HELP PLEASE DUE IN 3 MINUTES

Answer:

lane 3    Ф   = 450 vehicles,  ρ = 0.1 vehicle / s

lane 2    Ф_{average} = 300 vehicles,  ρ _{average} = 6.66 10⁻² vehicles/s

lane 1      Ф_{average} = 300 vehicles,  ρ_{average} = 2.66 10⁻² vehicle/s

Explanation:

Before solving this exercise we must clarify the concepts the flow is defined as the occurrence of an event in a time interval, in this case the passage of a car through time

Flux Density is the flux between unit area or unit time

Let's start by calculating the calculation for lane 3

the flow.

Let's use a direct rule of proportions (rule three) if the number of vehicles per unit of time (t₀ = 10s), for the observation time how many vehicles passed in the observation time (t_total = 75 * 60 = 4500 s)

            Ф = 4500 s (1 vehicle / 10 s)

            Ф   = 450 vehicles

The flux density is the flux per unit area, in this case the area is not indicated, so we can define the flux density as the flux per unit of time.

           ρ = 450/4500

           ρ = 0.1 vehicle / s

Lane 2

we look for the flow

we can have separates the interval into two parts

* for the first t₁1 = 30 * 60 = 1800 s

            Ф₁ = 1800 s (1 vehicle / 6s)

            Ф₁ = 300 vehicles during t₁

* for the rest of the time t₂ = 4500-1800 = 2700 s

           Ф₂ = 0

the average density is the total number of vehicles between the total time

           #_ {vehicle} = 300 +0

            Ф_{average) = # _vehicle

            Ф_{average} = 300 vehicles in all time

The density is

            ρ 1 = fi1 / t1

            ρ1 = 300/1800

            ρ1 = 1.66 10-1 vehicles / s

the average density is

            ρ_{average} = [tex]\frac{\phi_1 + \phi_2}{ t_{total}}[/tex]

            ρ _{average} = (300 +0) / 4500

            ρ _{average} = 6.66 10⁻² vehicles / s

Lane 1

flow

* first time interval t₁ = 10 * 60 = 600 s

             Ф₁ = 600 s (1 vehicle / 5s)

             Ф₁ = 120 vehicles in interval t₁

* second interval t₂ = 4500-600 = 3900 s

              Ф2 = 0

 average flow

             Ф = Ф1 + Ф2

             Ф = 120 vehicles at all time

Density

* first interval

          ρ₁ = 120/600

          ρ₁ = 0.2 vehicles / s

* second interval

          ρ₂ = 0

average density

         ρ+{average} = 120/4500

         ρ_{average} = 2.66 10⁻² vehicle/s

Answer:

true

Explanation:

Gravity (or the acceleration due to gravity) is 9.81 meters per second squared, on the surface of Earth, because of the size of Earth and the distance we are on its surface from its center.

Answer:

It is actually Albert Bandura! Hope this helps<3

Explanation:

In contrast to Skinner's idea that the environment alone determines behavior, Bandura (1990) proposed the concept of reciprocal determinism, in which cognitive processes, behavior, and context all interact, each factor simultaneously influencing and being influenced by the others.

The theorist who expanded the behavioristic perspective to include cognitive influences on personality was Albert Bandura. So, the correct option is B.

Personality is defined as a distinctive set of behavior, cognition, and emotional patterns that are formed by biological and environmental factors, and that change over time. This explained as the enduring characteristics and behavior which comprise a person's unique adjustment to life, including major traits, interests, drives, values, self-concept, abilities, and emotional patterns.

The four theory is also called as the proto-psychological theory which suggests that there are four fundamental personality types: sanguine, choleric, melancholic, and phlegmatic. Albert Bandura (1990) proposed the concept of reciprocal determinism in which cognitive processes, behavior, and context all interact, with each factor simultaneously affecting and being affected by the others.

Thus, the theorist who expanded the behavioristic perspective to include cognitive influences on personality was Albert Bandura. So, the correct option is B.

Learn more about Personality development, here:

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Answer:

T =  30.42°C

Explanation:

According to the conservation of energy principle:

[tex]Energy\ Given\ by\ Resistor = Heat\ Gain\ by\ Copper + Heat\ Gain\ by\ Water\\E = m_{c}C_{c}(T_{2c} - T_{1c}) + m_{w}C_{w}(T_{2w} - T_{1w})[/tex]

E = 120 KJ

mc = mass of copper = 13 kg

Cc = specific heat capacity of copper = 0.385 KJ/kg.°C

T2c = T2w = Final Equilibrium Temperature = T = ?

