Which best describes velocity?

A. 15 miles an hour
B. 15 miles an hour going west.
C. Speeding up from 10 to 15 miles an hour.
D. Going 15 miles an hour for 2 hours.

Answer:

below

Explanation:

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:

After 15 seconds, the green car will catch up with the blue car

Explanation:

Let the time for the green car to catch up with the blue car be T

When the green car catches up to the blue car, the distances covered by each car after time T will be equal. Also, their velocities at that instant will be equal

Distance covered by blue car after time T is given by: s = ut + 0.5 at²

Where u = 0, a = 0.2 m/s², t = T

S = 0.5 × 0.2 × T² = 0.1 T²

Velocity of blue car, v = u+ at

v = 0.2T

Distance covered by green car at T is given as: S = Velocity × time

Where v = 0.2T, t = T - 7.5 (since the blue car started 7.5 seconds earlier)

S = 0.2T (T - 7.5)

S = 0.2 T² - 1.5T

Equating the distance covered by the two cars

0.2T² - 1.5T = 0.1T²

0.1T² - 1.5T = 0

T(0.1T - 1.5) = 0

T = 0 or

T = 1.5/0.1 = 15 secs

Therefore, after 15 seconds, the green car will catch up with the blue car

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

(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

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

OUM I THINK IS WIND

CORRECT ME IF IM WRONG

Answer:

13 behind o2

Explanation:

answer is in photo above

Answer:

12

Explanation:

Answer:

First of all the formula is F= uR,( force= static friction× reaction)

mass= 5+25=30

F= 50

R= mg(30×10)=300

u= ?

F=UR

u= F/R

u= 50/300=0.17N

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:

Explanation:

Distance=10km

Time=30min=0.5hr

So,speed=10/0.5

So,speed=20km/hr

Answer:

(a)Magnitude=28.81 m/s

Direction=33.3 degree below the horizontal

(b) No, it is not perfectly elastic collision

Explanation:

We are given that

Mass of stone, M=0.150 kg

Mass of bullet, m=9.50 g=[tex]9.50\times 10^{3} kg[/tex]

Initial speed of bullet, u=380 m/s

Initial speed of stone, U=0

Final speed of bullet, v=250m/s

a. We have to find the magnitude and direction of the velocity of the stone after it is struck.

Using conservation of momentum

[tex]mu+ MU=mv+ MV[/tex]

Substitute the values

[tex]9.5\times 10^{-3}\times 380 i+0.150(0)=9.5\times 10^{-3} (250)j+0.150V[/tex]

[tex]3.61i=2.375j+0.150V[/tex]

[tex]3.61 i-2.375j=0.150V[/tex]

[tex]V=\frac{1}{0.150}(3.61 i-2.375j)[/tex]

[tex]V=24.07i-15.83j[/tex]

Magnitude of velocity of stone

=[tex]\sqrt{(24.07)^2+(-15.83)^2}[/tex]

|V|=28.81 m/s

Hence, the magnitude and direction of the velocity of the stone after it is struck, |V|=28.81 m/s

Direction

[tex]\theta=tan^{-1}(\frac{y}{x})[/tex]

=[tex]tan^{-1}(\frac{-15.83}{24.07})[/tex]

[tex]\theta=tan^{-1}(-0.657)[/tex]

=33.3 degree below the horizontal

(b)

Initial kinetic energy

[tex]K_i=\frac{1}{2}mu^2+0=\frac{1}{2}(9.5\times 10^{-3})(380)^2[/tex]

[tex]K_i=685.9 J[/tex]

Final kinetic energy

[tex]K_f=\frac{1}{2}mv^2+\frac{1}{2}MV^2[/tex]

=[tex]\frac{1}{2}(9.5\times 10^{-3})(250)^2+\frac{1}{2}(0.150)(28.81)^2[/tex]

[tex]K_f=359.12 J[/tex]

Initial kinetic energy is not equal to final kinetic energy. Hence, the collision is not perfectly elastic collision.

Answer:

the correct answer is C   v = 60 cm / s

Explanation:

The speed of a wave is related to the frequency and the wavelength

         v = λ f

They indicate that the object performs 20 oscillations every second, this is the frequency

         f = 20 Hz

the wavelength is the distance until the wave repeats, the distance between two consecutive peaks corresponds to the wavelength

         λ = 3 cm = 0.03 m

let's calculate

       v = 20 0.03

       v = 0.6 m / s

       v = 60 cm / s

the correct answer is C

Answer:

I dnt know that language

Explanation:

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.

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]

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]

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:

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:

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

Explanation:

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:

9.6352× 10²⁰ C atoms

Explanation:

From the given information,

The molar mass of Carbon = 12 g/mol

number of moles = 0.020g/ 12 g/mol

number of moles = 0.0016 mol

If 1 mole of C = 6.022 × 10²³ C atoms

∴

0.0016 mol of C = (6.022 × 10²³ C atoms/ 1 mol of C)×0.0016 mol of C

= 9.6352× 10²⁰ C atoms

Hence, the number of carbon atoms present in 0.020 g of carbon = 9.6352× 10²⁰ C atoms

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:

13.2

Explanation:

I would say this cause this is a reasonable answer

The length of cotton is wound closely around a pen, it goes around six times. The distance once around the pen is 13.2 cm. The correct option is c.

Distance is the sum of an object's movements, regardless of direction. The distance can be defined as the amount of space an object has covered, regardless of its starting or ending position.

Measuring units are units that are used to measure the magnitude, amount, or quantity of any object. There are 7 basic measuring units that are used in worldwide and everyday life.

Given, the piece of cotton is tied around the pen, it goes around six times around the pen. The diameter of the pen will the length of the cotton. The actual length can be for six rounds is 13.2 cm.

Therefore, the correct option is с. 13.2 cm.

To learn more about length, refer to the link:

https://brainly.com/question/16188698

#SPJ2

Answer:

if we want to find force by using newton's law equation ( f = ma ) we have to use mass and acceleration not velocity ,but in this question they did not mention about acceleration but speed so the answer is D

Answer:

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

Explanation:

Is this what you want?

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.