A fluid flows through a pipe consisting of two segments. If the cross sectional area increases, which will decrease?
a. velocity
b. mass
c. flow rate d. density

In order to convert the maximum speed to miles per hour, we must follow a series of steps, which will lead us to the conclusion that 360 km/h is equal to 223.7 miles per hour.

For this, we only really have to make one conversion. Since both metrics will be using hours as the denominator, that can remain unchanged. Therefore, we must only convert the kilometers into miles. To do this, we use the conversion metric which is that one mile is equal to 1.6 kilometers. So we can take 360 and divide that by 1.6 to reach our answer of 223.7 miles.

Therefore, through using the metrics for converting kilometers into miles, we can confirm that 360 km/h is equal to 223.7 miles per hour.

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Answer: C

Explanation: An object that has less mass will help it move more quickly and the more intertia it has the faster it accelerates.

Hope this helps.

Answer:

10kg

Explanation:

it is 10kg bc  the energy is false kenetic. The greater mass a moving object has, the more kinetic energy it has it has has kinetic energy because of its position Friction causes some mechanical energy to change to thermal energy. Doubling an object's velocity will double its kinetic energy

We have that for the Question "(a) what multiple of the maximum charge is on the capacitor and (b) what multiple of the maximum current is in the inductor"  

Answer:

From the question we are told

In an oscillating LC circuit, when 86.6% of the total energy is stored in the inductor's magnetic field

A) When 86.6\% energy is stored in inductor

\(\%\)of energy stored in electric field = \(1 - 0.866 = 13.4\%\)

\(\frac{V_E}{V} = \frac{\frac{q^2}{2c}}{\frac{Q^2}{2c}} = 0.134\\\\\frac{q}{Q} = \sqrt0.134\\\\\frac{q}{Q} = 0.366\\\\q = 0.366Q_{max\)

B)

\(\frac{V_B}{V} = \frac{\frac{Li^2}{2}}{\frac{LI^2}{2}} = 0.866\\\\\frac{i}{I} = \sqrt0.866\\\\\frac{i}{I} = 0.931\\\\i = 0.931I_{max\)

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The mass of a material with density is required to make a hollow spherical shell having an inner radius of r1 and an outer radius of r2 is given as \(\frac{4\pi.p(r^3_2-r^3_1)}{3}\)

The amount of anything in relation to a unit of volume, area, or length: as. : the substance's mass in relation to its volume. grams per cubic centimeter is a unit of density.

The volume of a spherical shell can be calculated from:

V= \(V_o - V_1 = \frac{4}{3}\pi(r^3_2 - r^3_1)\)

From the definition of density, ρ = m/v

So,

m = pv = p\((\frac{4}{3} \pi)\)\((r^3_2 - r^3_1)\) = \(\frac{4\pi.p(r^3_2-r^3_1)}{3}\)

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The average velocity of the particle from rest to 9 seconds would be 2 meters/sec.

This average velocity can be calculated with the following equation:

v = dx / dt

where:

dx = displacement = 18 meters

dt = interval of time = 9 seconds

Substitute;

v = dx / dt

v = 18 / 9

v = 2 meters/sec

Hence, the average velocity of the particle from rest to 9 seconds would be 2 meters/sec.

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Answer:

Explanation:

The speed of the second car after the collision can be found using the law of conservation of momentum. The law states that the total momentum of a system remains constant if no external forces act on it.The initial momentum of the first car is (15g)(22 cm/s) = 330 g cm/s to the right.

The initial momentum of the second car is (22g)(-31 cm/s) = -682 g cm/s to the left.

The total initial momentum of the system is 330 g cm/s - 682 g cm/s = -352 g cm/sAfter the collision, the final momentum of the first car is (15g)(-42 cm/s) = -630 g cm/s to the left.

The final momentum of the second car is (m)(v) where m is the mass of the second car and v is the speed after the collision.

The total final momentum of the system is -630 g cm/s + (m)(v) = -352 g cm/s (since it remains constant)Therefore, m*v = -630 g cm/s + 352 g cm/s = -278 g cm/sTo find v, we need to divide the momentum by the mass

v = -278 g cm/s / 22 g = -12.6 cm/sSo the speed of the second car after the collision is -12.6 cm/s to the left.

