____________is obtained from the fleece of animals.​

Answer: see below explanation, should be straight forward from there? ;)

Explanation: 1 watt = 1 joule per second

Watt is a measure of energy over time

So 10 seconds... u got this :)

Answer:

S = 1/2 gt² = 1/2 × 9.8 × 3² = 4.9×9 = 44.1 m

Explanation:

Answer:

9.965 nF

Explanation:

The capacitance of the axon C = εA/d where ε = dielectric constant = 24.78 × 10⁻¹² F/m, A = surface area of axon = 2πrL where r = radius of axon = 8 μm = 8 × 10⁻⁶ m and L = length of axon = 8 cm = 8 × 10⁻² m and d = thickness of membrane = 0.01 μm = 0.01 × 10⁻⁶ m

So, C = εA/d

C = ε2πrL/d

Substituting the of the values variables into the equation, we have

C = ε2πrL/d

C =  24.78 × 10⁻¹² F/m × 2π × 8 × 10⁻⁶ m × 8 × 10⁻² m/0.01 × 10⁻⁶ m

C =  9964.63 × 10⁻²⁰ Fm/0.01 × 10⁻⁶ m

C =  996463 × 10⁻¹⁴ F

C = 9.96463 × 10⁻⁹ F

C = 9.96463 nF

C ≅ 9.965 nF

Answer:

below

Explanation:

Explanation:

In the parallel combination, the equivalent resistance is given by :

[tex]\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+....[/tex]

4. When three 150 ohms resistors are connected in parallel, the equivalent is given by :

[tex]\dfrac{1}{R}=\dfrac{1}{150}+\dfrac{1}{150}+\dfrac{1}{150}\\\\R=50\ \Omega[/tex]

5. Three resistors of 20 ohms, 40 ohms and 100 ohms are connected in parallel, So,

[tex]\dfrac{1}{R}=\dfrac{1}{20}+\dfrac{1}{40}+\dfrac{1}{100}\\\\=11.76\ \Omega[/tex]

Hence, this is the required solution.

Answer:

52.006 Kilograms

.............................................

The mass of an object that experience a gravitional force of 510 N near earths surface in 52.0 kg

Answer:

a)  T = 2838.6 N,  b)   W = 1003.2 J,  c) W = 6.22 10⁴ J,  d) W = 2.79 10³ J

e) v_f = 2.65  m / s

Explanation:

a) To find the tension of the cable let's use Newton's second law

        T - W = m a

         T = W + ma

        T = m (g + a)

let's calculate

        T = 285 (-9.8 - 0.160)

        T = 2838.6 N

b) net work is stress work minus weight work

        W = F d

        W = (T-W) d

        W = (m a) d

        W = (285 0.160) 22

        W = 1003.2 J

c) the work done by the cable

         W = T d cos 0

          W = 2838.6 22.0

          W = 6.22 10⁴ J

d) The work done by the weight

the displacement is upwards and the weight points downwards, so the angle is 180º

        W = F. d

         W = F d cos 180

         W = -285 22.0

         W = 2.79 10³ J

e) the final speed of the load. Let's use the relationship between work and the change in kinetic energy

         W = ΔK

as part of rest K₀ = 0

          W = ½ m v_f²

          v_f = [tex]\sqrt{ \frac{2W}{m} }[/tex]

          v_f = [tex]\sqrt{\frac{2 \ 1003.2}{285} }[/tex]

          v_f = 2.65  m / s

Solution :

 Given :

Angle q = angle between the thread supporting the ball with the vertical.

Let mass [tex]$m_1 >>>m_2$[/tex].

Then [tex]$m_1+m_2=m_1$[/tex]

In this case, acceleration can be found out by applying Newton's law of motion.

Thus,

Acceleration, [tex]$a=\frac{m_1}{m_1+m_2}. g$[/tex]

                      [tex]$a=\frac{m_1}{m_1}. g$[/tex]

                       [tex]$a=g$[/tex]

Therefore, [tex]$\tan \theta =\frac{a}{g}$[/tex]

or                [tex]$\tan \theta =\frac{a}{a}$[/tex]

or                [tex]$\tan \theta =1$[/tex]

                   [tex]$\theta = \tan ^{-1}(1)$[/tex]

                   [tex]$\theta = 45^\circ$[/tex]

Therefore the largest angle q is  [tex]$\theta = 45^\circ$[/tex]

Complete question is;

Use a variation model to solve for the unknown value.

The stopping distance of a car is directly proportional to the square of the speed of the car.

a. If a car traveling 50 mph has a stopping distance of 170 ft, find the stopping distance of a car that is traveling 70 mph.

b. If it takes 244.8 ft for a car to stop, how fast was it traveling before the brakes were applied?

