A car of mass 1800 kg can be just be lifted. What is the least force that the electromagnet must use to lift the car? (1 g = 10 N/kg)

Answer:

[tex]0.005847\ \text{s}[/tex]

Explanation:

Radio waves travel at the speed of light

[tex]c[/tex] = Speed of light = [tex]3\times 10^8\ \text{m/s}[/tex]

[tex]d_r[/tex] = Distance between two radios = 46 km

[tex]v[/tex] = Speed of sound in air = 348 m/s

[tex]d_s[/tex] = Distance sound travels across the newsroom = 2.1 m

Time taken by the radio signal to reach the required location is

[tex]t_r=\dfrac{d_r}{c}\\\Rightarrow t_r=\dfrac{46\times 10^3}{3\times 10^8}\\\Rightarrow t_r=0.000153\ \text{s}[/tex]

Time taken by sound to reach the required location is

[tex]t_s=\dfrac{d_s}{v}\\\Rightarrow t_s=\dfrac{2.1}{348}\\\Rightarrow t_s=0.006\ \text{s}[/tex]

The time difference is

[tex]t_s-t_r=0.006-0.000153=0.005847\ \text{s}[/tex]

The difference in time that the message is received is [tex]0.005847\ \text{s}[/tex].

Answer:

Advantages of mercury as a thermometric liquid.

-It is a good conductor of heat and therefore the whole liquid reaches the temperature of the surroundings quickly.

-It does not wet (cling to the sides of) the tube.

-It has a high boiling point

-It expands uniformly (linear expansion) and responds quickly to temperature changes, hence is sensitive.

-It has a visible meniscus.

Disadvantages

-Mercury is very poisonous.

-its expansively is fairly low

-it is expensive

-It has a high freezing point therefore it cannot be used in places where the temperature gets very low.

Alcohol has a thermometric fluid

-Alcohol expands uniformly.

-It has a low freezing point (-115 degreecentigrade) therefore it is very suitable for place where the temperature gets very low.

-It has a large expansively

-It is an easily available cheap liquid, which is safe to use

Disadvantages of alcohol

-it wets the tube

-it has a low boiling point (cannot be used in places with high temperatures)

-it does not react quickly to changes in temperature

-It needs to be dyed, since it's colourless.

Disadvantages of water

-Water has high specific heat capacity. So it cannot be used for measuring small temperature differences.

- Water will wet the surface of the glass tube. It is a sticky substance.

- Water is transparent

Explanation:

Answer:

a. i) The pressure drop per unit length is 52,151.89 Pa

ii) The atmospheric pressure ≈ 19.175 × The pressure drop along 10 cm length of aorta

b i) The power required for the heart to push blood along a 10 cm length of aorta, is 2.6075945 Watts

ii) The basal metabolic rate ≈ 38.35 × The power to push the blood along a 10 cm length of aorta

c. i) Please find attached the drawing for the velocity profile created with Microsoft Excel

ii) The velocity at the center is approximately 2.04 m/s

Explanation:

The given diameter of the aorta, D = 2.5 cm = 0.025 m

The maximum flow rate, Q = 500 cm³/s = 0.0005 m³/s

Assumptions;

The blood flow is laminar

The blood is a Newtonian fluid

The viscosity of water ≈ 0.01 poise = 1 cp

a. i) The pressure drop per unit length of pipe ΔP/L is given by the Hagen Poiseuille equation as follows;

[tex]Q = \dfrac{\pi \cdot R^4}{8 \cdot \mu} \cdot \left(\dfrac{\Delta p}{L} \right)[/tex]

Where;

Q = The flow rate = A·v

A = The cross sectional area

R = The radius = D/2

Δp/L = The pressure drop per unit length of the pipe

Therefore, we have;

[tex]\dfrac{\Delta p}{L} = \dfrac{Q\cdot 8 \cdot \mu }{\pi \cdot R^4} = \dfrac{0.0005 \times 8 \times 1}{\pi \times 0.0125^4 } = 52151.89[/tex]

The pressure drop per unit length ΔP/L = 52,151.89 Pa

ii) The pressure, ΔP, drop along 10 cm (0.1 m) length of aorta = ΔP/L × x;

∴ ΔP = 52,151.89 Pa × 0.1 m = 5,215.189 Pa

Given that the atmospheric pressure, [tex]P_{atm}[/tex] = 10⁵ Pa, we have;