T1c = Initial Temperature of Copper = 27°C

T1w = Initial Temperature of Water = 50°C

mw = mass of water = 4 kg

Cw = specific heat capacity of water = 4.2 KJ/kg.°C

Therefore,

[tex]120\ KJ = (13\ kg)(0.385\ KJ/kg^oC)(T-27^oC) + (4\ kg)(4.2\ KJ/kg^oC)(T-50^oC)\\120\ KJ - 135.135\ KJ - 840\ KJ = (- 5.005T - 16.8 T)\ KJ/^oC\\T = \frac{-855.135\ KJ}{-28.105\ KJ/^oC}\\[/tex]

T =  30.42°C

Answer:

a) Δx = 11.6 m

b) t = 3.9 s

Explanation:

a)

       [tex]a = \frac{F_{fr} }{m} = \frac{-195N}{128kg} = -1.5 m/s2 (1)[/tex]

       [tex]v_{f} ^{2} -v_{o} ^{2} = 2* a* \Delta x (2)[/tex]

       [tex]\Delta x =- \frac{v_{o}^{2}}{2*a} =- \frac{(5.90m/s)^{2}}{2*(-1.5m/s2)} = 11.6 m (3)[/tex]

b)

       [tex]v_{f} = v_{o} + a*\Delta t (4)[/tex]

       [tex]\Delta t = \frac{-v_{o} }{a} = \frac{-5.9m/s}{-1.5m/s2} = 3.9 s (5)[/tex]

Answer:

B). half-ton car

Explanation:

We can find Kinectic using below expresion.

m= mass of the object

v= velocity

mass (1) 1 ton= 1000kg

Mass(2)0.5= 500kg

K.E= 1/2 mv^2

For 1 ton car

Substitute the values

Kinectic energy= 1/2( 1000) 30^2

= 450,000J

For 0.5 ton car

Kinectic energy= 1/2(500)60^2

= 900,000J

Hence, half-ton car has greater Kinectic energy

Answer:

v = 28.98 ft / s

Explanation:

For this problem we must solve it in parts, let's start by looking for the speed of the two cars after the collision

In the exercise they indicate the weight of each car

          Wₐ = 1500 lb

          W_b = 1125 lb

Car B's velocity from v_b = 42.0 mph westward, car A travels east

let's find the mass of the vehicles

             W = mg

             m = W / g

             mₐ = Wₐ / g

             m_b = W_b / g

             mₐ = 1500/32 = 46.875 slug

             m_b = 125/32 = 35,156 slug

Let's reduce to the english system

             v_b = 42.0 mph (5280 foot / 1 mile) (1h / 3600s) = 61.6 ft / s

We define a system formed by the two vehicles, so that the forces during the crash have been internal and the moment is preserved

we assume the direction to the east (right) positive

initial instant. Before the crash

           p₀ = mₐ v₀ₐ - m_b v_{ob}

final instant. Right after the crash

           p_f = (mₐ + m_b) v

the moment is preserved

           p₀ = p_f

           mₐ v₀ₐ - m_b v_{ob} = (mₐ + m_b) v

           v = [tex]\frac{ m_a \ v_{oa} - m_b \ v_{ob} }{ m_a +m_b}[/tex]

we substitute the values

           v = [tex]\frac{ 46.875}{82.03} \ v_{oa} - \frac{35.156}{82.03} \ 61.6[/tex]

           v = 0.559 v₀ₐ - 26.40                  (1)

Now as the two vehicles united we can use the relationship between work and kinetic energy

the total mass is

              M = mₐ + m_b

              M = 46,875 + 35,156 = 82,031 slug

starting point. Jsto after the crash

              K₀ = ½ M v²

final point. When they stop

             K_f = 0

The work is

             W = - fr x

the negative sign is because the friction forces are always opposite to the displacement

Let's write Newton's second law

Axis y

           N-W = 0

           N = W

the friction force has the expression

            fr = μ N

we substitute

            -μ W x = Kf - Ko

            -μ W x = 0 - ½ (W / g) v²

            v² = 2 μ g x  

            v = [tex]\sqrt{ 2 \ 0.750 \ 32 \ 17.5}[/tex]Ra (2 0.750 32 17.5  

            v = 28.98 ft / s

Answer:

Tectonic Plate Movement

Explanation:

Each continent and ocean sits on its own tectonic plate which floats on the Earths upper mantle. They move very little over time.