Answer:

A) d = 333.2 ft

B) 60 mph

Explanation:

Let the stopping distance be d

Let the speed of the car be v

We are told that the stopping distance is directly proportional to the square of the speed of the car. Thus;

d ∝ v²

Therefore, d = kv²

Where k is constant of variation.

A) Speed is 50 mph and stopping distance of 170 ft.

v = 50 mph

d = 170 ft = 0.032197 miles

Thus,from d = kv², we have;

0.032197 = k(50²)

0.032197 = 2500k

k = 0.032197/2500

k = 0.0000128788

If the car is now travelling at 70 mph, then;

d = 0.0000128788 × 70²

d = 0.06310612 miles

Converting to ft gives;

d = 333.2 ft

B) stopping distance is now 244.8 ft

Converting to miles = 0.046363636 miles

Thus from d = kv², we have;

0.046363636 = 0.0000128788(v²)

v² = 0.046363636/0.0000128788

v² = 3599.99658

v = √3599.99658

v ≈ 60 mph

Answer:

x = 132.7 m

y = 51.34 m

Explanation:

Given :

Initial speed, u = 49.5 m/s²

Angle of projection, θ = 40°

Time, t = 3.50 seconds

The distance, x = horizontal component ;

Distance = speed * time

Distance = uCosθ * 3.50

Distance = 49.5 * Cos40° * 3.50

Distance = 49.5 * Cos40° * 3.50

Horizontal distance = 132.7 m

Vertical distance, y :

Sy = ut + 1/2gt²

Sy = Vertical distance ; g = 9.8 m/s²

Sy = 49.5 * sin40 * 3.5 - (0.5 * 9.8 * 3.5²)

Sy = 111.36295 - 60.025

Sy = 51.33795 m

x = 132.7 m

y = 51.34 m

Answer:

[tex]T_2=39.5N[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=5kg[/tex]

Length [tex]L=0.67m[/tex]

Tension [tex]T=88N[/tex]

Generally the equation for Tension is mathematically given by

 [tex]T = m * ( g + v^2 /l)[/tex]

Therefore

 [tex]T_1 = m * ( g + \frac{v^2}{l})[/tex]

 [tex]88 = 5 * ( 9.8 + \frac{v^2}{0.67})[/tex]

 [tex]v^2=5.2[/tex]

 [tex]v=2.4m/s[/tex]

The uniform velocity is

 [tex]v=2.4m/s[/tex]

Therefore

The tension at the side of the circle halfway between the top and bottom is

 [tex]T_2=5*\frac{2.3^2}{0.67}[/tex]

 [tex]T_2=39.5N[/tex]

Answer:

Distance = speed * time

55*5

275 meters.

The train would have covered a distance of 275 m

We can define distance as to how much ground an object has covered despite its starting or ending point.

Distance = speed * time

given

speed= 55 m/s

time = 5 sec

Distance = 55 * 5 = 275 m

The train would have covered a distance of 275 m

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Answer:

Work done applied = 12 newton-meter

Explanation:

Given examples:

Force applied = 6 newton

Distance of book = 2 meter

Find from the given data:

Work done

Computation:

The equation can be used to compute work.

Work done applied = Force applied x Distance of book

Work done applied = Force x Distance

Work done applied = 6 x 2

Work done applied = 12 newton-meter

OUM I THINK IS WIND

CORRECT ME IF IM WRONG

Answer:

cm/s

6

128.2

96.0

7

145.8

Table B

(a) standard deviation = σ = 4.9996

(b) variance = σ² = 24.996

Given frequency table (find attached as Table 1);

(a) To find the sample standard deviation and sample variance, follow these steps;

i. Calculate the mid-point c for each group by using the mid-point formula;

c = (lower bound + upper bound) / 2

=> c = (6.51 + 8.50) / 2 = 7.505

=> c = (8.51 + 10.50) / 2 = 9.505

=> c = (10.51 + 12.50) / 2 = 11.505

=> c = (12.51 + 14.50) / 2 = 13.505

=> c = (14.51 + 16.50) / 2 = 15.505

So the new table becomes (find attached as Table 2);

ii. Calculate the total number of samples (n) which is the sum of all the frequencies.

n = 50+18+42+20+46

n = 176

iii. Calculate the mean (M)

This is done by first multiplying the midpoints by the corresponding frequencies and then dividing the result by the total number of samples (n).