[tex]P_{atm}[/tex]/ΔP = 10⁵/5,215.189 ≈ 19.175

Therefore, the atmospheric pressure is approximately 19.175 times the pressure drop along 10 cm length of aorta

b. i) The power, P = Q × ΔP

Therefore, the power required for the heart to push blood along a 10 cm length of aorta, is P₁₀ = 0.0005 m³/s × 5,215.189 Pa = 2.6075945 Watts

ii) Therefore compared to the basal metabolic rate of, 'P', 100 W, we have;

P/P₁₀ = 100 W/2.6075945 Watts = 38.349521 ≈ 38.35

The basal metabolic rate is approximately 38.35 times more powerful than the power to push the blood along a 10 cm length

c. i) The velocity profile across the aorta is given as follows;

[tex]v_m = \dfrac{1}{4 \cdot \mu} \cdot \dfrac{\Delta P}{L} \cdot R^2[/tex]

Where;

[tex]v_m[/tex] = The velocity at the center

We get;

[tex]v_m = \dfrac{1}{4 \times 1} \times 52,151.89 \times 0.0125^2 \approx 2 .04[/tex]

The velocity at the center, [tex]v_m[/tex] ≈ 2.04 m/s

ii) The velocity profile, v(r), is given by the following formula;

[tex]v(r) = v_m \cdot \left[1 - \dfrac{r^2}{R^2} \right][/tex]

Therefore, we have;

[tex]v(r) = 2.04 - \dfrac{2.04 \cdot r^2}{0.0125^2} \right] = 2.04 - 163\cdot r^2[/tex]

The velocity profile of the pipe is created with Microsoft Excel

Answer:

a)  v = 2.9 m / s, b) A = 0.350 m, c)    v = 2.9 m / s, d)   A = 1.00 m

Explanation:

The oscillatory motion is described by the expression

          x = A cos (wt + Ф)

the wavelength which is the distance for the wave to repeat and the frequency which is the number of times a wave oscillates per unit of time

a) In this part they ask us for the speed of the wave.

Let's use the relationship between speed, wavelength and frequency

          v = λ f

For the wavelength they indicate that the distance between two crest is 6.1 m

        λ / 2 = 6.10

        λ = 12.20 m

They give us the period of the wave is the time it takes to return to the same point, in this case they give half a period

       A / 2 = 2.10

       A = 4.20 me

        f = 1 / t

        f = ¼, 2

        f = 0.238 Hz

let's calculate

         v = 12.20 0.238

         v = 2.9 m / s

b) the amplitude of the wave, is the distance from zero to some maximum

                 2A = 0.700

                   A = 0.350 m

c) the speed of the wave is not function of the amplitude, so the speed is the same

           v = 2.9 m / s

d) the amplitude is

           2A = 0.50

             A = 1.00 m

This question is incomplete, the complete question is;

A crude oil storage tank initially contains 1000 m³ of crude oil. Oil is pumped into the tank through a pipe at a rate of 2 m³/min and out of the tank at a velocity of 1.5 m/s through another pipe having a diameter of 0.15m. The crude oil has a specific volume of 0.0015 m³/kg .

Determine : a) The mass of oil in the tank, in kg, after 24 hours b) The volume of oil in the tank, in m³, at that time ( after 24hrs )  

Answer:

a) The mass of oil in the tank after 24hrs is 1059852.6667 kg  

b) The volume of oil in the tank, in m³, at that time is 1589.779 m³

Explanation:

Given that;

volume of tank v = 1000 m³

Discharge  (AV)₁ = 2 m³/min

specific volume (v) = 0.0015 m³/kg

a)

for one inlet, one exit control volume, we have]

d(mc[tex]_v[/tex])/dt = "m[tex]_i[/tex] - "m[tex]_e[/tex]

"m[tex]_i[/tex]  = (AV)₁  / v

"m[tex]_i[/tex]  = 2 / 0.0015

"m[tex]_i[/tex] = 1333.33 kg/min = ( 1333.33 kg/min × 60 ) = 80,000 kg/hr

"m[tex]_e[/tex] =  AV₂/v

AREA A = πr² = π(0.15m/2)²

so

"m[tex]_e[/tex] =  ( π(0.15m/2)² × 1.5 m/s ) / 0.0015m³/kg

"m[tex]_e[/tex] = 17.6714 kg/s = (17.6714 kg/s × 60 × 60 ) = 63,617.25 kg/hr

 d(m[tex]_{cv}[/tex])/dt = 80000 - 63,617.25 =  16382.75 kg/hr

now after 24 hours, the mass of the crude oil tank is;

mc[tex]_v[/tex]( 24 ) = ( 16382.75 × 24 ) + mass initially contained

so initial mass m[tex]_i[/tex] = volume / specific volume

initial mass m[tex]_i[/tex] = V / v = 1000 / 0.0015 = 666,666.6667 kg

Total mass = m[tex]_i[/tex] + mc[tex]_v[/tex]

