Water, in a 100-mm-diameter jet with speed of 30 m/s to the right, is deflected by a cone that moves to the left at 14 m/s. Determine (a) the thickness of the jet sheet at a radius of 230 mm. and (b) the external horizontal force needed to m

Answer:

a)Car E = Car D  > (Car F = Car B = Car C) > Car A

b)Car E = Car D  > (Car F = Car B = Car C) > Car A

Explanation:

Car A: mass = 500 kg; speed = 10 m/s

Car B: mass = 2000 kg;speed = 5 m/s

Car C:mass = 500 kg; speed = 20 m/s

Car D: mass = 1000 kg; speed = 20 m/s

Car E:mass = 4000 kg; speed = 5 m/s

Car F: mass = 1000 kg; speed = 10 m/s

Part a) Now we know that momentum of each car is product of mass and velocity , so we will have

CarA:

[tex]P_1 = m \times v\\P_1 = (500)(10)\\P_1 = 5 \times 10^3 kg m/s[/tex]

Car B:

[tex]P_2 = m v\\P_2 = (2000)(5)\\P_2 = 10^4 kg m/s[/tex]

Car C:

[tex]P_3 = m v\\P_3 = (500)(20)\\P_3 = 10^4 kg m/s[/tex]

Car D:

[tex]P_4 = m v\\P_4 = (1000)(20)\\P_4 = 2\times 10^4 kg m/s[/tex]

Car E:

[tex]P_5 = m v\\P_5 = (4000)(5)\\P_5 = 2\times 10^4 kg m/s[/tex]

Car F:

[tex]P_6 = m v\\P_6 = (1000)(10)\\P_6 = 10^4 kg m/s[/tex]

So the momentum is given as ,

Car E = Car D  > (Car F = Car B = Car C) > Car A

Part b)Impulse is given as change in momentum so here we can say that final momentum of all the cars will be zero as they all stops and hence the impulse is same as initial momentum of the car

so the order of impulse from largest to least is given as

Car E = Car D  > (Car F = Car B = Car C) > Car A

Answer:

Coefficient of friction.

Explanation:

The amount of friction divided by the weight of an object is equal to the coefficient of friction. It is a dimensional less number. It can be given by :

[tex]F=\mu N[/tex]

N is normal force.

[tex]\mu[/tex] = coefficient of friction

[tex]\mu=\dfrac{F}{N}[/tex]

Explanation:

A) Draw free body diagrams of both blocks.

Force P is pushing right on block A, which will cause it to move right along the incline.  Therefore, friction forces will oppose the motion and point to the left.

There are 5 forces acting on block A:

Applied force P pushing to the right,

Normal force N pushing up and left 10° from the vertical,

Friction force Nμ pushing down and left 10° from the horizontal,

Reaction force Fab pushing down,

and friction force Fab μ pushing left.

There are 2 forces acting on block B:

Reaction force Fab pushing up,

And elastic force kx pushing down.

(There are also horizontal forces on B, but I am ignoring them.)

Sum of forces on A in the x direction:

∑F = ma

P − N sin 10° − Nμ cos 10° − Fab μ = 0

Solve for N:

P − Fab μ = N sin 10° + Nμ cos 10°

P − Fab μ = N (sin 10° + μ cos 10°)

N = (P − Fab μ) / (sin 10° + μ cos 10°)

Sum of forces on A in the y direction:

N cos 10° − Nμ sin 10° − Fab = 0

Solve for N:

N cos 10° − Nμ sin 10° = Fab

N (cos 10° − μ sin 10°) = Fab

N = Fab / (cos 10° − μ sin 10°)

Set the expressions equal:

(P − Fab μ) / (sin 10° + μ cos 10°) = Fab / (cos 10° − μ sin 10°)

Cross multiply:

(P − Fab μ) (cos 10° − μ sin 10°) = Fab (sin 10° + μ cos 10°)

Distribute and solve for Fab:

P (cos 10° − μ sin 10°) − Fab (μ cos 10° − μ² sin 10°) = Fab (sin 10° + μ cos 10°)

P (cos 10° − μ sin 10°) = Fab (sin 10° + 2μ cos 10° − μ² sin 10°)

Fab = P (cos 10° − μ sin 10°) / (sin 10° + 2μ cos 10° − μ² sin 10°)

Sum of forces on B in the y direction:

∑F = ma

Fab − kx = 0

kx = Fab

x = Fab / k

x = P (cos 10° − μ sin 10°) / (k (sin 10° + 2μ cos 10° − μ² sin 10°))

Plug in values and solve.

x = 500 N (cos 10° − 0.4 sin 10°) / (12000 (sin 10° + 0.8 cos 10° − 0.16 sin 10°))

x = 0.0408 m

x = 4.08 cm

B) Draw free body diagrams of both blocks.

