cellus
An object ends up at a position of
327 m after a displacement of -144 m.
What was its initial position?
(Unit = m)

Answer:

0.1575 m/s^2

Explanation:

Solution:-

- Acceleration ( a ) is expressed as the rate of change of velocity ( v ).

- We are given that the trains starts from rest i.e the initial velocity ( vo ) is equal to 0. Then the train travels from reference point ( so = 0 ) to ( sf = 5.6 km ) from the reference.

- During the travel the train accelerated uniformly to a speed of ( vf =42 m/s ).

- We will employ the use of 3rd kinematic equation of motion valid for constant acceleration ( a ) as follows:

                         [tex]v_f^2 = v_i^2 + 2*a*( s_f - s_o )[/tex]

- We will plug in the given parameters in the equation of motion given above:

                         [tex]42^2 = 0^2 + 2*a* ( 5600 - 0 )\\\\1764 = 11,200*a\\\\a = \frac{1,764}{11,200} \\\\a = 0.1575 \frac{m}{s^2}[/tex]

Answer: the acceleration during the first 5.6 km of travel is 0.1575 m / s^2

Answer:

a. 78 degree

Explanation:

According to Snell's Law, we have:

(ni)(Sin θi) = (nr)(Sin θr)

where,

ni = Refractive index of medium on which light is incident

ni = Refractive index of ethyl alcohol = 1.361

nr = Refractive index of medium from which light is refracted

nr = Refractive index of ethyl alcohol = 1.333

θi = Angle of Incidence

θr = Angle of refraction

So, the Angle of Incidence is know as the Critical Angle (θc), when the refracted angle becomes 90°. This is the case of total internal reflection. That is:

θi = θc

when, θr = 90°

Therefore, Snell's Law becomes:

(1.361)(Sin θc) = (1.333)(Sin 90°)

Sin θc = 1.333/1.361

θc = Sin⁻¹ (0.9794)

θc = 78.35° = 78° (Approximately)

Therefore, correct answer will be:

a. 78 degree

The angle relative to the normal interface of the two liquids at which the light is totally reflected is 78 degrees.

From the information given;

According to Snell's Law of refraction;

[tex]\mathbf{n_1 sin \theta _c = n_2 sin \theta_r}[/tex]

[tex]\mathbf{1.361 \times sin \theta _c = 1.333 \times sin 90}[/tex]

[tex]\mathbf{ sin \theta _c =\dfrac{ 1.333 \times sin 90}{1.361}}[/tex]

[tex]\mathbf{ sin \theta _c =\dfrac{ 1.333 \times 1}{1.361}}[/tex]

[tex]\mathbf{ \theta _c = sin^{-1} (0.9794)}[/tex]

[tex]\mathbf{ \theta _c =78.35^0}[/tex]

[tex]\mathbf{ \theta _c \simeq78^0}[/tex]

Therefore, we can conclude that the angle relative to the normal interface of the two liquids at which the light is totally reflected is 78 degrees.

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Answer:

Explanation:

The velocity at the inlet and exit of the control volume are same [tex]V_i=V_e=V[/tex]

Calculate the inlet and exit velocity of water jet

[tex]V=V_j+V_e\\\\V=30+14\\\\V=44m/s[/tex]

The conservation of mass equation of steady flow

[tex]\sum ^e_i\bar V. \bar A=0\\\\(-V_iA_i+V_eA_e)=0[/tex]

[tex]A_i\ \texttt {is the inlet area of the jet}[/tex]

[tex]A_e\ \texttt {is the exit area of the jet}[/tex]

since inlet and exit velocity of water jet are equal so the inlet and exit cross section area of the jet is equal

The expression for thickness of the jet

[tex]A_i=A_e\\\\\frac{\pi}{4} D_j^2=2\pi Rt\\\\t=\frac{D^2_j}{8R}[/tex]

R is the radius

t is the thickness of the jet

D_j is the diameter of the inlet jet

[tex]t=\frac{(100\times10^{-3})^2}{8(230\times10^{-3}} \\\\=5.434mm[/tex]

