Water enter the horizontal, circular cross-sectional, sudden-contraction nozzle sketched below at section (1) with a uniformly distributed velocity of 30 ft/s and a pressure of 80 psi. The water exits from the nozzle into the atmosphere at section (2) where the uniformly distributed velocity is 100 ft/s. Determinethe axial component of the anchoring force required to hold the contraction in place.

Answer:

It's plastic.

trust me it's plastic, i've rad it somewhere.

All of them have something that's not like the others.

-- Rubber is the only one on the list that has two repeated letters.

-- Plastic is the only one on the list thagt has no repeated letters.

-- Plastic is the only one on the list that has no 'r' in its name.

-- Copper is the only one on the list that is an element, not a compound.

-- Copper is the only good electrical conductor on the list.

-- Plastic is the only one on the list with more than six letters in its name.

-- Rubber is the only one on the list with no 'p' in its name.

-- Plastic is the only one on the list that doesn't end in "-er".

1. false

2. true

I hope this helps ^-^

Answer:

Centripetal Acceleration = 88826.44 m/s²

RCF = 9054.7

Explanation:

First, we will find the value of the centripetal acceleration by using the following formula:

[tex]Centripetal\ Acceleration = \frac{v^2}{r}\\[/tex]

where,

v = linear speed of liquid or separator = rω

ω = angular speed of liquid or separator = (9000 rpm)(2π rad/rev)(1 min/60 s) = 942.48 rad/s

r = radius of seperator = diameter/2 = 20 cm/2 = 10 cm = 0.1 m

Therefore,

[tex]Centripetal\ Acceleration = \frac{(r\omega)^2}{r}\\Centripetal\ Acceleration = r\omega^2\\Centripetal\ Acceleration = (0.1\ m)(942.48\ rad/s)^2\\[/tex]

Centripetal Acceleration = 88826.44 m/s²

Now, for the RCF:

[tex]RCF = \frac{Centripetal\ Acceleration}{g}\\RCF = \frac{88826.44\ m/s^2}{9.81\ m/s^2}\\[/tex]

RCF = 9054.7

Answer:

48kg

Explanation:

Answer:  I'm not sure, but I think it would be a total lunar eclipse

When the moon passes through the penumbra of Earth’s shadow it is referred to as partial lunar eclipse. The correct option is C.

A partial lunar eclipse occurs when the moon is not completely immersed in the umbra of the earth's shadow.

During a partial solar eclipse, the Moon, Sun, and Earth do not align perfectly straight, and the Moon casts only the penumbra of its shadow on Earth. From our vantage point, it appears that the Moon has eaten the Sun.

A shadow's penumbra is the lighter outer edge. Partial solar eclipses are caused by the Moon's penumbra, while penumbral lunar eclipses are caused by the Earth's penumbra. The penumbra is a type of lighter shadow.

The Penumbra is a half-shadow region that occurs when an object only partially covers a light source.

Thus, the correct option is C.

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Answer:

35 N

Explanation:

F = ma

centripetal force = 10(3.5) = 35 N

Answer:

White hole is an impossible object in universe. ... This means that in a hypothetical universe where there is a black and a white hole, in a short time after their first interaction the white hole will become another black hole so that the system will end up with two black holes.

Answer: if weight affects how fast they go?

Explanation:

Answer:

How can we change the speed of a toy car on a racetrack to describe the car’s motion?

Explanation:

thats the sample respond

Answer:

In metals there are free electrons at normal temperature so when we increase temperature it resistivity gets increases,so conductivity decreases,while in semiconductor the electrons are not free so when we increase the temperature the covalent bonds begin to break and the electron becomes free so conductivity get.

Explanation:

Answer:

An Oven

Explanation:

The heat is higher, so it moves faster. Shile in a freezer the particles are extremely slow!

Answer:

0.002 m/s

Explanation:

27.13(60) = 1,627.8 seconds

3.12/1,627.8 = 0.00191 ≈ 0.002 = s

Are you sure you're not looking for cm/s?

Explanation:

15m/s

acceleration= (+)

so, 15m/s +32m/s=47m/s

42m/s. X 8 = 336

The framers wanted senators to be more mature and more experienced than representatives. (C)

Answer:

C -> The framers wanted senators to be more mature and more experienced than representatives.