Force P is pushing block A to the right relative to the ground C, so friction force points to the left.

Block A moves right relative to block B, so friction force on A will point left.  Block B moves left relative to block A, so friction force on B will point right (opposite and equal).

Block B moves up relative to the wall D, so friction force on B will point down.

There are 5 forces acting on block A:

Applied force P pushing to the right,

Normal force Fc pushing up,

Friction force Fc μ₁ pushing left,

Reaction force Fab pushing down and left 15° from the vertical,

and friction force Fab μ₂ pushing up and left 15° from the horizontal.

There are 5 forces acting on block B:

Weight force 750 n pushing down,

Normal force Fd pushing left,

Friction force Fd μ₁ pushing down,

Reaction force Fab pushing up and right 15° from the vertical,

and friction force Fab μ₂ pushing down and right 15° from the horizontal.

Sum of forces on B in the x direction:

∑F = ma

Fab μ₂ cos 15° + Fab sin 10° − Fd = 0

Fd = Fab μ₂ cos 15° + Fab sin 15°

Sum of forces on B in the y direction:

∑F = ma

-Fab μ₂ sin 15° + Fab cos 10° − 750 − Fd μ₁ = 0

Fd μ₁ = -Fab μ₂ sin 15° + Fab cos 15° − 750

Substitute:

(Fab μ₂ cos 15° + Fab sin 15°) μ₁ = -Fab μ₂ sin 15° + Fab cos 15° − 750

Fab μ₁ μ₂ cos 15° + Fab μ₁ sin 15° = -Fab μ₂ sin 15° + Fab cos 15° − 750

Fab (μ₁ μ₂ cos 15° + μ₁ sin 15° + μ₂ sin 15° − cos 15°) = -750

Fab = -750 / (μ₁ μ₂ cos 15° + μ₁ sin 15° + μ₂ sin 15° − cos 15°)

Sum of forces on A in the y direction:

∑F = ma

Fc + Fab μ₂ sin 15° − Fab cos 15° = 0

Fc = Fab cos 15° − Fab μ₂ sin 15°

Sum of forces on A in the x direction:

∑F = ma

P − Fab sin 15° − Fab μ₂ cos 15° − Fc μ₁ = 0

P = Fab sin 15° + Fab μ₂ cos 15° + Fc μ₁

Substitute:

P = Fab sin 15° + Fab μ₂ cos 15° + (Fab cos 15° − Fab μ₂ sin 15°) μ₁

P = Fab sin 15° + Fab μ₂ cos 15° + Fab μ₁ cos 15° − Fab μ₁ μ₂ sin 15°

P = Fab (sin 15° + (μ₁ + μ₂) cos 15° − μ₁ μ₂ sin 15°)

First, find Fab using the given values.

Fab = -750 / (0.25 × 0.5 cos 15° + 0.25 sin 15° + 0.5 sin 15° − cos 15°)

Fab = 1151.9 N

Now, find P.

P = 1151.9 N (sin 15° + (0.25 + 0.5) cos 15° − 0.25 × 0.5 sin 15°)

P = 1095.4 N

Answer:

Make a hypothesis, conduct an experiment, Analyze the experimental data..

Answer:

The velocity is  [tex]v_2= 0.45 \ m/s[/tex]

Explanation:

From the question we are told that

      The initial speed of the hot water is  [tex]v_1 = 0.85 \ m/s[/tex]

     The pressure from the heater  [tex]P_1 = 450 \ KPa = 450 *10^{3} \ Pa[/tex]

      The height of the hot water before flowing is  [tex]h_1 = 0 \ m[/tex]

      The height of bathtub above the heater is [tex]h_2 = 3.70 \ m[/tex]

       The pressure in the pipe is [tex]P_2 = 414 KPa = 414 *10^{3} \ Pa[/tex]

       The density of water is [tex]\rho = 1000 \ kg/m^3[/tex]

Apply Bernoulli equation

      [tex]P_1 + \rho gh_1 +\frac{1}{2} \rho v_1^2 = \rho g h_2 + \frac{1}{2}\rho v_2 ^2[/tex]

Substituting values

     [tex](450 *10^{3}) + (1000 * 9.8 * 0) + (0.5 * 1000 * 0.85^2) = (1000 * *9.8*3.70) + (0.5*1000*v_2^2 )[/tex]