(b)

[tex]R-x=\rho(AV_r)[-(V_i)+(V_c)\cos 60^o]\\\\=\rho(V_j+V_c)A[-(V_i+V_c)+(V_i+V_c)\cos 60^o]\\\\=\rho(V_j+V_c)(\frac{\pi}{4}D_j^2 )[V_i+V_c](\cos60^o-1)][/tex]

[tex]1000kg/m^3=\rho\\\\44m/s=(V_j+V+c)\\\\100\times10^{-3}m=D_j[/tex]

[tex]R_x=[1000\times(44)\frac{\pi}{4} (10\times10^{-3})^2[(44)(\cos60^o-1)]]\\\\=-7603N[/tex]

The negative sign indicate that the direction of the force will be in opposite direction of our assumption

Therefore, the horizontal force is -7603N

Answer:

10 Sig Figs

Explanation:

Just start counting at the first non zero after the decimal so in this case the nine, and count all of the numbers including zeros after that.

Answer:

2000 m

Explanation:

since the boat is 1000 m from the river bank, the beam must be focused at this point. This indicates that the focal length is 1000 m

for a spherical mirror, the focal length is given by

f = R/2

where R is the radius of curvature

1000 = R/2

R = 2000 m

R = 2000 m

this means that the radius of curvature must be 2000 m

Answer:

a. global warming

Explanation:

that's the definitain of global warming

Answer:

A climate

Explanation:

Answer:

Maximum weight that can be lifted = 18,000 N

Explanation:

Given:

Cross-sectional area of input (A1) = 0.004 m²

Cross-sectional area of the output (A2) = 1.2 m ²

Force (F) = 60 N

Computation:

Pressure on input piston (P1) = F / A1

Assume,

Maximum weight lifted by piston = W

Pressure on output piston (P2) = W / A2

We, know that

P1 = P2

[F / A1]  = [W / A2]

[60 / 0.004] = [W / 1.2]

150,00 = W / 1.2

Weight = 18,000 N

Maximum weight that can be lifted = 18,000 N

Answer:

a) 5.143 m/s^2

b) T = 15.43 N

c) Fr = 10.29 N

d) 5.143 m/s^2 , T = 15.43 N

Explanation:

Given:-

- The mass of left most block, m1 = 1.0 kg

- The mass of center block, m2 = 2.0 kg

- The mass of right most block, m3 = 4.0 kg

- A force that acts on the right most block, F = 36 N

Solution:-

a)

- For the first part we will consider the three blocks with masses ( m1 , m2 , and m3 ) as one system on which a force of F = 36 N is acted upon. The masses m1 and m3 are connected with a string with tension ( T ) and the m1 and m2 are in contact.

- We apply the Newton's second law of motion to the system with acceleration ( a ) and the combined mass ( M ) of the three blocks as follows:

                       [tex]F = M*a\\\\36 = ( 1 + 2 + 4 )*a\\\\a = \frac{36}{7}\\\\a = 5.143 \frac{m}{s^2}[/tex]

Answer: The system moves in the direction of external force ( F ) i.e to the right with an acceleration of 5.143 m/s^2

b)

- The blocks with mass ( m1 and m3 ) are connected with a string with tension ( T ) with a combined acceleration of ( a ).

- We will isolate the massive block ( m3 ) and notice that two opposing forces ( F and T ) act on the block.

- We will again apply the Newton's 2nd law of motion for the block m3 as follows:

                       [tex]F_n_e_t = m_3 * a\\\\F - T = m_3 * a\\\\36 - T = 4*5.143\\\\T = 36 - 20.5714\\\\T = 15.43 N[/tex]

Answer:- A tension of T = 15.43 Newtons acts on both blocks ( m1 and m3 )

c)

- We will now isolate the left most block ( m1 ) and draw a free body diagram. This block experiences two forces that is due to tension ( T ) and a reaction force ( Fr ) exerted by block ( m2 ) onto ( m3 ).