M = [(7.505 x 50) + (9.505 x 18) + (11.505 x 42) + (13.505 x 20) + (15.505 x 46)] / 176

M = [375.25 + 171.09 + 483.21 + 270.1 + 713.23] / 176

M = [2012.88] / 176

M = 11.44

iv. Find the variance (σ²);

The variance is calculated using the following formula

σ² = [Σ(f x c²) - (n x M²)] / (n - 1)                ------------(i)

Where;

f = frequency of each boundary data point

=>  Let's first calculate Σ(f x c²).

This is done by finding the sum of the product of the frequency (f) of each boundary point and the square of their corresponding mid-points(c)

Σ(f x c²) = [(50 x 7.505²) + (18 x 9.505²) + (42 x 11.505²) + (20 x 13.505²) + (46 x 15.505²)]

Σ(f x c²) = [(2816.25125) + (1626.21045) + (5559.33105) + (3647.7005) + (11058.63115)]

Σ(f x c²) = 24708.1244

=> Now calculate (n x M²)

n x M² = 176 x 11.44²

n x M² = 23033.7536

=> Now substitute these values into equation (i) to calculate the variance

σ² = [Σ(f x c²) - (n x M²)] / (n - 1)

σ² = [24708.1244 - 23033.7536] / (176 - 1)

σ² = [4374.3708] / (175)

σ² = 24.996

Therefore, the variance is 24.996

v. Find the standard deviation (σ)

The standard deviation is the square root of the variance. i.e

σ = √σ²

σ = √24.996

σ = 4.9996

Therefore, the standard deviation is 4.9996

Answer:

1. 10³ m²

2. 32.4 ml

3. 1.91 × 10²¹ molecules

Explanation:

Here is the complete question

1. Estimate the size of a one-molecule-thick oil film formed by spreading 1 ml of oil on the surface of the water. Assume that an oil molecule is roughly 10 nm in size. Show your work.

2. If we will be conducting the experiments in tanks with dimensions of roughly 10 in by 10 in. Estimate the volume of oil you would need to cover half the total area. Again assume a molecule is 10 nm in size. Show your work.

3. Estimate the number of molecules in 1 ml of oil assuming they’re 10 nm in size. What assumptions do you have to make? Show your work.

Solution

1. Since the film would cover an area, A, and would have a height which is the thickness of the molecule, h = 10 nm = 1 × 10⁻⁹ m, its volume is V = Ah. This volume also equals the volume of the oil. Since the volume of oil is 1 ml = 1 × 10⁻⁶ m³, V = 1 × 10⁻⁶ m³.

The size of the oil drop is its area. So, A = V/h

= 1 × 10⁻⁶ m³ ÷ 1 × 10⁻⁹ m

= 10³ m²

2. The area of the tank is 10 in by 10 in = 100 in². Since we want to cover half the area, we require 100 in²/2 = 50 in² = 50 in² × (0.0254)² m²/in² = 0.0324 m².

If the thickness of oil is one molecule thick which is 10nm = 1 × 10⁻⁹ m, the volume of oil is then thickness × area = 1 × 10⁻⁹ m × 0.0324 m²

= 0.0324 × 10⁻⁹ m³

= 32.4 × 10⁻⁶ m³

= 32.4 ml since 1 × 10⁻⁶ m³ = 1 ml

3. Since the volume of oil is 1 ml = 1 × 10⁻⁶ m³, we need to find the volume of one molecule. Since it is assumed to be a sphere, its volume is V' = πd³/6 where d = size of oil molecule = 10 nm = 1 × 10⁻⁹ m.

Let n be the number of molecules present in 1 ml, then nV' = 1 ml = 1 × 10⁻⁶ m³. So, n = 1 × 10⁻⁶ m³/V' = 1 × 10⁻⁶ m³ ÷ πd³/6 = 6 × 10⁻⁶ m³/πd³

Substituting d into the equation, we have

n = 6 × 10⁻⁶ m³/π(1 × 10⁻⁹ m)³

n = 6 × 10⁻⁶ m³/π × 10⁻²⁷ m³

n = 1.91 × 10²¹ molecules

Answer:

(a), (c) and (e) s correct.

Explanation:

a. the power used by a circuit is the resistance times the current squared.

The power is given by P = I^2 R, so the statement is correct.  

b. electric and magnetic fields are transporting the energy.

false

c. electrons are transporting the energy.

The energy is transferred by flow of electrons. It is correct.  

d. the power used by a circuit is the voltage times the current squared.

The power is given by P  = V I, the statement is wrong.  

e. the power used by a circuit is the current times the voltage.

The power is given by P  = V I, the statement is correct.  