Answer:

2025m

Explanation:

Since all materials of the sphere is made to a cylindrical wire, it implies the volume of the sphere material is same as that of the cylinder. This is expressed mathematically thus.

Volume of Sphere= volume of cylinder

4/3 ×π×R^3= π× r2× L

4/3 ×R^3= r^2×L

Hence

L = 3/4 × R^3/ r^2

But R = 6.0/2 = 3.0cm{ Diameter is twice raduis}

r= 0.2/2 = 0.1mm=>0.01cm{ Diameter is twice raduis and unit converted by dividing by 10 since 10mm = 1cm}

Substituting R and r into the expression for L, we have :

L = 3/4 × 3^3/ 0.01^2= 0.75 ×27/0.0001 = 202500cm

202500/100= 2025m{ we divide by 100 because 100cm=1m}

The total mechanical energy at point A is 617,400 J, and the velocity of the coaster at point A is 25 m/s, the velocity of the coaster at point B is 34.4 m/s. and the highest hill the coaster could have gotten over before point A with no additional mechanical energy is 63 m.

To solve this problem, we can use the conservation of mechanical energy, which states that the sum of kinetic energy and potential energy is constant in a closed system where there is no work done by non-conservative forces like friction. We can use this principle to find the answers to the questions.

a) The Total Mechanical Energy of the rollercoaster at Point A

The total mechanical energy of the rollercoaster at point A is the sum of its potential energy and kinetic energy. At point A, the rollercoaster is at its highest point, so its kinetic energy is zero. Therefore, the total mechanical energy at point A is equal to the potential energy, which is given by:

mgh = 1000 kg × 9.8 m/s^2 × 63 m = 617,400 J

b) The velocity of the coaster at point A

To find the velocity of the coaster at point A, we can use the conservation of mechanical energy principle again. At point A, all of the potential energy is converted into kinetic energy. Therefore, we can write:

1/2 mv^2 = mgh

where m is the mass of the rollercoaster, v is the velocity at point A, g is the acceleration due to gravity, and h is the height of point A relative to point E. Solving for v, we get:

v = sqrt(2gh) = sqrt(2 × 9.8 m/s^2 × 63 m) = 25 m/s

Therefore, the velocity of the coaster at point A is 25 m/s.

c) The velocity of the coaster at point B

To find the velocity of the coaster at point B, we can use the conservation of mechanical energy principle again. At point B, the rollercoaster is at a height of 42 m above the ground. Therefore, the potential energy at point B is:

mgh = 1000 kg × 9.8 m/s^2 × 42 m = 411,600 J

At point B, some of the potential energy is converted into kinetic energy, so we can write:

1/2 mv^2 + mgh = mgh_D

where v is the velocity at point B, h is the height of point B relative to point E, and h_D is the height of point D relative to point E. Solving for v, we get:

v = sqrt(2(mgh_D - mgh)/m) = sqrt(2(8780 J)/1000 kg + 2gh) = 34.4 m/s

Therefore, the velocity of the coaster at point B is 34.4 m/s.

d) The highest hill the coaster could have gotten over before point A with no additional mechanical energy

If the rollercoaster had no additional mechanical energy, its total mechanical energy at point A would be equal to its potential energy, which we calculated in part (a) to be 617,400 J. Therefore, the maximum height that the rollercoaster could reach without any additional mechanical energy is given by:

mgh_max = 617,400 J

Solving for h_max, we get:

h_max = 617,400 J / (1000 kg × 9.8 m/s^2) = 63 m

So, the highest hill the coaster could have gotten over before point A with no additional mechanical energy is 63 m.

Therefore, The coaster has a total mechanical energy of 617,400 J at point A, a velocity of 25 m/s, a velocity of 34.4 m/s, and a maximum hill height of 63 m that it could have traversed without using any further mechanical energy.

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While we run on a circular path and returns to the starting point, the distance will be obviously greater than 0, but as the initial and final point are same, the value of displacement is equal to 0.