- Again we will apply the the Newton's 2nd law of motion for the block m3 as follows:

                         [tex]F_n_e_t = m_1*a\\\\T - F_r = m_1*a\\\\15.43 - F_r = 1*5.143\\\\F_r = 15.43 - 5.143\\\\F_r = 10.29 N[/tex]

- The reaction force ( Fr ) is contact between masses ( m1 and m2 ) exists as a pair of equal magnitude and opposite direction acting on both the masses. ( Newton's Third Law of motion )

Answer: The block m2 experiences a contact force of ( Fr = 10.29 N ) to the right.

d)

- If we were to stack the block ( m2 ) on-top of block ( m1 ) such that block ( m2 ) does not slip we the initial system would remain the same and move with the same acceleration calculated in part a) i.e 5.143 m/s^2

- We will check to see if the tension ( T ) differs or not as the two block ( m1 and m2 ) both experience the same Tension force ( T ) as a sub-system. with a combined mass of ( m1 + m2 ).

- We apply the Newton's 2nd law of motion for the block m3 as follows:

                            [tex]T = ( m_1 + m_2 ) *a\\\\T = ( 1 + 2 ) * 5.143\\\\T = 15.43 N[/tex]

Answer: The acceleration of the whole system remains the same at a = 5.143 m/s^2 and the tension T = 15.43 N also remains the same.

Explanation:

If the kinetic energy of an object is given and we need to find its velocity of motion, then we can find it by using the formula of kinetic energy as :

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

m is mass of the object

We can rearrange the above equation such that,

[tex]v=\sqrt{\dfrac{2K}{m}}[/tex]

Hence, this is the velocity at the end of a rollercoaster ride.

Answer:

= 0.0532 cm^3

Explanation:

The computation of volume of the sphere is shown below:-

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

Where,

Density = 13.6 g/cm^3

Mass of sphere = 0.723 g

now we will put the values into the above formula to reach volume of the sphere which is here below:-

[tex]Volume = \frac{0.723}{13.6}[/tex]

= 0.0532 cm^3

Therefore for computing the volume of the sphere we simply applied the above formula.

Answer:

the arrow symbol ⇒ in irreversible reactions and doble arrow symbol in reversible reactios⇔

Explanation:

i hope this will help you

Answer:

It will be A. So since its 2 times more the kinetic energy. But then you have to square it 2^2 = 4

Answer:

The location are [tex]x_1 = 2 \ and \ x_2 = 3[/tex]

Explanation:

From the question we are told that

    The potential energy is  [tex]U(x) = (2.0 \ J/m^3) * x^3 - (15 \ J/m) * x^2 + (36 \ J/m) * x - 23 \ J[/tex]

The force on the mass can be mathematically evaluated as  

      [tex]F = - \frac{d U(x)}{d x } = -( 6 x^2 - 30x +36)[/tex]

The negative sign shows that the force is moving in the opposite  direction of the potential energy

       [tex]F = - 6 x^2 + 30x - 36[/tex]

At critical point

      [tex]\frac{d U(x)}{dx} = 0[/tex]

So  

     [tex]- 6 x^2 + 30x - 36 = 0[/tex]

     [tex]- x^2 + 5x - 6 = 0[/tex]

Using quadratic equation formula to solve this we have that

       [tex]x_1 = 2 \ and \ x_2 = 3[/tex]

Answer:

3.2 m/s

Explanation:

Given:

Δx = 1000 m

v₀ = 23 m/s

a = -0.26 m/s²

t = 76 s

Find: v

This problem is over-defined.  We only need 3 pieces of information, and we're given 4.  There are several equations we can use.  For example:

v = at + v₀

v = (-0.26 m/s²) (76 s) + (23 m/s)

v = 3.2 m/s

Or:

Δx = ½ (v + v₀) t

(1000 m) = ½ (v + 23 m/s) (76 s)

v = 3.3 m/s

Or:

v² = v₀² + 2aΔx

v² = (23 m/s)² + 2(-0.26 m/s²)(1000 m)

v = 3.0 m/s

Or:

Δx = vt − ½ at²

(1000 m) = v (76 s) − ½ (-0.26 m/s²) (76 s)²

v = 3.3 m/s

As you can see, you get slightly different answers depending on which variables you use.  Since 1000 m has 1 significant figure, compared to the other variables which have 2 significant figures, I recommend using the first equation.

