A bicycle wheel has a diameter of 63.0 cm and a mass of 1.75 kg. Assume that the wheel is a hoop with all of the mass concentrated on the outside radius. The bicycle is placed on a stationary stand and a resistive force of 121 N is applied tangent to the rim of the tire. (a) What force must be applied by a chain passing over a 8.96-cm-diameter sprocket in order to give the wheel an acceleration of 4.40 rad/s2

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:

tectonic plate movement

Explanation:

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:

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:

#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:

The temperature of the steam during the heat rejection process is 42.5°C

Explanation:

Given the data in the question;

the maximum temperature T[tex]_H[/tex] in the cycle is twice the minimum absolute temperature T[tex]_L[/tex] in the cycle

T[tex]_H[/tex]  = 0.5T[tex]_L[/tex]

now, we find the efficiency of the Carnot cycle engine

η[tex]_{th[/tex] = 1 - T[tex]_L[/tex]/T[tex]_H[/tex]

η[tex]_{th[/tex] = 1 - T[tex]_L[/tex]/0.5T[tex]_L[/tex]

η[tex]_{th[/tex] = 0.5

the efficiency of the Carnot heat engine can be expressed as;

η[tex]_{th[/tex] = 1 - W[tex]_{net[/tex]/Q[tex]_H[/tex]

where W[tex]_{net[/tex] is net work done, Q[tex]_H[/tex] is is the heat supplied

we substitute

0.5 = 60 / Q[tex]_H[/tex]

Q[tex]_H[/tex] = 60 / 0.5

Q[tex]_H[/tex] = 120 kJ

Now, we apply the first law of thermodynamics to the system

W[tex]_{net[/tex] = Q[tex]_H[/tex] - Q[tex]_L[/tex]

60 = 120 -  Q[tex]_L[/tex]

Q[tex]_L[/tex] = 60 kJ

now, the amount of heat rejection per kg of steam is;

q[tex]_L[/tex] = Q[tex]_L[/tex]/m

we substitute

q[tex]_L[/tex] = 60/0.025

q[tex]_L[/tex] = 2400 kJ/kg

which means for 1 kilogram of conversion of saturated vapor to saturated liquid , it takes 2400 kJ/kg of heat ( enthalpy of vaporization)

q[tex]_L[/tex] = h[tex]_{fg[/tex] = 2400 kJ/kg

now, at  h[tex]_{fg[/tex]  = 2400 kJ/kg from saturated water tables;

T[tex]_L[/tex] = 40 + ( 45 - 40 ) ( [tex]\frac{2400-2406.0}{2394.0-2406.0}\\}[/tex] )

T[tex]_L[/tex] = 40 + (5) × (0.5)

T[tex]_L[/tex] = 40 + 2.5

T[tex]_L[/tex] = 42.5°C  

Therefore, The temperature of the steam during the heat rejection process is 42.5°C  

Answer:

Centripetal force formula

Explanation:

F = mv²

----

r

Answer:

c

Explanation:

mark brainleyest plz if right

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]

Wave speed = (wavelength) x (frequency)

Wavelength = (wave speed) / (frequency)

Wavelength = (9 m/s) / (0.5 Hz)

Wavelength = 18 m

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:

https://brainly.com/question/1452486

#SPJ6

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:

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

Explanation

Answer:

The answer is "[tex]- 0.5747\ \frac{rad}{s^2}[/tex]"

Explanation:

Let

[tex]M_L = 12\ kg\\\\M_R = 5.8\ kg[/tex]

While it is in balance, its net force mostly on a machine is zero where the board will rotate an upward torques were given by:

[tex]\to M_L \times g \times R_L - M_R \times g \times R_R = \tau[/tex]

[tex]\to 12 \times 9.8 \times 0.6 - 5.8 \times 9.8 \times 1.4 = \tau[/tex]

[tex]\to \tau= 70.56-79.576\\\to \tau = -9.016[/tex]

let,

[tex]\tau = I \alpha[/tex]

where

[tex]\alpha =[/tex]angular acceleration

I = moment of inertia of the system

[tex]\to I = M_L \times r \times L_2 + M_R \times r \times R_2\\\\\to I = 12 \times 0.6 \times 0.6 + 5.8 \times 1.4 \times 1.4\\\\\to I= 4.32+11.368\\\\\to I = 15.688\ kg\ m2\\\\[/tex]

 Calculating the angular acceleration:

[tex]\alpha = \frac{\tau}{I}\\\\[/tex]

    [tex]= \frac{-9.016}{15.688}\\\\=- 0.5747\ \frac{rad}{s^2}\\\\[/tex]

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:

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

Answer:

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

Explanation:

:)

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:

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

Explanation:

Answer:

the frame rate of the time-lapse video if no aliasing occurs is 10min per frame

Explanation:

Given the data in the question;

No aliasing effect, meaning the clock with missing hour hand will not have any effect

Time required for one frame = 10 minutes

frame rate of time = Time required / number of frame

= 10 min / 1 frame

= 10min per frame

Therefore, the frame rate of the time-lapse video if no aliasing occurs is 10min per frame

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:

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:

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:

Explanation:

m₂ is hanging vertically and m₁ is placed on inclined plane . Both are in limiting equilibrium so on m₁ , limiting friction will act in upward direction as it will tend to slip in downward direct . Tension in cord connecting the masses be T .

For equilibrium of m₁

m₁ g sinα= T + f where f is force of friction

m₁ g sinα= T + μsx m₁ g cosα

m₁ g sinα -  μs x m₁ g cosα = T

For equilibrium of m₂

T = m₂g

Putting this value in equation above

m₁ g sinα -  μs x m₁ g cosα = m₂g

m₂ = m₁ sinα -  μs x m₁ cosα

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:

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:

a

Explanation:

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