Total mass = 666,666.6667 kg + (16382.75 × 24)  

Total mass = 1059852.6667 kg  

Therefore, The mass of oil in the tank after 24hrs is 1059852.6667 kg  

b)

Volume will be;

V = total mass × density

V = 1059852.6667 kg  × 0.0015 m³/kg

V = 1589.779 m³

Therefore, The volume of oil in the tank, in m³, at that time is 1589.779 m³

Answer:

b i hope this is correct answee

Answer:

a)     v₀ = 2.5 m / s,   heading south.

b)  W = 1,219 10⁵ J

Explanation:

a) For this exercise we can use the conservation of the moment, we create a system formed by all the 4 cars, in this case when the last one separates the forces are intense and the moment is conserved

initial instant. Before separation

        p₀ = M v₀

final instant. When uncoupling the last car

        p_f = 3m v₁ + m v₂

where they indicate that the speed of the wagons is v₁ = 2.00 m / s and the speed of the last wagon is v₂ = 4.00 m / s

the total mass is M = 4m

how the moment is preserved

           p₀ = p_f

         4m v₀ = 3m v₁ + mv₂

         v₀ = ¾ v₁ + v₂ / 4

let's calculate

           v₀ = ¾ 2 + ¼ 4

           v₀ = 2.5 m / s

heading south.

b) work is equal to the change in kinetic energy

              W = ΔK = K_f -K_o

              W = ½ m v_f² - ½ m v₀²

               W = ½ m (v_f² -v₀²)

               W = ½ 2.50 10⁴ (4² - 2.5²2)

               W = 1,219 10⁵ J

Answer:

The Cell Membrane is found in both plant and animal cells.

Explanation:

Chloroplast, Large Vacuole and Cell Wall are all found in plant cells.

Answer:

E = 216.76 J

Explanation:

The amount of energy absorbed by the goalkeeper will be equal to the difference between the initial and the final kinetic energy of the ball.

[tex]Energy\ Absorbed = Initial\ Kinetic\ Energy - Final\ Kinetic\ Energy\\Energy\ Absorbed = E = \frac{1}{2}mv_{i}^2 - \frac{1}{2}mv_{f}^2\\E = \frac{1}{2}m(v_{i}^2 - v_{f}^2)\\[/tex]

where,

m = mass of ball = (16 ounce)(0.0283495 kg/1 ounce) = 0.4535 kg

vf = final velocity = 2.25 m/s

vi = initial velocity = 31 m/s

Therefore,

[tex]E = \frac{1}{2}(0.4535\ kg)[(31\ m/s)^2 - (2.25\ m/s)^2][/tex]

E = 216.76 J

Answer:

Once we place a positive test at a point close to the sphere, we find that an electrostatic force is applied to the outside of the sphere. Therefore, at any point around the sphere, the electric field vector is radially outward.

Answer:

determination

Explanation:

Answer:

I think it's theoretical.

From the given options the best example of an ascribed status will be poor.  Hence, option 2 is the correct one.

In sociology, the term "ascribed status" refers to a human's social standing as it is either allocated to them at birth or implicitly assumed in later life. The position of status is one that a person neither chooses for themselves nor earns. Instead, the position is decided upon in accordance with cultural and societal norms and standards. No matter how hard you try or how much you want to, these slots are filled. These fixed social designators are independent of the favorable or unfavorable stereotypes associated with one's assigned statuses and stay unchanged throughout a person's life.

Every society engages in the process of classifying people into categories based on factors including gender, race, family history, and ethnicity. This process is cross-cultural.

To know more about ascribed status:

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Answer:

e)

Explanation:

i)

        [tex]V = I*r_{i} + I*R_{1} (1)[/tex]

      [tex]I_{i} = \frac{V}{r_{int} + R_{i}} = \frac{1.5V}{0.1 \Omega + 1.4 \Omega} = 1.0 A (2)[/tex]

ii)

       [tex]I_{ii} =\frac{V_{term} }{R_{ii} } =\frac{3.6V}{1.8 \Omega} = 2.0 A (3)[/tex]

iii)

      [tex]I_{iii} =\frac{V-V_{term}}{r_{int} } =\frac{12.0 V- 11.0 V}{0.2 \Omega} =\frac{1.0V}{0.2\Omega} = 5.0 A (4)[/tex]

Answer:

Did you ask the question and answer it for yourself?

Answer:

Your amazing thank you, A was right.