Answer:

107 N, option d

Explanation:

Given that

mass of the ball, m = 0.2 kg

initial velocity of the ball, u = 20 m/s

final velocity of the ball, v = -12 m/s

time taken, Δt = 60 ms

Solving this question makes us remember "Impulse Theorem"

It states that, "that the product between the average force applied and the duration of the collision is equal to the change in momentum of the object"

Mathematically, it is represented as

FΔt = m(v - u), where

F = the average force

Δt = time taken

m = mass of the ball

v = final velocity of the ball

u = initial velocity of the ball

From the question we were given, if we substitute the values in it, we have

F = ?

Δt = 60 ms = 0.06s

m = 0.2 kg

v = -12 m/s

u = 20 m/s

F = 0.2(-12 - 20) / 0.06

F = (0.2 * -32) / 0.06

F = -6.4 / 0.06

F = -106.7 N

Thus, the magnitude is 107 N

Question: The circular handle of a faucet is attached to a rod that opens and closes a valve when the handle is turned. If the rod has a diameter of 1cm and the IMA of the machine is 6, what is the radius of the handle?

Answer:

Radius of the handle = 3 cm = 0.03 m

Explanation:

Mechanical Accuracy, MA = (Radius of the handle)/(Radius of the rod).......(1)

Diameter of the rod = 1 cm

Radius of the rod = Diameter/2

Radius of the rod = 1/2

Radius of the rod = 0.5 cm

Mechanical Accuracy of the machine, MA = 6

Substitute the values into equation (1)

6 =  (Radius of the handle)/0.5

Radius of the handle = 6 * 0.5

Radius of the handle = 3 cm

The Radius of the handle is = 3 cm = 0.03 m

Since

Mechanical Accuracy, MA = (Radius of the handle)/(Radius of the rod)

Here,

Diameter of the rod = 1 cm

We know that

The radius of the rod = Diameter/2

So,

Radius of the rod = 1/2

So,

Radius of the rod = 0.5 cm

Now

Mechanical Accuracy of the machine, MA = 6

Now

6 =  (Radius of the handle)/0.5

Radius of the handle = 6 * 0.5

Radius of the handle = 3 cm

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Answer:

St Lawrence Sea way

Explanation:

The great lake connects the middle of North America which is at the Canada-United states border connecting to the Atlantic Ocean through the St Lawrence River.

Answer:

  More than 48%

Explanation:

If the interest is computed monthly on the outstanding balance, it has an effective annual rate of ...

  (1 +4%)^12 -1 = 60.1% . . . .  more than 48%

The effective annual or yearly interest rate would be=30.56% which is Between 4% and 48%

The formula used to calculate annual Interest rate =

[tex](1+ \frac{i}{n} ) {}^{n} - 1[/tex]

where i= nominal interest rate = 4%

n= number of periods= 12 months

Annual Interest rate=

[tex](1 + \frac{4\%}{12} ) {}^{12} - 1[/tex]

= (1+0.333)^12 -1

= (1.333)^12-1

= 31.56 - 1

= 30.56%

Therefore, the effective annual or yearly interest rate would be= 30.56%

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Answer:

The speed of the freight car decreases.

Explanation:

According to the law of conservation of momentum indicates that for colliding in an isolated system, the total momentum pre and post collision is same for the two objects this is done because the momentum that one item has lost is same for the momentum that the other received

In the given situation, the freight car travels at constant speed along a frictionless railroad line. The top floor freight car is open. Then a huge load of coal is dumped inside the car.