So, correct choice is :

The distance travelled would be greater than 0, and displacement would be 0.

Athletes training for the 100-meter dash will mostly rely on phosphocreatine for this event's rapid energy.

Numerous creatine kinases catalyze the reversible phosphorylation of creatine, which includes both the forward and backward reactions. It is used to detect tissue injury when creatine kinase (CK-MB, creatine kinase myocardial band) is present in blood plasma.

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The initial speed of the car was approximately 24.9 m/s.

We can use the equation for the work done by kinetic friction to find the initial speed of the car. The work done by kinetic friction is given by:

W_friction = F_friction * d

where;

The force of kinetic friction is given by:

F_friction = m * g * μ

where;

The work done by kinetic friction is equal to the change in kinetic energy of the car:

W_friction = (1/2) * m * v^2

where;

Setting these two equations equal to each other and solving for v, we get:

(1/2) * m * v^2 = F_friction * d

(1/2) * m * v^2 = m * g * μ * d

v^2 = 2 * g * μ * d

v = √(2 * g * μ * d)

Substituting the given values, we get:

v = √(2 * 9.81 m/s^2 * 0.85 * 45 m)

v = 24.9 m/s

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Answer:

The below diagram shows pure spectrum with white light

Explanation:

Answer:

Potential energy

Explanation:

Potential energy is stored energy

Answer: deceleration is considered to describe a decrease or negative change of speed or velocity. ... Three of these namely time, distance and speed are scalar quantities, whereas the remaining three attributes namely displacement, velocity and acceleration are vectors.

Explanation:

Read it online

Answer:very hard

Explanation:

a) The speed of the 6.00 kg block after descending 0.800 m is 2.07 m/s.

b) We cannot calculate the work done by the friction force.

c) The net work done on the system of two blocks during the 0.800 m downward displacement of the 6.00 kg block is 29.13 J. The work done by gravity is 47.04 J, the work done by friction is unknown, and the work done by the tension in the rope is zero.

(a) The work done on the 6.00 kg block by gravity can be calculated using the formula:

Work_gravity = force_gravity * displacement * cos(theta),

where force_gravity is the weight of the block, displacement is the downward displacement of the block, and theta is the angle between the force and displacement vectors (which is 0 degrees in this case).

The weight of the block is given by:

force_gravity = mass * acceleration_due_to_gravity = 6.00 kg * 9.8 m/s^2 = 58.8 N.

Plugging in the values, we get:

Work_gravity = 58.8 N * 0.800 m * cos(0) = 47.04 J.

The work done on the 6.00 kg block by the tension in the rope is given by:

Work_tension = tension * displacement * cos(theta).

Plugging in the values, we get:

Work_tension = 37.0 N * 0.800 m * cos(180) = -29.6 J.

The negative sign indicates that the tension is in the opposite direction of the displacement.

Using the work-energy theorem, we can find the speed of the 6.00 kg block after descending 0.800 m:

Work_net = change_in_kinetic_energy.

Since the block starts from rest, its initial kinetic energy is zero. Therefore:

Work_net = Final_kinetic_energy - Initial_kinetic_energy = 1/2 * mass * velocity^2.

Solving for velocity, we get:

velocity = sqrt(2 * Work_net / mass).

The net work done on the block is the sum of the work done by gravity and the tension:

Work_net = Work_gravity + Work_tension = 47.04 J - 29.6 J = 17.44 J.

Plugging in the values, we get:

velocity = sqrt(2 * 17.44 J / 6.00 kg) = 2.07 m/s.

Therefore, the speed of the 6.00 kg block after descending 0.800 m is 2.07 m/s.

(b) The total work done on the 8.00 kg block during the 0.800 m displacement can be calculated using the work-energy theorem:

Work_net = change_in_kinetic_energy.

Since the 8.00 kg block is not moving vertically, its initial and final kinetic energies are zero. Therefore:

Work_net = Final_kinetic_energy - Initial_kinetic_energy = 0.

The work done on the 8.00 kg block by the tension in the rope is given by:

Work_tension = tension * displacement * cos(theta).

