 0.0018 m^3/s, or 1.8 liters/sec.

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Answer:

23 cm

Explanation:

m = 64.3 kg, diameter = 0.910 cm

radius, r = 0.455 cm = 4.55 x 10^-3 m

A = 3.14 x (4.55 x 10^-3)^2 = 6.5 x 10^-5 m^2

L = 32.2 m

Y = 1.35 x 10^9 Pa

Let the stretched length is l.

Use the formula for Young's modulus

Y = mg L / A l

l = m g L / A Y

l = (64.3 x 9.8 x 32.2) / (6.5 x 10^-5 x 1.35 x 10^9)

l = 0.23 m

l = 23 cm

Answer:

y = 43.2 cm

Explanation:

As we know by the formula of diffraction we have

[tex]a sin\theta = N\lambda[/tex]

here we have

[tex]\theta = 17.4[/tex]

N = 4

[tex]\lambda = 602 nm[/tex]

now we have

[tex]a sin(17.4) = 4(602\times 10^{-9})[/tex]

[tex]a = 8.05 \times 10^{-6} m[/tex]

now the distance of screen is 138 cm

now we can say

[tex]\frac{y}{L} = tan\theta[/tex]

[tex]y = L tan(\theta)[/tex]

[tex]y = 138 tan(17.4) = 43.2 cm[/tex]

Answer:

Part a)

[tex]Q = 2.47 \times 10^6[/tex]

Part b)

t = 4950.3 s

Explanation:

As we know that heat required to raise the temperature of container and water in it is given as

[tex]Q = m_1s_2\Delta T_1 + m_2s_2\Delta T_2[/tex]

here we know that

[tex]m_1 = 0.750[/tex]

[tex]s_1 = 900[/tex]

[tex]m_2 = 2.50 kg[/tex]

[tex]s_2 = 4186[/tex]

[tex]\Delta T_1 = \Delta T_2 = 100 - 30 = 70^oC[/tex]

now we have

[tex]Q_1 = 0.750(900)(70) + (2.5)(4186)(70) = 779800 J[/tex]

now heat require to boil the water

[tex]Q = mL[/tex]

here

m = 0.750 kg

[tex]L = 2.25 \times 10^6 J/kg[/tex]

now we have

[tex]Q_2 = 0.750(2.25 \times 10^6) = 1.7 \times 10^6 J[/tex]

Now total heat required is given as

[tex]Q = Q_1 + Q_2[/tex]

[tex]Q = 779800 + 1.7 \times 10^6 = 2.47 \times 10^6 J[/tex]

Part b)

Time taken to heat the water is given as

[tex]t = \frac{Q}{P}[/tex]

here we know that

power = 500 W

now we have

[tex]t = \frac{2.47 \times 10^6}{500} = 4950.3 s[/tex]

Answer:

[tex]v = \sqrt{v_0^2 - \frac{2k}{3}(s^3 - s_0^3)}[/tex]

v = 9.67 m/s

Explanation:

As we know that acceleration is rate of change in velocity

so it is defined as

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

[tex]a = v\frac{dv}{ds}[/tex]

here we know that

[tex]a = - ks^2 = v\frac{dv}{ds}[/tex]

now we have

[tex]vdv = - ks^2ds[/tex]

integrate both sides we have

[tex]\int vdv = -k \int s^2ds[/tex]

[tex]\frac{v^2}{2} - \frac{v_0^2}{2} = -k(\frac{s^3}{3} - \frac{s_0^3}{3})[/tex]

[tex]v^2 = v_0^2 - \frac{2k}{3}(s^3 - s_0^3)[/tex]

here we know that

[tex]v_0 = 10 m/s[/tex]

[tex]s_0 = 3 m[/tex]

[tex]v^2 = 10^2 - \frac{2(0.10)}{3}(5^3 - 3^3)[/tex]

[tex]v = 9.67 m/s[/tex]

Answer:

Power, P = 600 watts

Explanation:

It is given that,

Mass of sprinter, m = 54 kg

Speed, v = 10 m/s

Time taken, t = 3 s

We need to find the average power generated. The work done divided by time taken is called power generated by the sprinter i.e.

[tex]P=\dfrac{W}{t}[/tex]

Work done is equal to the change in kinetic energy of the sprinter.