Therefore the speed of the freight car decreased by applying the law of conservation of momentum i

Answer:

(A) 4* 6 ^ ⁻6 T m² (B) 2 * 10 ^ ⁻6 v

Explanation:

Solution

Given that:

A refrigerator magnet about = 2 mm

The estimated magnetic field strength of the magnet is = 5 m T

The Area = 8 cm²

Now,

(A) The magnetic flux ΦB = BA

Thus,

ΦB  = (5 * 10^⁻ 3) ( 4 * 10 ^⁻2) * ( 2 * 10^ ⁻2) Tm²

So,

ΦB =  4* 6 ^ ⁻6 T m²

(B)By applying Faraday's Law we have the following formula given below:

Ε = Bℓυ

Here,

ℓ = 2 cm the same as 2 * 10 ^⁻2 m

B = 5 m T = 5 * 10 ^ ⁻3 T

υ = 2 cm/s  = 2 * 10 ^ ⁻2 m/s

Thus,

Ε = (5 * 10 ^ ⁻3 T) *  (2 * 10 ^ ⁻2) (2 * 10 ^ ⁻2) v

E =2 * 10 ^ ⁻6 v

A) The magnetic flux ΦB into the refrigerator door behind the magnet :

B) The estimated emf induced under the rectangle on the door's surface ;

Given data :

magnetic field strength of magnet ( B )  = 5 mT

size of refrigerator magnet = 2 mm

Area of magnet ( A )  = 4 * 2 = 8 cm²

where ; ΦB  = BA

ΦB = ( 5 * 10⁻³ ) * ( 4 * 10⁻² ) * ( 2 * 10⁻² ) Tm²

      =  4 * 6⁻⁶ Tm²

To determine the estimated emf we will apply Faraday's law

Ε = Bℓυ ---- ( 2 )

where : B = 5 * 10⁻³ T,  ℓ = 2 * 10⁻² m,  υ = 2 * 10⁻² m/s

insert values into equation 2

E = ( 5 * 10⁻³ ) * ( 2 * 10⁻² ) * ( 2 * 10⁻² )

  = 2 * 10⁻⁶ v

Hence we can conclude that The magnetic flux ΦB is 4 * 6⁻⁶ Tm² and The estimated emf induced is  2 * 10⁻⁶ v

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Explanation:

The relationship between electric force and distance between charged objects is given by the formula as follows :

[tex]F=\dfrac{kq_1q_2}{d^2}[/tex]

k is electrostatic constant and d is distance between charges

The electric force between charges is inversely proportional to the square of distance between them.

Answer:

i) A dimension is the physical nature of a quantity. The two limitations of dimensional analysis is as following:

ii) Given:

Velocity = v

Time = t

Force = F

Force = mass x acceleration

         = mass x velocity/time

So, mass= (force x time) / velocity

[mass] = Ftv^-1

Hence, dimesnion of mass is Ftv^-1.

Answer:

the correct answer is B

Explanation:

We analyze this exercise a little, the block goes into the air and is under the acceleration of gravity. The ball is fired by the hand and is describing a parabolic movement, subjected to the acceleration of gravity.

For the ball to hit the block we must have the distance the ball goes up equal to the distance the block moves, therefore we must shoot the ball at the block at its highest point.

Let's write the kinematic equation for the two bodies

The block. At the highest point of the path

      y = - ½ g t2

The ball, in its vertical movement

     y = vo t - ½ g t2

therefore the correct answer is B

Answer:

Energy Stored = 36000 J = 36 KJ

Explanation:

The power of a battery is given by the formula:

P = IV

where,

P = Power delivered by the battery

I = Current Supplied to the battery

V = Potential Difference between terminals of battery = 12 volt

Now, we multiply both sides by the time period (t):

Pt = VIt

where,

Pt = (Power)(Time) = Energy Stored = E = ?