Plugging in the values, we get:

Work_tension = 37.0 N * 0.800 m * cos(0) = 29.6 J.

The work done on the 8.00 kg block by the friction force can be calculated using the formula:

Work_friction = force_friction * displacement * cos(theta),

where force_friction is the frictional force on the block. However, the problem statement does not provide the value of the friction force. Therefore, we cannot calculate the work done by the friction force.

(c) The net work done on the system of two blocks during the 0.800 m displacement of the 6.00 kg block can be found using the work-energy theorem:

Work_net = change_in_kinetic_energy.

Since the system starts from rest, the initial kinetic energy of the system is zero. Therefore:

Work_net = Final_kinetic_energy - Initial_kinetic_energy = 1/2 * (6.00 kg + 8.00 kg) * velocity^2.

Simplifying, we get:

Work_net = 1/2 * 14.00 kg * velocity^2.

Using the value of velocity calculated in part (a), we get:

Work_net = 1/2 * 14.00 kg * (2.07 m/s)^2 = 29.13 J.

The work done on the system of two blocks by gravity is the sum of the work done on the individual blocks by gravity:

Work_gravity_system = Work_gravity_6kg + Work_gravity_8kg = 47.04 J + 0 J = 47.04 J.

The work done on the system of two blocks by the tension in the rope is the sum of the work done on the individual blocks by the tension:

Work_tension_system = Work_tension_6kg + Work_tension_8kg = -29.6 J + 29.6 J = 0 J.

Therefore, the net work done on the system of two blocks during the 0.800 m downward displacement of the 6.00 kg block is 29.13 J. The work done by gravity is 47.04 J, the work done by friction is unknown, and the work done by the tension in the rope is zero.

Note: The calculations for part (b) and (c) were based on the given information, but the value of the friction force was not provided in the problem statement.

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Answer:

D. Because they are using space technology on a shirt so people can wear it on earth as well

Answer:

The area under the speed - time graph denotes the distance travelled by the object

In the given graph, we just have to think about the first 4 seconds, we also notice that the velocity at 4 seconds is 20 m/s

The distance travelled by the object in 4 seconds is the area of the triangle in the graph with a base of 4 units and height of 20 units (image included)

Distance Travelled = Area of triangle =  1/2 * base * height

Distance Travelled = 1/2 * 4 * 20

Distance travelled = 4 * 10

Distance travelled = 40 m

Answer:

Multiply the friction factor by pipe length and divide by pipe diameter. Multiply this product by the square of velocity. Divide the answer by 2.

Explanation:

Multiply the friction factor by pipe length and divide by pipe diameter. Multiply this product with the square of velocity. Divide the answer by 2.

Hope this answer helps you

Answer:

Explanation:

the answer is b

Answer:

Explanation:

We can use the equation for the wavelength of light that passes through a double-slit interference pattern

λ = (d * sinθ) / m

where λ is the wavelength of light

d is the distance between the two slits

θ is the angle between the m=0 and m=1 interference lines

m is the order of the interference line.

Given that

d = 6 X 10^-5 meters

θ = 10 degrees

Interference between the m=0 and m=1 interference lines, so m = 1

Converting θ to radians:

θ = 10 degrees = 0.174533 radians

Put the values in given equation

λ = (6 X 10^-5 meters * sin(0.174533 radians)) / 1

λ ≈ 6.89 X 10^-7 meters

Therefore, the wavelength of light that shines on two slits that are 6 X 10^-5 meters apart and that angle between m=0 and m=1 interference lines is 10 degrees is approximately 6.89 X 10^-7 meters.

Answer:

s = 2250 m = 2.25 km

Explanation:

First we need to find the time traveled by the friend. So, we use:

s₁ = v₁ t₁

where,

s₁ = distance traveled by friend = 630 m

v₁ =  speed of friend = 2.1 m/s

t₁ = time taken by friend = ?