Answer:

[tex]3.52\times 10^{25}\ \text{J}[/tex]

Explanation:

G = Gravitational constant = [tex]6.674\times 10^{-11}\ \text{Nm}^2/\text{kg}^2[/tex]

M = Mass of Sun = [tex]1.989\times 10^{30}\ \text{kg}[/tex]

m = Mass of Earth = [tex]5.972\times 10^{24}\ \text{kg}[/tex]

[tex]r_i[/tex] = Initial radius of orbit = [tex]1.5\times 10^{11}\ \text{m}[/tex]

[tex]r_f[/tex] = Final radius of orbit = [tex]((1.5\times 10^{11})+2\times 10^3)\ \text{m}[/tex]

Energy required is given by

[tex]E=\dfrac{1}{2}\Delta U\\\Rightarrow E=\dfrac{GMm}{2}(\dfrac{1}{r_i}-\dfrac{1}{r_f})\\\Rightarrow E=\dfrac{6.674\times 10^{-11}\times 1.989\times 10^{30}\times 5.972\times 10^{24}}{2}(\dfrac{1}{1.5\times 10^{11}}-\dfrac{1}{(1.5\times 10^{11})+2\times 10^3})\\\Rightarrow E=3.52\times 10^{25}\ \text{J}[/tex]

The energy required would be [tex]3.52\times 10^{25}\ \text{J}[/tex].

Answer:

1.61 N/kg

Explanation:

Take the universal gravitational constant G as 6.67 × 10^(-11) Nm²/kg²

The required gravitational field strength

= 6.67 × 10^(-11) × 7.3 × 10^(22) / (1738000)²

= 1.61194103 N/kg

= 1.61 N/kg (corr. to 3 sig. fig.)

Answer:

Regular reflection

Explanation:

Regular reflection occurs when light reflects off a very smooth surface and forms a clear image.

i hope this helps a bit.

According to the context, a sharp image is formed when light reflects from a regular reflection.

It is reflection without diffusion that obeys the laws of geometrical optics, as in mirrors.

This reflection of light happens when the angles that the two rays determine with the surface are equal.

Therefore, we can conclude that according to the context, a sharp image is formed when light reflects from a regular reflection.

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Answer:

The right solution is "126 Psi".

Explanation:

The given values are:

P₁ = 130 psig

i.e.,

   = [tex]130\times 6.894[/tex]

   = [tex]896.22 \ Kpa[/tex]

or,

   = [tex]896.22\times 10^3 \ Pa[/tex]

Z₂ = 10ft

    = 3.05 m

[tex]\delta[/tex] = 1000 kg/m³

According to the question,

Z₁ = 0

V₁ = V₂

As we know,

⇒  [tex]\frac{P_1}{\delta_g} +\frac{V_1^2}{2g} +Z_1=\frac{P_2}{\delta_g} +\frac{V_2^2}{2g} +Z_2[/tex]

On substituting the values, we get

⇒  [tex]\frac{P_1}{\delta_g} +0+0=\frac{P_2}{\delta_g} +0+Z_2[/tex]

⇒  [tex]\frac{896.22\times 10^3}{1000\times 9.8} =\frac{P_2}{1000\times 9.8} +3.05[/tex]

⇒  [tex]P_2=866330 \ P_a[/tex]

i.e.,

⇒       [tex]=866330\times 0.000145[/tex]

⇒       [tex]=126 \ Psi[/tex]

Answer:

Q = 636.48 J

Explanation:

Given that,

The mass of gold, m = 0.052 kg

The temperature increase from 30°C to 120°C.

The specific heat of gold is 136  J/kg °C.

We need to find the heat needed to warm the gold. The formula for heat needed is given by :

[tex]Q=mc\Delta T\\\\Q=0.052\times 136\times (120-30)\\\\Q=636.48\ J[/tex]

So, 636.48 J of heat is needed to warm gold.

using hooke's relation

F = K . d

50N = k . 0.09m

k = 50N / 0.09m

k = 5555.56 N/m

Calculating the potential energy of the spring

Ep = 1/2 k . x²

Ep = 1/2 (5555.56 N/m) (0.09m)²

Ep = 22.5 Joules

the spring constant? =

k =  5555.56 N/m

potential energy of the spring?​

Ep = 22.5 Joules

The Potential energy of the spring is 2.25 J

This is the energy stored in spring due to its elastic properties.

To calculate the potential energy of the spring, we use the formula below.

Formula:

Where:

From the question,

Given:

Substitute these values into equation 1

Hence, The Potential energy of the spring is 2.25 J

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Answer:

B. in the direction of the force

Explanation:

Sana nakatulong

Answer:

athletic

Explanation:

because internet system has been down since we were in few days

An inventor makes a clock using a brass rod and a heavy mass as a pendulum. when the clock gets colder then the time clock would gain time

The expansion of any material due to the variation of the temperature is known as thermal expansion. It varied differently for different materials according to their corresponding values of the coefficient of the thermal expansion.

As given in the problem statement that an inventor makes a clock using a brass rod and a heavy mass as a pendulum, when the clock gets colder then the length of the brass decreases due to thermal expansion.