Answer:

tectonic plates movement

Answer:

The right approach is "0.85 m".

Explanation:

According to the question, the diagram will is provided below.

So that as per the diagram,

The values will be:

My height,

AO = 1.70 m

My eyes at,

AB = 12 cm

i.e.,

     = 0.12 m

As we can see, the point of incidence lies between the feet as well as the eyes, then

BO = 1.58 m

Now,

⇒ [tex]O'D = \frac{1.58}{2}[/tex]

           [tex]=0.79 \ m[/tex]

The point of incidence of the ray will be:

⇒ [tex]CO'=1.70-\frac{0.12}{2}[/tex]

           [tex]=1.70-0.06[/tex]

           [tex]=1.64 \ m[/tex]

Hence,

The smallest length of the mirror will be:

= [tex]CO'-O'D[/tex]

On substituting the values, we get

= [tex]1.64-0.79[/tex]

= [tex]0.85 \ m[/tex]

Answer:

Required heat Q = 11,978 KJ

Explanation:

Given:

Mass = 5.3 kg

Latent heat of vaporization of water = 2,260 KJ / KG

Find:

Required heat Q

Computation:

Required heat Q = Mass x Latent heat of vaporization of water

Required heat Q = 5.3 x 2260

Required heat Q = 11,978 KJ

Required heat Q = 12,000 KJ (Approx.)

Answer:

0.6

Explanation:

Given that :

Radius, R = 7m

Period, T = 6.9s

The Coefficient of static friction, μs can be obtained using the relation :

μs = v² / 2gR

Recall, v = 2πR/T

μs becomes ;

μs = (2πR/T)² / 2gR

μs = (4π²R² / T²) ÷ 2gR

μs = (4π²R² / T²) * 1/ 2gR

μs = 4π²R / T²g

μs = 4π²*7 / 6.9^2 * 9.8

μs = 28π² / 466.578

μs = 276.34892 / 466.578

μs = 0.5922887

μs = 0.6

Answer:

I thinks its a, but its really about gravity im not sure

Explanation:

:)

Answer:

speed of car after collision, v2 =16.1 m/s and of the truck, v1 = 4.6 m/s

Explanation:

Given:

mass of truck M = 1370 kg

speed of truck = 12.0 m/s

mass of car m = 593 kg

collision is elastic therefore,

Applying law of momentum conservation we have

momentum before collision = momentum after collision

1370×12 + 0( initially car is at rest) = 1370×v1+ 593×v2               ....(i)

Also for a collision to be elastic,

velocity of approach = velocity of separation

12 -0 = v2-v1                  ....(ii)

using (i) and (ii) we have

So speed of car after collision, v2 =16.1 m/s and of the truck, v1 = 4.6 m/s

Answer:

A) a_max = 4.9 m/s²

B) the metal cabinet will not slip.

C) a_max = 9.8 m/s². The cabinet will slip.

Explanation:

A) We are given;

Coefficient of static friction; μ_s = 1

Coefficient of kinetic friction; μ_k = 0.7

Formula for maximum static friction;

F_s = μ_s•N

We are dealing with half of the weight of a flatbed truck. Thus;

N = mg/2

Thus;

F_s(max) = ½μ_s•mg

Now, the maximum acceleration it can achieve on dry concrete will be when the Maximum static friction is reached.

Because after maximum static friction, the cabinet will slip.

Thus; F_s(max) = ma

Therefore,

ma = ½μ_s•mg

m will cancel out to give;

a = ½μ_s•g

a = ½ × 1 × 9.8

a_max = 4.9 m/s²

B) we are told that the coefficient of static friction is now 0.55

Thus, the friction between the metal and the wood is;

F = μ_s•g = 0.55 × 9.8

F = 5.39 N

The acceleration gotten in the first part is less than μ_s•g = 5.39 N, then the metal cabinet will not slip.

C) Now, if we are considering a 4 wheel drive, then we will not divide the mass by 2 and so; N = mg

Like we did in A above;

ma = μ_s•mg

a = μ_s•g

a = 1 × 9.8

a = 9.8 m/s²

Now, this value of a_max is greater than μ_s•g in answer A above.

Thus, a_max > μ_s•g and thus, the cabinet will slip.

Answer:

Tuesday bc instead of running he/she was walking bc he/she might not have as much energy

Explanation:

Answer:

The answer is C. time. Please mark as brainliest.