=>   [tex]v_2^2 = \frac{ (450 *10^{3}) + (1000 * 9.8 *0 ) + (0.5 * 1000 * 0.85^2) -[ (1000 * *9.8*3.70) ]}{0.5*1000}[/tex]

=>   [tex]v_2= \sqrt{ \frac{ (450 *10^{3}) + (1000 * 9.8 * 0) + (0.5 * 1000 * 0.85^2) -[ (1000 * *9.8*3.70) ]}{0.5*1000}}[/tex]

=>    [tex]v_2= 0.45 \ m/s[/tex]

Answer:

If, instead, that rocket was on a planet with the same mass as Earth but half the diameter, the escape velocity would be 15.8 km/s Any object that is smaller than its Schwarzschild radius is a black hole – in other words, anything with an escape velocity greater than the speed of light is a black hole.

Explanation:

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Answer:

F = ILB

Explanation:

To find the net force on the conducting bar you take into account the following expression:

[tex]\vec{F}=I( \vec{L}X \vec{B})[/tex]

I: current in the conducting bar

L: length of the bar

B: magnitude of the magnetic field

In this case the direction of the magnetic field and the motion of the bar are perpendicular between them. The direction of the bar is + i, and the magnetic field poits upward + k. The cross product of these vector give us the direction of the net force:

+i X +k = +j

The direction of the force is to the right and its magnitude is F = ILB

Answer: 10m

Explanation:

The magnitude of the vector would be 10

[tex]\sqrt{6^{2}+8^{2} } =10[/tex]

Answer:

Explanation:

Ex(z,t) = Eocos(kz - ω t + φ)

k = 2π/λ  , ω = 2π f

φ = +30° , E₀ = 10³ V .

z/λ = 0.25 , ft = 0.125

Ex(z,t) = Eocos(2πz/λ - 2πf t + φ)

Putting the values given above

Ex(z,t) = 10³ cos ( 2π / 4 - 2π x .125 + 30⁰ )

= 1000cos (90⁰ - 45+30)

= 1000 cos 75

=258.8  V .

Answer:

a) 463.29 K

b) 8065.65 Pa

c) 0 J

Explanation:

The parameters given are;

Volume of the tank, V = 3.72 m³

Number of moles of gas present in the tank, n = 22.1 moles

Temperature of the gas before heating, T₁ = 300 k

Heat added to the gas, ΔQ = 4.5 × 10⁴ J

Specific heat capacity at constant volume, [tex]c_v[/tex], for monatomic gas = 12.47 J/K/mole

Avogadro's number = 6.022 × 10²³ particles per mole

a) ΔQ = n × [tex]c_v[/tex] × ΔT

Where:

ΔT = T₂ - T₁

T₂ = Final temperature of the gas

Hence, by plugging in the values, we have;

4.5 × 10⁴ = 22.1 × 12.47 × (T₂ - 300)

[tex]T_{2} - 300 = \frac{4.5\times 10^{4}}{22.1\times 12.47}[/tex]

T₂ = 300 + 163.29 = 463.29 K

b) The pressure of the gas is found from the relation;

P×V = n×R×T

[tex]P = \dfrac{n \times R \times T}{V}[/tex]

Where:

P = Pressure of the gas

R = Universal gas constant = 8.3145 J/(mol·K)

T = Temperature of the gas

V = Volume of the gas = 3.72 ³ (constant)

n = Number of moles of gas present = 22.1 moles (constant)

Hence the change in pressure is given by the relation;

[tex]\Delta P = \dfrac{n \times R \times (T_2 - T_1)}{V} = \dfrac{n \times R \times \Delta T}{V}[/tex]

Plugging in the values, we have;

[tex]\Delta P = \dfrac{22.1 \times 8.3145 \times 163.29}{3.72} = 8065.65 \, Pa[/tex]

c) Work done, W, by the gas is given by the area under the pressure to volume graph which gives;

W = f(P) × ΔV

The volume given in the question is constant

∴ ΔV = 0

Hence, W =  f(P) × 0 = 0 J

No work done by the gas during the process.

Answer:

rod end A is strongly attracted towards the balls

rod end B is weakly repelled by the ball as it is at a greater distance

Explanation:

When the ball with a negative charge approaches the A end of the neutral bar, the charge of the same sign will repel and as they move they move to the left end, leaving the rod with a positive charge at the A end and a negative charge of equal value at end B.

Therefore rod end A is strongly attracted towards the balls and

rod end B is weakly repelled by the ball as it is at a greater distance

An object's resistance to any change in motion is the Inertia of the object.