It = Battery Current Rating = 50 A.min

Converting this to A.sec;

It = Battery Current Rating = (50 A.min)(60 sec/min) = 3000 A.sec

Therefore,

E = (12 volt)(3000 A.sec)

E = 36000 J = 36 KJ

Answer:

(a) t = 3.74 s

(b) H = 136.86 m

(c) Vₓ = 41.83 m/s,  Vy = 0 m/s

(d) ax = 0 m/s²,  ay = 9.8 m/s²

Explanation:

(a)

Time to reach maximum height by the projectile is given as:

t = V₀ Sinθ/g

where,

V₀ = Launching Speed = 55.6 m/s

Angle with Horizontal = θ = 41.2°

g = 9.8 m/s²

Therefore,

t = (55.6 m/s)(Sin 41.2°)/(9.8 m/s²)

t = 3.74 s

(b)

Maximum height reached by projectile is:

H = V₀² Sin²θ/g

H = (55.6 m/s)² (Sin²41.2°)/(9.8 m/s²)

H = 136.86 m

(c)

Neglecting the air resistance, the horizontal component of velocity remains constant. This component can be evaluated by the formula:

Vₓ = V₀ₓ = V₀ Cos θ

Vₓ = (55.6 m/s)(Cos 41.2°)

Vₓ = 41.83 m/s

Since, the projectile stops momentarily in vertical direction at the highest point. Therefore, the vertical component of velocity will be zero at the highest point.

Vy = 0 m/s

(d)

Since, the horizontal component of velocity is uniform. Thus there is no acceleration in horizontal direction.

ax = 0 m/s²

The vertical component of acceleration is always equal to the acceleration due to gravity during projectile motion:

ay = 9.8 m/s²

Answer:

The thermal conductivity of the insulated metal rod is [tex]202.92\,\frac{W}{m\cdot K}[/tex].

Explanation:

This is a situation of one-dimensional thermal conduction of a metal rod in a temperature gradient. The heat transfer rate through the metal rod is calculated by this expression:

[tex]\dot Q = \frac{k_{rod}\cdot A_{c, rod}}{L_{rod}}\cdot \Delta T[/tex]

Where:

[tex]\dot Q[/tex] - Heat transfer due to conduction, measured in watts.

[tex]L_{rod}[/tex] - Length of the metal rod, measured in meters.

[tex]A_{c,rod}[/tex] - Cross section area of the metal rod, measured in meters.

[tex]k_{rod}[/tex] - Thermal conductivity, measured in [tex]\frac{W}{m\cdot K}[/tex].

Let assume that heat conducted to melt some ice was transfered at constant rate, so that definition of power can be translated as:

[tex]\dot Q = \frac{Q}{\Delta t}[/tex]

Where Q is the latent heat required to melt the ice, whose formula is:

[tex]Q = m_{ice}\cdot L_{f}[/tex]

Where:

[tex]m_{ice}[/tex] - Mass of ice, measured in kilograms.

[tex]L_{f}[/tex] - Latent heat of fussion, measured in joules per gram.

The latent heat of fussion of water is equal to [tex]330000\,\frac{J}{g}[/tex]. Hence, the total heat received by the ice is:

[tex]Q = (6.15\,g)\cdot \left(330\,\frac{J}{g} \right)[/tex]

[tex]Q = 2029.5\,J[/tex]

Now, the heat transfer rate is:

[tex]\dot Q = \frac{2029.5\,J}{(10\,min)\cdot \left(60\,\frac{s}{min} \right)}[/tex]

[tex]\dot Q = 3.382\,W[/tex]

Turning to the thermal conduction equation, thermal conductivity is cleared and computed after replacing remaining variables: ([tex]L_{rod} = 0.75\,m[/tex], [tex]A_{c,rod} = 1.25\times 10^{-4}\,m^{2}[/tex], [tex]\Delta T = 100\,K[/tex], [tex]\dot Q = 3.382\,W[/tex])

[tex]\dot Q = \frac{k_{rod}\cdot A_{c, rod}}{L_{rod}}\cdot \Delta T[/tex]

[tex]k_{rod} = \frac{\dot Q \cdot L_{rod}}{A_{c,rod}\cdot \Delta T}[/tex]

[tex]k_{rod} = \frac{(3.382\,W)\cdot (0.75\,m)}{(1.25\times 10^{-4}\,m^{2})\cdot (100\,K)}[/tex]

[tex]k_{rod} = 202.92\,\frac{W}{m\cdot K}[/tex]

The thermal conductivity of the insulated metal rod is [tex]202.92\,\frac{W}{m\cdot K}[/tex].