Answer:

A)  a_c = 1.75 10⁴ m / s², B) a = 4.43 10³ m / s²

Explanation:

Part A) The relation of the test tube is centripetal

               a_c = v² / r

the angular and linear variables are related

              v = w r

we substitute

               a_c = w² r

let's reduce the magnitudes to the SI system

              w = 4000 rpm (2pi rad / 1 rev) (1 min / 60s) = 418.88 rad / s

               r = 1 cm (1 m / 100 cm) = 0.10 m

let's calculate

              a_c = 418.88² 0.1

               a_c = 1.75 10⁴ m / s²

part B) for this part let's use kinematics relations, let's start looking for the velocity just when we hit the floor

as part of rest the initial velocity is zero and on the floor the height is zero

                v² = v₀² - 2g (y- y₀)

                v² = 0 - 2 9.8 (0 + 1)

                v =√19.6

                v = -4.427 m / s

now let's look for the applied steel to stop the test tube

                v_f = v + a t

                0 = v + at

                a = -v / t

                a = 4.427 / 0.001

                a = 4.43 10³ m / s²

Answer:

#1 Yes

Explanation: #1: The rest of them are used mainley by farmers, and crops are used by common citizens in the world.

Question 1: Crops.

Question 2: Diagnostic Services.

Question 3: A cable company needs to lay new fiber optic cable to reach its customers across a large lake.

Question 4: A bachelor's degree in energy research.

Question 5: Environmental Resources.

If any of these answers are incorrect, please tell me, so I can fix my mistake. Thank you.

Answer:

a

Explanation:

I think don't get mad if I'm wrong

Answer:

i don't know but i hope you get it right

Explanation

Answer:

"0.25 kg-m²" is the appropriate answer.

Explanation:

The diagram of the question is missing. Find the attachment of the diagram below.

According to the diagram, the values are:

m₁ = 0.2

m₂ = 0.3

m₃ = 0.3

m₄ = 0.2

d₁ = d₂ = d₃ = d₄ = 0.5 m

As we know,

The moment of inertia is:

⇒  [tex]I=\Sigma M_id_i^2[/tex]

then,

⇒  [tex]I=m_1d_1^2+m_2d_2^2+m_3d_3^2+m_4d_4^2[/tex]

⇒     [tex]=d^2(m_1+m_2+m_3+m_4)[/tex]

On substituting the values, we get

⇒     [tex]=0.5^2\times (0.2+0.3+0.3+0.2)[/tex]

⇒     [tex]=0.25\times 1[/tex]

⇒     [tex]=0.25 \ Kg-m^2[/tex]

Answer:

[tex]\boxed{\text{\sf \Large 42 m}}[/tex]

Explanation:

Use height formula

[tex]\displaystyle \sf H=\frac{u^2 sin(\theta)^2}{2g}[/tex]

u is initial velocity

θ = 90° (fired vertically upward)

g is acceleration of gravity

[tex]\displaystyle \sf H=\frac{29^2 \times sin(90 )^2}{2 \times 10}=42.05[/tex]

Answer:

The mass of water boiled off is [tex]0.0 \overline{185}[/tex] kg

Explanation:

The given percentage by weight of protein solids in raspberries = 10 weight%  

The ratio of sugar to raspberries in ja-m = 45:55

The mass of the mixture after boiling = 0.4 weight fraction water

Let 's' represent the mass of sugar in the mixture, and let 'r' represent the mass of raspberry

The mass of raspberry, r = 1 kg

The percentage by weight of water in raspberry = 90 weight %

The mass of water in 1 kg of raspberry =  90/100 × 1 kg = 0.9 kg

The ratio of the mass of sugar to the mass of raspberry in jam = r/s = 45/55

∴ s = 1 kg × 55/45 = 11/9 kg

The mass of the mixture before boiling = 1 kg + 11/9 kg = 20/9 kg

The weight fraction of water in the remaining mixture after boiling = 0.4 weight fraction

Let 'w' represent the mass of water boiled off, we have;

(0.9 - w)/(20/9 - w) = 0.4

(0.9 - w) = 0.4 × (20/9 - w)

0.9 - w = 8/9 - 0.4·w

9/10 - 8/9 = w - 0.4·w = 0.6·w = (6/10)·w

(81 - 80)/(90) = (6/10)·w

1/90 = (6/10)·w

w = ((10/6) × 1/90) = 1/54

w = 1/54  

The mass of water boiled off, w = (1/54) kg = [tex]0.0 \overline{185}[/tex] kg

Answer:A)u =4.295m/s  , B)a = 29.746m/s²   C) F=3,153N

Explanation:

Using the kinematic expression  

v² = u² - 2as

where

u = initial velocity

v = final velocity

s = distance

g = acceleration due to gravity .