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:

1)copper wire

Explanation:

it is the best electric conductor

Answer:

a) a = 3.09 m/s²

b) aₓ = 2.60 m/s²

Explanation:

a) The magnitude of her acceleration can be calculated using the following equation:

[tex] V_{f}^{2} = V_{0}^{2} + 2ad [/tex]

Where:

[tex]V_{f}[/tex]: is the final speed = 8.89 m/s

[tex]V_{0}[/tex]: is the initial speed = 0 (since she starts from rest)

a: is the acceleration

d: is the distance = 12.8 m    

[tex] a = \frac{V_{f}^{2}}{2d} = \frac{(8.89 m/s)^{2}}{2*12.8 m} = 3.09 m/s^{2} [/tex]

Therefore, the magnitude of her acceleration is 3.09 m/s².              

b) The component of her acceleration that is parallel to the ground is given by:

[tex] a_{x} = a*cos(\theta) [/tex]

Where:

θ: is the angle respect to the ground = 32.6 °

[tex] a_{x} = 3.09 m/s^{2}*cos(32.6) = 2.60 m/s^{2} [/tex]

Hence, the component of her acceleration that is parallel to the ground is 2.60 m/s².

I hope it helps you!

A skateboarder, starting from rest, rolls down a 12.8-m ramp the magnitude of the skateboarder's acceleration is approximately 3.07 [tex]m/s^2[/tex], the component of her acceleration that is parallel to the ground is approximately 1.66 [tex]m/s^2[/tex].

(a) The following kinematic equation can be used to calculate the skateboarder's acceleration:

[tex]v^2 = u^2 + 2as[/tex]

[tex](8.89)^2 = (0)^2 + 2a(12.8)[/tex]

78.72 = 25.6a

a = 78.72 / 25.6

a = 3.07 [tex]m/s^2[/tex]

(b) Trigonometry can be used to calculate the part of her acceleration that is parallel to the ground. We are aware that the ramp's angle with the ground is 32.6°.

[tex]a_{parallel }= a * sin(\theta)[/tex]

Plugging in the values:

[tex]a_{parallel[/tex] = 3.07  [tex]m/s^2[/tex]* sin(32.6°)

[tex]a_{parallel[/tex]≈ 1.66  [tex]m/s^2[/tex]

Therefore, the component of her acceleration that is parallel to the ground is approximately 1.66  [tex]m/s^2[/tex].

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Answer:

At thestratosphere: it 20- 25km

Answer:

2C

Explanation:

The equivalent capacitance of a parallel combination of capacitors is the sum of their capacitance.

So, if the capacitance of each capacitor is half the previous one, we have a geometric series with first term = C and rate = 0.5.

Using the formula for the sum of the infinite terms of a geometric series, we have:

Sum = First term / (1 - rate)

Sum = C / (1 - 0.5)

Sum = C / 0.5 = 2C

So the equivalent capacitance of this parallel connection is 2C.

Answer:  If the woodpecker drums upon a tree 5 times in one second, then the frequency is 5 Hz; each drum must endure for one-fifth a second, so the period is 0.2 s.

Answer:

The correct answer is C 1, 4, and 5 tie, then 2 and 3 tie

Explanation:

Solution

The electric field due to sheets E₁ positive =б/2E₀

E₂ is negative = б/2E₀

Now,

At the point 1, 4, 5 the electric field due to the sheets are in the opposite direction

At the point 1, the net field = -E₁ + E₂ =0

At the point A, the net field = -E₁ - E₂ = 0

Now,

At nay point inside between them, the electric field is seen to be at the same direction.

At the 2, 3 points the field is seen at the right

Thus,

E net = E₁ + E₂

= б/2E₀ + σ/2E₀

=б/E₀

Note: Kindly find an attached copy of the complete question to the solution

The correct answer is option C

The rank of the points according to the magnitude of the electric field is 1, 4, and 5 tie, then 2 and 3 tie

Let sheet 1 has positive surface charge density and sheet 2 has a negative surface charge density

The electric field (without direction) due to sheets will be

E₁ =σ/2E₀

E₂= σ/2E₀

Now,

At the point 1, 4, 5 the electric field due to the sheets is given by:

E = E₁ - E₂

E = σ/2E₀ - σ/2E₀

since the positive charge plate will have electric field lines away from the sheet and the negative charge plate will have electric field lines towards the sheet

E = 0

Now,

At points 2, 3 which are between the plates,

The net electric field is:

E = E₁ + E₂

since the electric field due to both the plates will be from positive to negative ( towards the negatively charged plate)

E = σ/2E₀ + σ/2E₀

E = σ/E₀

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Answer: A chemical change results from a chemical reaction, while a physical change is when matter changes forms but not chemical identity. Examples of chemical changes are burning, cooking, rusting, and rotting. Examples of physical changes are boiling, melting, freezing, and shredding. Often, physical changes can be undone, if energy is input.