[tex]P=\dfrac{\dfrac{1}{2}mv^2}{t}[/tex]

[tex]P=\dfrac{\dfrac{1}{2}\times 54\ kg\times (10\ m/s)^2}{3\ s}[/tex]

P = 900 watts

So, the average power generated by the sprinter is 900 watts. Hence, this is the required solution.

Final answer:

The average power generated is 900 Watts, which is approximately 1.21 horsepower.

Explanation:

The question involves calculating the average power generated by a sprinter during acceleration. Power is defined as the work done per unit time. To find the power, we need to first calculate the work done, which, in the case of the sprinter, is the kinetic energy gained as they accelerate to the final speed.

In this case, we can calculate the kinetic energy (KE) of the sprinter using the formula KE = 0.5 × mass ×[tex]velocity^2[/tex]. The mass of the sprinter is given as 54 kg and the final velocity is 10 m/s. Thus, KE = 0.5 × 54 kg × [tex](10 m/s)^2[/tex] = 2700 Joules.

Now, the power can be found by dividing the work done by the time it takes to do that work. The sprinter takes 3 seconds to achieve this final velocity. So, the average power P is P = work/time = 2700 Joules / 3 s = 900 Watts.

Converting this power to horsepower (1 horsepower = 746 Watts), we get approximately P = 900 W × 1 hp/746 W = 1.21 horsepower.

Answer:

1.85 x 10^10 cycles per second

Explanation:

B = 0.66 T, theta = 90 degree, q = 1.6 x 10^-19 C,

The time period of electron is given by

T = 2 π m / B q

Frequency is teh reciprocal of time period.

f = 1 /T

f = B q / (2 π m)

f = (0.66 x 1.6 x 10^-19) / (2 x 3.14 x 9.1 x 10^-31)

f = 1.85 x 10^10 cycles per second

Answer:

Rate of heat is 45.1 kJ/min

Explanation:

Heat required to evaporate the water is given by

Q = mL

here we know that

[tex]L = 2.25 \times 10^6 J/kg[/tex]

now we have

[tex]Q = (0.200)(2.25 \times 10^6 J/kg)[/tex]

[tex]Q = 452.1 kJ[/tex]

now the power is defined as rate of energy

[tex]P = \frac{Q}{t}[/tex]

[tex]P = \frac{452.1 kJ}{10}[/tex]

[tex]P = 45.1 kJ/min[/tex]

Answer:

[tex]\omega _{f}=0.3107rad/sec[/tex]

partb) [tex]\omega _{f}=14.93rad/sec[/tex]

Explanation:

Since the system is isolated it's angular momentum shall be conserved

[tex]L_{system}=I_{system}\omega \\\\I_{system}=I_{rod}+2\times m\times r^{2}\\\\I_{system}=\frac{1}{12}ml^{2}+2\times 0.2\times (4.8\times 10^{-2})^{2}\\\\I_{system}=1.22845\times 10^{-3}kgm^{2}\\\\L_{system}=1.22845\times 10^{-3}kgm^{2}\times 3.66rad/sec[/tex]

[tex]\L_{system}=4.496\times 10^{-3}kgm^{2}[/tex]

Now the final angular momentum of the system

[tex]L_{fianl}=I_{final}\omega \\\\I_{final}=I_{rod}+2\times m\times r_{f}^{2}\\\\I_{final}=\frac{1}{12}ml^{2}+2\times 0.2\times (0.19)^{2}\\\\I_{final}=0.01447kgm^{2}\\\\L_{final}=0.01447kgm^{2}\times \omega _{final}rad/sec[/tex]

Thus equating initial and final angular momentum we solve for final angular velocity of rod as

[tex]\omega _{f}=\frac{4.496\times 10^{-3}}{0.01447}\\\\\omega _{f}=0.3107rad/sec[/tex]

part b)

When the rings leave the system we again conserve the angular momentum just before the rings leave the system and the instant when they just leave

[tex]\therefore 0.01447=\frac{1}{12}ml^{2}w\\\\\therefore w=\frac{4.582\times 10^{-3}}{3.068\times10^{-4} }\\\\w_{final}=14.93rad/sec[/tex]

Answer:

A)[tex]\omega_{f}=2.92\frac{rev}{min}[/tex].

B)[tex]\omega_{f}=140\frac{rev}{min}[/tex].

Explanation:

A)

For this problem, we will use the conservation of angular momentum.

[tex]L_{0}=L_{f}\\[/tex].