Therefore,

630 m = (2.1 m/s)(t₁)

t₁ = 300 s

Since, you traveled 10 mintes = 600 s more than the friend. Hence,

s₂ = v₂ t₂

where,

s₂ = distance traveled by friend = ?

v₂ =  speed of you = 1.8 m/s

t₂ = time taken by you = 300 s + 600 s = 900 s

Therefore,

s₂ = (1.8 m/s)(900 s)

s₂ = 1620 m

Now, the total distance between house will be:

s = s₁ + s₂ = 630 m + 1620 m

s = 2250 m = 2.25 km

The average acceleration of the motorcycle over the given distance is approximately 9.39 m/s².

To calculate the average acceleration of the motorcycle, we can use the formula:

Average acceleration = (final velocity - initial velocity) / time

First, let's convert the final velocity from km/h to m/s since the distance is given in meters. We know that 1 km/h is equal to 0.2778 m/s.

Converting the final velocity:

Final velocity = 78 km/h * 0.2778 m/s = 21.67 m/s

Since the motorcycle starts from rest (initial velocity is zero), the formula becomes:

Average acceleration = (21.67 m/s - 0 m/s) / time

To find the time taken to reach this velocity, we need to use the formula for average speed:

Average speed = total distance/time

Rearranging the formula:

time = total distance / average speed

Plugging in the values:

time = 50 m / 21.67 m/s ≈ 2.31 seconds

Now we can calculate the average acceleration:

Average acceleration = (21.67 m/s - 0 m/s) / 2.31 s ≈ 9.39 m/s²

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Answer:

Chemical weathering demands chemical reactions with minerals inside the rock and causes changes in rock composition. Sometimes this process will produce a different kind of product due to the reaction. Mechanical weathering only involves the physical breakage of rocks to smaller pieces of fragments.

Explanation:

The potential energy U of an object with mass m located a distance h above the ground is given by the expression:

Where g=9.81m/s^2 is the gravitational acceleration on the surface of Earth.

Since the 2.85kg object is located 12.7 meters below the ground, we can consider h to be negative. Replace m=2.85kg, g=9.81m/s^2 and h=-12.7m to find the potential energy of the object measured from the ground level:

Therefore, the potential energy of the object measured from the ground level is approximately 355J.

Answer:

d = 1.51 x 10⁻⁵ m = 15.1 μm

Explanation:

We will use young's Double Slit formula here:

\(Y = \frac{\lambda L}{d}\\\\d = \frac{\lambda L}{Y}\)

where,

d = width of strand = ?

λ = wavelength = 631.8 nm = 6.318 x 10⁻⁷ m

L = Screen to hair distance = 1.2 m

Y = fringe spacing = 5.02 cm = 0.0502 m

Therefore,

\(d = \frac{(6.318\ x\ 10^{-7}\ m)(1.2\ m)}{0.0502\ m}\)

d = 1.51 x 10⁻⁵ m = 15.1 μm

In "The Argonauts," Robinson effectively utilizes language techniques such as metaphor, repetition, and sentence structure to develop his argument to his younger self and engage young readers in the present. Through these techniques, Robinson creates a powerful and relatable narrative that resonates with his audience.

Robinson employs metaphors to convey complex ideas in a compelling and accessible manner. For instance, he compares his struggle with identity and gender to the mythical journey of the Argonauts, making it relatable and captivating for young readers. This metaphorical language enables readers to grasp the profound emotions and challenges he faced during his own personal journey.

Repetition is another technique Robinson employs to reinforce key ideas and create a rhythmic and memorable reading experience. By repeating certain phrases or concepts, he emphasizes their significance and invites readers to reflect on them. This repetition serves to engage young readers by encouraging them to contemplate their own experiences and perspectives.

Furthermore, Robinson carefully structures his sentences to create a sense of rhythm and flow, enhancing the overall readability and impact of his argument. Short, concise sentences create moments of clarity and emphasis, while longer, more descriptive sentences evoke a contemplative and introspective tone. This varied sentence structure adds depth and nuance to his narrative, captivating young readers and keeping them engaged throughout.

In conclusion, through the effective use of metaphor, repetition, and sentence structure, Robinson engages and captivates young readers, inviting them to reflect on their own identities and experiences. His language choices not only develop his argument to his younger self but also establish a connection with present-day young readers, making his work both impactful and relatable.

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