Answer:

the magnitude is 7 and sign of the point charge on the surface shell is -13

Explanation:

Answer:

The angular momentum is  [tex]L = 8440.32 \ kg \cdot m^2 \cdot s^{-1}[/tex]

Explanation:

From the question we are told that

    The mass of the woman is  [tex]m = 50 \ kg[/tex]

     The angular  speed of the rim is  [tex]w = 0.80 \ rev/s = 0.8 * [\frac{2 \pi}{1} ] = 5.024 \ rad \cdot s^{-1}[/tex]

      The mass of the disk is  [tex]m_d = 110 \ kg[/tex]

       The radius of the disk is [tex]r_d = 4.0 \ m[/tex]

The moment of inertia of the disk is mathematically represented as

        [tex]I_D = \frac{1}{2} m_d r^2_d[/tex]

substituting values

          [tex]I_D = \frac{1}{2} * 110 * 4^2[/tex]

          [tex]I_D = 880 \ kg \cdot m^2[/tex]

The moment of inertia of the woman is  

          [tex]I_w = m * r_d^2[/tex]

substituting values

        [tex]I_w = 50 * 4^2[/tex]

       [tex]I_w =800\ kg[/tex]

The moment of inertia of the system (the woman + the large disk ) is  

        [tex]I_t = I_w + I_D[/tex]

substituting values  

      [tex]I_t = 880 +800[/tex]

     [tex]I_t =1680 \ kg \cdot m^2[/tex]

The angular momentum of the system is

      [tex]L = I_t w[/tex]

substituting values  

      [tex]L = 1680 * 5.024[/tex]

      [tex]L = 8440.32 \ kg \cdot m^2 \cdot s^{-1}[/tex]

Answer:

Explanation:

angular momentum of the putty about the point of rotation

= mvR   where m is mass , v is velocity of the putty and R is perpendicular distance between line of velocity and point of rotation .

= .045 x 4.23 x 2/3 x .95 cos46

= .0837 units

moment of inertia of rod = ml² / 3 , m is mass of rod and l is length

= 2.95 x .95² / 3

I₁ = .8874 units

moment of inertia of rod + putty

I₁ + mr²

m is mass of putty and r is distance where it sticks

I₂  = .8874 + .045 x (2 x .95 / 3)²

I₂ = .905

Applying conservation of angular momentum

angular momentum of putty = final angular momentum of rod+ putty

.0837 = .905 ω

ω is final angular velocity of rod + putty

ω = .092 rad /s .

Answer:

P1 = 0 gage

P2 = 87.9 lb/ft³

Explanation:

Given data

Airplane flying = 200 mph = 293.33 ft/s

altitude height = 5000-ft

air velocity relative to the airplane = 273 mph = 400.4 ft/s

Solution

we know density at height 5000-ft is 2.04 × [tex]10^{-3}[/tex] slug/ft³

so here P1 + [tex]\frac{\rho v1^2}{2}[/tex]  = P2 + [tex]\frac{\rho v2^2}{2}[/tex]

and here

P1 = 0 gage

because P1 = atmospheric pressure

and so here put here value and we get

P1 + [tex]\frac{\rho v1^2}{2}[/tex]  = P2 + [tex]\frac{\rho v2^2}{2}[/tex]

0 + [tex]\frac{2.048 \times 10^{-3} \times 293.33^2}{2}[/tex]  [tex]= P2 + \frac{2.048 \times 10^{-3} \times 400.4^2}{2}[/tex]  

solve it we get

P2 = 87.9 lb/ft³