The length of the pendulum gets reduced which further results in the reduction in the time period, as per the formula of the time period for the pendulum

T = 2π√(L/g)

As the length of the brass gets reduced. This means the pendulum of the clock moves faster and the clock would gain time

Thus, if a  pendulum made of a heavy mass and a brass rod is used to create a clock by an inventor. The time clock would advance in time as the clock get colder

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Answer:

a) y = 2.21 10⁻³ m, b)  y = 5.528 10⁻³ m

Explanation:

In the double-slit experiment the interferences occur at the positions

         d sin θ = m λ                    constructive interference

         d sin θ = (m + ½) λ           destructive interference

let's use trigonometry for the angle

          tan θ = y / L

as in this experiment the angles are very small

          tan θ = sin θ/cos θ = sin θ

          sin θ = y / L

we substitute

         d y / L = m λ                   constructive interference

         d y / L = (m + ½) λ          destructive interference

Let's answer the questions

a) first line of constructive interference me = 1

          y = m  λ L / d

          y = 1 647 10⁻⁹ 3.93 /1.15 10⁻³

          y = 2.21 10⁻³ m

b) second dark band m = 2

            y = (m + ½) λ L / d

            y = (2 + ½)  647 10⁻⁹ 3.93 /1.15 10⁻³

            y = 5.528 10⁻³ m

Answer:

e. the number of active motor units increases.

Explanation:

There is a direct relationship between the number of active motor units and the grip strength in a given scenario. For example, increase in the grip strength leads to increase in the number of active motor units. In the other-hand, the decrease in grip strength leads to the decrease in the number of active motor units.

Answer:

true

Explanation:

force or powerbecause he pushes a disk

Answer:

The magnification of an image is equal to the ratio of the image height to the object height.

Answer:

Resistance = 1.5 ohms

Explanation:

Given:

Potential difference = 3v

Current flow = 2 A

Find:

Resistance

Computation:

Resistance = Potential difference / Current flow

Resistance = 3 / 2

Resistance = 1.5 ohms

The image of this hollow sphere and uniform rod is missing, so i have attached it.

Answer:

A) J = 0.7443 kg•m²

B) T = 1.9169 N•m CCW

C) α = 2.5754 rad/s²

D) a = 3.966 m/s²

Explanation:

A) The moment of inertia J of the contraption around the fulcrum is given by the formula;

J = Jℓ + Jr

Let's calculate Jℓ

Jℓ = [((0.34²/3) × 0.50 × 0.34)/1.78] + (0.48 × (0.34 + 0.64)²)

Jℓ = 0.4647 kg•m²

Now, let's Calculate Jr

Jr = ((1.78 - 0.34)²/3) × ((1.78 - 0.34)/1.78) × 0.50

Jr = 0.2796 kg•m²

Thus;

J = 0.4647 + 0.2796

J = 0.7443 kg•m²

(b) Using CCW as positive, Torque in Nm is calculated as;

T = Tℓ - Tr

Let's calculate Tℓ

Tℓ = [(0.48 × (0.64 + 0.34)) + (0.50 × 0.34/1.78) × 0.34/2)] × 9.81

Tℓ = 4.7739 N•m CCW

Now, let's Calculate Tr;

Tr = [(0.50 × (1.78 - 0.34)/1.78) × (1.78 - 0.34)/2)] × 9.81

Tr = 2.857 N•m CW

Thus;

T = 4.7739 - 2.857

T = 1.9169 N•m CCW

(c) The angular acceleration α of the contraption, using CCW is gotten from the formula;

α = T/J

α = 1.9169/0.7443

α = 2.5754 rad/s²

(d) The linear acceleration a of the right end of the rod, using up as positive is given by;

a = α*(1.78 - 0.34)

a = 2.5754 × 1.54

a = 3.966 m/s²

A) the moment of inertia of the contraption is 0.7443 kgm²

B) The torque about the fulcrum is 1.9169 Nm

C) Angular acceleration of the contraption is 2.5754 rad/s²

D) The linear acceleration of the contraption is 3.966 m/s²

(A) The moment of inertia I of the contraption around the fulcrum is given by :

[tex]I = [(0.34^2/3) \times 0.50 \times 0.34)/1.78 + (0.48 \times (0.34 + 0.64)^2)] + [(1.78 - 0.34)^2/3) \times (1.78 - 0.34)/1.78) \times 0.50][/tex]

I = 0.4647 + 0.2796

I = 0.7443 kgm²

(B) Using CCW as positive, Torque in Nm is given by;

T = [(0.48 × (0.64 + 0.34)) + (0.50 × 0.34/1.78) × 0.34/2)] × 9.81 - [(0.50 × (1.78 - 0.34)/1.78) × (1.78 - 0.34)/2)] × 9.81

T = 4.7739 - 2.857

T = 1.9169 Nm

(C) The angular acceleration (α) of the contraption is given by:

α = T/I

since, torque is defined as T = Iα

α = 1.9169/0.7443

α = 2.5754 rad/s²

(D) The linear acceleration (a) of the right end of the rod

a = αr

where r is the distance from the pivot

a = α × (1.78 - 0.34)

a = 2.5754 × 1.54

a = 3.966 m/s²

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Answer:

work done= force × displacement

=800×5

=4000J

Explanation:

The amount of work done is the result of the magnitude of force applied and the displacement of the body due to the force applied. Therefore, work done is defined as the product of the applied force and the displacement of the body.