In the beginning, we have that

[tex]L_{0}=I_{0}\omega_{0}\\[/tex]

where [tex]I_{0}[/tex] is the inertia moment of all the system at the starting position, this is the inertia moment of the rod plus the inertia moment of each ring ([tex]mr^{2}[/tex], with [tex]r[/tex] the distance from the ring to the fixed axis and, [tex]m[/tex] its mass) at the starting position and, [tex]\omega_{0}[/tex] is the initial angular velocity. So

[tex]L_{0}=(\frac{1}{12}Ml^{2}+2mr_{0}^{2})\omega_{0}[/tex].

When the rings are at the ends of the rod the angular momentum becomes

[tex]L_{f}=(\frac{1}{12}Ml^{2}+2mr_{f}^{2})\omega_{f}[/tex],

where [tex]r_{f}[/tex] is the distance from the fixed axis to the end of the rod (the final position of the rings).

Using conservation of angular momentum we get

[tex](\frac{1}{12}Ml^{2}+2mr_{0}^{2})\omega_{0}=(\frac{1}{12}Ml^{2}+2mr_{f}^{2})\omega_{f}[/tex].

thus

[tex]\omega_{f}=\frac{(\frac{1}{12}Ml^{2}+2mr_{0}^{2})\omega_{0}}{(\frac{1}{12}Ml^{2}+2mr_{f}^{2})}[/tex]

computing this last expresion we get

[tex]\omega_{f}=\frac{(\frac{1}{12}(2.55*10^{-2})(0.380)^{2}+2(0.200)(4.80*10^{-2})^{2})(35.0)}{(\frac{1}{12}(2.55*10^{-2})(0.380)^2+2(0.200)(0.19)^{2})}[/tex]

[tex]\omega_{f}=2.92\frac{rev}{min}[/tex].

B)

Again we use the conservation of angular momentum. The initial angular momentum if the same as before. The final angular momentum will be

[tex]L_{f}=(\frac{1}{12}Ml^{2})\omega_{f}[/tex],

this time we will not take into account the inertia moment of the rings because they are no longer part of the system (they leave the rod).

[tex](\frac{1}{12}Ml^{2}+2mr_{0}^{2})\omega_{0}=(\frac{1}{12}Ml^{2})\omega_{f}[/tex].

thus

[tex]\omega_{f}=\frac{(\frac{1}{12}Ml^{2}+2mr_{0}^{2})\omega_{0}}{(\frac{1}{12}Ml^{2})}[/tex]

computing this last expresion we get

[tex]\omega_{f}=\frac{(\frac{1}{12}(2.55*10^{-2})(0.380)^{2}+2(0.200)(4.80*10^{-2})^{2})(35.0)}{(\frac{1}{12}(2.55*10^{-2})(0.380)^2})[/tex]

[tex]\omega_{f}=140\frac{rev}{min}[/tex].

Answer:

Option B is the correct answer.

Explanation:

Speed of car = 70mph [tex]=70\times \frac{1609}{3600}=31.28m/s[/tex]

Deceleration of car = 2 m/s²

We have equation of motion

          v = u + at

          0 = 31.28 - 2 t

           t = 15.64 seconds.

We also have

          [tex]s=ut+\frac{1}{2}at^2=31.28\times 15.64-\frac{1}{2}\times 2\times 15.64^2=244.70m[/tex]

Option B is the correct answer.

Answer: The mass of the object will be 11250 kg.

Explanation:

Density is defined as the mass contained per unit volume.

[tex]Density=\frac{mass}{Volume}[/tex]

Given :

Density of the object= [tex]12500kg/m^3[/tex]

Mass of object = ?

Volume of the object = [tex]0.9m^3[/tex]

Putting in the values we get:

[tex]12500kg/m^3=\frac{mass}{0.9m^3}[/tex]

[tex]12500kg/m^3=\frac{mass}{0.9m^3}[/tex]

[tex]mass=11250kg[/tex]

Thus the mass of the object is 11250 kg.

Answer:

Acceleration of gravity on Noveria = 4.4 m/s²

Explanation:

Commander Shepard, an N7 spectre for Earth, weighs 799 N on the Earth's surface.

We have weight, W = mg

Acceleration due to gravity, g = 9.81m/s²    

799 = m x 9.81

Mass of Shepard, m = 81.45 kg            

She lands on Noveria, a distant planet in our galaxy, she weighs 356 N.

We have weight, W = mg'

                 356 = 81.45 xg'

Acceleration of gravity on Noveria, g' = 4.4 m/s²