Answer:

1) thinner, diverge

2) thicker, converge

Explanation:

i got it right thanks to the other user :)

A concave lens is thinner in the middle than at the edges, and causes light rays to diverge.

A convex lens is thicker in the middle than at the edges, and causes light rays to converge.

The lenses used to form images of the object placed a distant apart.

The concave mirror form images by virtual meeting of the light rays and appear to meet. They are thin at the middle portion.

The convex mirror form images by real meeting of the light rays. They converge the rays coming from object. They are thick at the middle portion.

Thus, A concave lens is thinner in the middle than at the edges, and causes light rays to diverge.

A convex lens is thicker in the middle than at the edges, and causes light rays to converge.

Learn more about concave and convex lens.

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Answer:

a) quantity to be measured is the height to which the body rises

b) weighing the body , rule or fixed tape measure

c)   Em₁ = m g h

d) deformation of the body or it is transformed into heat during the crash

Explanation:

In this exercise of falling and rebounding a body, we must know the speed of the body when it reaches the ground, which can be calculated using the conservation of energy, since the height where it was released is known.

a) What quantities must you know to calculate the energy after the bounce?

The quantity to be measured is the height to which the body rises, we assume negligible air resistance.

So let's use the conservation of energy

starting point. Soil

          Em₀ = K = ½ m v²

final point. Higher

          Em_f = U = mg h

         Em₀ = Em_f

         Em₀ = m g h₀

b) to have the measurements, we begin by weighing the body and calculating its mass, the height was measured with a rule or fixed tape measure and seeing how far the body rises.

c) We use conservation of energy

starting point. Soil

          Em₁ = K = ½ m v²

final point. Higher

          Em_f = U = mg h

         Em₁ = Em_f

         Em₁ = m g h

d) to determine if the energy is conserved, the arrival energy and the output energy must be compared.

There are two possibilities.

* that have been equal therefore energy is conserved

* that have been different (most likely) therefore the energy of the rebound is less than the initial energy, it cannot be stored in the possible deformation of the body or it is transformed into heat during the crash

Answer:

27: 85

28:75%

Explanation:

27:68=80

?=100 hence (68×100)÷80

=85

28:18/24× 100

=75%

Answer:

it would be 83% in a fatal crash.

Answer:

Work= Fcos∆×S

W=60N×cos 30⁰×100

W=60×0.866×100=5196.1J

PLEASE GIVE BRAINLIEST

molecular weights are written in the picture.

CH4<NH3<H2O<Cl2

sorry bro but i don't know his i am also bad

Explanation:

but maybe you can you the formula of volatage as well as

Answer:

1.11 m.

Explanation:

Why?

The speed for a wave is done by the equation:   v = f * w. Because the frecuency tells us about how many cycles the wave makes each time, but for each cycle the wave runs certain distance, given for the wavelenght. If you isolate the letter w you get the value just doing a ratio.

v = speed

f = frecuency

w = wavelenght

w = v / f

Answer:

A) F_g = 26284.48 N

B) v = 7404.18 m/s

C) E = 19.19 × 10^(10) J

Explanation:

We are given;

Mass of satellite; m = 3500 kg

Mass of the earth; M = 6 x 10²⁴ Kg

Earth circular orbit radius; R = 7.3 x 10⁶ m

A) Formula for the gravitational force is;

F_g = GmM/r²

Where G is gravitational constant = 6.67 × 10^(-11) N.m²/kg²

Plugging in the relevant values, we have;

F_g = (6.67 × 10^(-11) × 3500 × 6 x 10²⁴)/(7.3 x 10⁶)²

F_g = 26284.48 N

B) From the momentum principle, we have that the gravitational force is equal to the centripetal force.

Thus;

GmM/r² = mv²/r

Making v th subject, we have;

v = √(GM/r)

Plugging in the relevant values;

v = √(6.67 × 10^(-11) × 6 x 10²⁴)/(7.3 x 10⁶))

v = 7404.18 m/s

C) From the energy principle, the minimum amount of work is given by;

E = (GmM/r) - ½mv²

Plugging in the relevant values;

E = [(6.67 × 10^(-11) × 3500 × 6 × 10²⁴)/(7.3 x 10⁶)] - (½ × 3500 × 7404.18)

E = 19.19 × 10^(10) J

Answer:

it is for sure B. Quantitative data !

Explanation:

As I learned from quizlet!. there your welcome

Answer:

B. Quantity

Explanation:

This is easy

Answer: 13.9 m

Explanation:

Answer:

13.9m

Explanation:

Answer on Edge

Answer:

ask internet?

Explanation:

easy .....

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