Given that he reaches a height of 0.940 m above the floor,

the final velocity  = 0

Here, acceleration due to gravity is acting in  opposite the initial direction of motion. So, a=-9.81 m/s.

v² = u² + 2as

0² - u² = 2 (- 9.81) × 0.940

- u² = 2 × - 9.81 × 0.920

- u² = -18.4428

cancelling the minus in both sides , we have that  

u² = 18.4428

u = √18.4428

u =4.295m/s

(b) His acceleration (in m/s2) while he is straightening his legs. He goes from zero to the velocity found in part (a) in a distance of 0.310 m. m/s2

Using v² = u² + 2as

where u = initial speed of basketball player before lengthening = 0 m/s,

v = final speed of basketball player after lengthening =  4.295m/s,

a = acceleration while  straightening his legs

s = distance moved during lengthening = 0.310m

v² = u² + 2as  

 a = (v² - u²)/2s

a = (4.29m/s)² - (0 m/s)²)/(2 × 0.310m)

a = (18.4428 m²/s² - 0 m²/s²)/(0.62 m)

a = (18.4428 m²/s²/(0.62 m)

a = 29.746m/s²

c) The force (in N) he exerts on the floor to do this, given that his mass is 106 kg. N

Force= mass x acceleration.

F = 106 kg X 29.746m/s²

 F = 3,153.076 rounded to  3,153N

Answer:

The amount of work required to empty the trough by pumping the water over the top is approximately 98,000 J

Explanation:

The length of the trough = 10 meters

The width of the through = 1 meter

The depth of the trough = 2 meters

The vertical cross section of the through = An isosceles triangle

The density of water in the through = 1000 kg/m³

Let 'x' represent the width of the water at a depth

x/y = 1/2

∴x = y/2

The volume of a layer of water, dV, is given as follows;

dV = 10 × y/2 × dy = 5·y·dy

The mass of the layer of water, m = ρ × dV

∴ m = 1000 kg/m³ × 5·y·dy m³ = 5,000·y·dy kg

The work done, W = m·g·h

Where;

h = The the depth of the trough from which water is pumped

g = The acceleration due to gravity ≈ 9.8 m/s²

[tex]\therefore \, W \approx \int\limits^2_0 {5,000 \times y \times 9.8 \, dy} = \left[24,500\cdot y^2 \right]^2_0 = 98,000[/tex]

The work done by the pump to pump all the water in the trough, over the top W ≈ 98,000 J

Answer:

[tex]\boxed{\text{\sf \Large 3.0 s}}[/tex]

Explanation:

Use distance formula

[tex]\displaystyle d=ut+\frac{1}{2} at^2[/tex]

[tex]u= \text{\sf initial velocity}\\d= \text{\sf distance}\\a= \text{\sf acceleration}\\t= \text{\sf time taken}[/tex]

[tex]\displaystyle 18=0 \times t+\frac{1}{2} \times 4 \times t^2[/tex]

[tex]t=3[/tex]

A block starting from rest slides down the length of an 18 m plank with an acceleration of 4.0 meters per second square. Time taken by the block  to reach at the bottom is 3 sec.

The rate at which an object changes its velocity is known as acceleration. Acceleration is a vector quantity. If an object's velocity is changing then it is accelerating and an object with a constant velocity is not accelerating.

The speed at which something moves in a specific direction is known as its velocity. As an illustration, think of the speed of a car travelling north on a highway or the speed at which a rocket takes off.

Given that in the question block slides down an 18 m plank length with an acceleration of 4.0 meters per second square when it begins at rest.

Using equation of motion,

S = ut + (1/2)at²

s is distance, s = 18 m

u is initial velocity, u = 0

a is acceleration, a = 4 m/sec²

t is time

18 = 0*t + (1/2)*4*t

solving we get t = 3 sec.

Time taken by the block  to reach at the bottom is 3 sec.

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The disadvantage of sending information over long distances wirelessly using digital signals is "the signals become weaker the farther the receiver is from the source."

Thus , The disadvantage of sending information over long distances wirelessly using digital signals is "the signals become weaker the farther the receiver is from the source."

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