Explanation: hope this helps have a good day

Answer:

If the reaction is a chemical change, new substances with different properties and identities are formed. This may be indicated by the production of an odor, a change in color or energy, or the formation of a solid.

Answer:

Since you would have to do work on the charge to bring it back to its original position, the charge moves to a position of lower potential and lower potential energy.

The positive charge is released from a point such that it will move along an uniform electric field to the position of lower potential and lower potential energy. Therefore, option (A) is correct,

When a positive charge (say +Q) is released from a point (say A) and moves in an uniform electric field to reach the point (say B), then some work is done on the charge. This work done is given as,

[tex]W=+Q(V_{A}-V_{B})[/tex]

Here, [tex]V_{A}[/tex] and [tex]V_{B}[/tex] are the potential differences between the points A and B respectively..

This means the charge is moving from higher potential to lower potential. And since it is moving along the uniform electric field, therefore the electric potential energy of charged system is decreased.

Thus, we conclude that on releasing the positive charge from a point, it starts moving along the electric field towards the direction of lower electric potential and lower electric potential energy. Hence, option (A) is correct.

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Complete Question

The complete question is shown on the first uploaded image

Answer:

The distance is [tex]r_2 = 4 \ AU[/tex]

Explanation:

From the question we are told that

    The size of Jupiter is  [tex]s_2 = 143,000 \ km[/tex]

    The  length of the International Space Station is [tex]r_1 = 380\ km[/tex]

    The  size of the International Space Station is  [tex]s_1 = 90 \ m =0.09 \ km[/tex]

The angular size where the same one night and this angular size is mathematically represented as

      [tex]\theta = \frac{s}{r}[/tex]

Since  [tex]\theta[/tex] is constant

        [tex]\frac{s_1}{r_1} = \frac{s_2}{r_2}[/tex]

substituting values

      [tex]\frac{0.09}{380} = \frac{143000}{r_2}[/tex]

=>   [tex]r_2 = 6.04 * 10^{9} \ km[/tex]

Now we are told to convert to AU and  1 AU  [tex]= 1.5 * 10^8 \ km[/tex]

  So

      [tex]r_2 = \frac{6.04 * 10^8}{1.5*10^{8}}[/tex]

      [tex]r_2 = 4 \ AU[/tex]

1)  100 ° C

2) 323 K

hope it helps youuuuuu

Answer:

f = 1.96 revolutions per minute

Explanation:

The formula for the the frequency of revolution of a satellite, to develop an artificial gravity, with the help of centripetal acceleration is given as follows:

f = (1/2π)√(ac/r)

where,

f = frequency of rotation = ?

ac = centripetal acceleration= apparent gravity or artificial gravity = 2.2 m/s²

r = radius of station or satellite = diameter/2 = 104 m/2 = 52 m

Therefore,

f = (1/2π)√[(2.2 m/s²)/(52 m)]

f = (0.032 rev/s)(60 s/min)

f = 1.96 revolutions per minute

Answer:

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

Explanation:

Answer:

[tex]\tau = 1\ ms[/tex]

Explanation:

First we need to find the capacitance of the capacitor.

The capacitance is given by:

[tex]C = \epsilon_0 * area / distance[/tex]

Where [tex]\epsilon_0[/tex] is the air permittivity, which is approximately 8.85 * 10^(-12)

The radius is 12/2 = 6 mm = 0.006 m, so the area of the capacitor is:

[tex]Area = \pi * radius^{2}\\Area = \pi * 0.006^2\\Area = 113.1 * 10^{-6}\ m^2[/tex]

So the capacitance is:

[tex]C = \frac{8.85 * 10^{-12} * 113.1 * 10^{-6}}{0.001}[/tex]

[tex]C = 10^{-12}\ F = 1\ pF[/tex]

The time constant of a rc-circuit is given by:

[tex]\tau = RC[/tex]

So we have that:

[tex]\tau = 10^{9} * 10^{-12} = 10^{-3}\ s = 1\ ms[/tex]

Answer:

it takes the shape and size of the container that it is in

Explanation:

Answer:

it takes the shape and size of a container

Answer:

20 Yards

Explanation:

|---20----|

|            |

| 50       |50

|---D--->|

Start      End

Total displacement(D)  20 yards (East).