Answer:

13 behind o2

Explanation:

answer is in photo above

Answer:

12

Explanation:

Answer:

A wave model of light is useful for explaining brightness,color, and the frequency-dependent bending of light at a surface between media. However, because light can travel through space, it cannot be a matter wave, like sound or water waves.

Answer:

I dnt know that language

Explanation:

Answer:

a) the cheetah's top speed is 64.4 miles/hr

b) time taken to reach top speed is 3.3 seconds

Explanation:

Given the data in the question;

Cheetahs can accelerate to a speed of 21.8 m/s in 2.50 s.

They can continue to accelerate to reach a top speed of 28.8 m/s.

a) Express the cheetah's top speed in mi/h. (Express your answer to three significant figures.)

The cheetah's top speed = 28.8 m/s = ( 28.8 × 2.237 ) miles/hr

= 64.4256 ≈ 64.4 miles/hr

Therefore, the cheetah's top speed is 64.4 miles/hr

b) Starting from a crouched position, how long does it take a cheetah to reach its top speed.

given that

v₁ = 21.8 m/s   and    t₁ = 2.50 s

let a represent the acceleration of the cheetah

From the First Equation of Motion;:

v = u + at

we substitute

21.8 = 0 + ( a × 2.50 )

21.8 = a × 2.50

a = 21.8 / 2.50

a = 8.72 m/s²

Now, let the time taken by cheetah to reach top speed ( 28.8 m/s ) be t

so from the first equation of motion;

v = u + at

we substitute

28.8 = 0 + ( 8.72 × t )

t = 28.8 / 8.72

t = 3.3 seconds

Therefore, time taken to reach top speed is 3.3 seconds

Answer:

3 ohm

Explanation:

Given :

V=9V

And according to given question same current is flowing in both resistance that means resistance will connected in series

So,

R= R+R=2R

Now,

Applying ohm's law

[tex]V=IR\\9=1.5*2R\\9=3R\\R=\frac{9}{3} \\R= 3ohm[/tex]

Therefore, answer is 3 ohm

Answer:

d. correctly described by all the statements above.

Explanation:

Kinetic molecular theory of gases states that gas particles exhibit a perfectly elastic collision and are constantly in motion.

According to the kinetic-molecular theory, the average kinetic energy of gas particles depends on temperature.

This ultimately implies that, the average kinetic energy of gas particles is directly proportional to the absolute temperature of an ideal gas. Thus, an increase in the average kinetic energy of gas particles would cause an increase in the absolute temperature of an ideal gas.

Temperature can be defined as a measure of the degree of coldness or hotness of a physical object. It is measured with a thermometer and its units are Celsius (°C), Kelvin (K) and Fahrenheit (°F).

Generally, the temperature of a quantity of an ideal gas is;

a. a measure of the ability of an ideal gas to transfer thermal energy to another body.

b. the average kinetic energy of gas particles is directly proportional to the absolute temperature of an ideal gas

c. proportional to the internal energy of the gas.

Hey again!

Ok..

Now... The melting Point of this solid is 90°C.

Meaning That as soon as it gets to this temp... It STARTS Melting.

So at that temp... It still has some solid parts in it.

You can say its a Solid Liquid Mixture.

Additional Heat being applied at that point is not raising the temperature;rather its used in breaking the bonds in the solid. This is the Fusion stage.

After Fusion...It'd then Be a Pure Liquid with no solids in it.

So

Q'=MC∆0----- This is the heat needed to take the solid's temp from 30°c - 90°c

Q"=ml ----- This is the heat used in breaking the bonds holding the solids in the solid-liquid phase.

So

Q= Q' + Q"

Q= mc∆0 + ml

∆0 = 90°c - 30°c = 60°c

Q= 2.5(390)(60) + (2.5)(4000)

Q=6.9 x 10⁴Joules

The heat required to change 2.5 kg of the solid at 30.0C to a liquid is 6.9 x 10⁴J.

The specific heat is the amount of heat energy required to change the temperature of 1kg of object by 1°C.

The heat needed to change the solid's temperature from 30°C - 90°C is

Q' = mC∆T

The heat used to change the phase solid-liquid phase .i.e.

Q'' =mL where L =latent heat of fusion

The total heat required is

Q= Q' + Q"

Q= mc∆T + ml

Q= 2.5(390)(90 - 30) + (2.5)(4000)

Q=6.9 x 10⁴Joules

Thus, the heat required to change the solid to liquid is 6.9 x 10⁴J.

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Answer:

Ohm's law is not applicable to semi-conductors and insulators.

Explanation:

Is this what you want?

RQ/PQ I think

rise/run