If you push on a door for 12 seconds with a force of 53 N, how much impulse
have you applied to the door?
A. 815 kg•m/s
B. 332 kg•m/s
C. 636 kg•m/s
D. 4.42 kg•m/s

The complete question is;

In electronic circuits it is not unusual to encounter currents in the microampere range. Assume a 35 μA current, due to the flow of electrons. What is the average number of electrons per second that flow past a fixed reference cross section that is perpendicular to the direction of flow?

Answer:

2.185 × 10^(14) electrons/seconds

Explanation:

We are given current as 35 μA = 35 × 10^(-6) Amperes

The value of the charge on one electron is; e = 1.602 × 10^(-19) coulombs

Now, from conversions;

One ampere = one coulomb/second =

Then, 35 × 10^(-6) Ampere's would give;

35 × 10^(-6) Coulombs/sec

Now,

1.602 × 10^(-19) coulombs = 1 electron

Thus;

35 × 10^(-6) Coulombs/sec gives;

(35 × 10^(-6))/(1.602 × 10^(-19)) = 2.185 × 10^(14) electrons /sec

The average number of electrons per second that flow past a fixed reference cross-section perpendicular to the direction of flow is 2.185 × 10⁻¹⁴ electrons /sec

The question is as given below:

In electronic circuits, it is not unusual to encounter currents in the microampere range. Assume a 35 μA current, due to the flow of electrons. What is the average number of electrons per second that flow past a fixed reference cross-section that is perpendicular to the direction of flow?

Given information:

Current I = 35 μA = 35 × 10⁻⁶ A

charge on electron e = 1.602 × 10⁻¹⁹C

We know that:

I = de/dt, rate of flow of charge.

1A = 1C/s

Thus,

35 × 10⁻⁶ A = 35 × 10⁻⁶ C/s

Now, one electron has a charge of  1.602 × 10⁻¹⁹C

So, the number of electrons for a current of 35 × 10⁻⁶ C/s will be:

n = (35 × 10⁻⁶ C/s) ÷ (1.602 × 10⁻¹⁹C)

n = 2.185 × 10⁻¹⁴ electrons /sec

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The reactive force when a fish swims through water is the force of the water on the fish. This is an example of Newton’s third law of motion which states that for every action, there is an equal and opposite reaction. The active force is the fish pushing against the water, so the reactive force would be the reverse, the equal force of the water pushing back on the fish.

The pressure of the tank, when filled with water at a depth of 6 m, is approximately 580.124 atmospheres (atm). To calculate the pressure of the tank, one can use the equation: Pressure (P) = Density (ρ) × g × Depth (h)

Pressure (P) = Density (ρ) × g × Depth (h)

Given: Density of water (ρ) = 1000 kg/m³

Acceleration due to gravity (g) = 9.8 m/s²

Depth (h) = 6 m

Using the given values, one can calculate the pressure:

Pressure = 1000 kg/m³ × 9.8 m/s² × 6 m Pressure

= 58800 kg·m⁻¹·s⁻²

Now, let's convert the units to pascals (Pa) using the conversion 1 atm = 1.013 x \(10^5\) Pa:

Pressure = 58800 kg·m⁻¹·s⁻² × (1 atm / 1.013 x\(10^5\) Pa)

Pressure = 580.124 atm

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Answer:

proton  nuetron electron

Answer:

The tiger will land at 3.71 meters from the base of the rock

Explanation:

Horizontal Motion

When an object is thrown horizontally with a speed vo from a height h, the range or maximum horizontal distance traveled by the object can be calculated as follows:

\(\displaystyle d=v\cdot\sqrt{\frac {2h}{g}}\)

The tiger leaps from a height of h=7.5 m with a speed of v=3 m/s. Substituting the values into the formula:

\(\displaystyle d=3\cdot\sqrt{\frac {2\cdot 7.5}{9.8}}=3.71\ m\)

The tiger will land at 3.71 meters from the base of the rock

Answer:

E

Explanation:

Since the definition of density is mass per unit volume, the surface density has to do with the area of the surface.

If a piece of plastic is uniformly charged with surface charge density n1, that means that the charge density will be charged per unit area.

The magnitude of the charge depends on the surface areas.

As the surface areas reduce, the magnitude of the charge will be increasing.

Ranking in order, from largest to smallest, the surface charge densities n1 to n3, the correct option will be option E which is n3>n2>n1

A wave period 10ms travels through a medium. The graph shows the variation of particle displacement with distance for the wave. The average speed of a particle in the medium during one cycle is (v)= 20 m/s.

The velocity is a physics term that means a matter that covers a distance in a particular time. its a vector quantity. It can be measured in m/s or cm/s.

To calculate the average velocity we are using the formula is,

v=d/t

Here we are given,

According to the picture the wave start from +4.0 displacement and goes decreasing at -4.0 and then comes back at +4.0 and completes the full cycle and complete the amount of distance is (d)= 20cm = 0.2m

The period of the wave is (t)= 10ms = 10*10⁻³ s

We have to calculate the  average speed of a particle in the medium during one cycle= v m/s.

Now we put the values in the above equation we get,

v=d/t

Or, v= 0.2/ 10*10⁻³

Or, v= 20 m/s

So from the calculation we can say that, The average speed of a particle in the medium during one cycle is (v)=20 m/s.

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Answer:

Distance = πr

Displacement = 2r

Explanation:

First we need to find the distance covered by the car. As the car is travelling on a circular path and it traveled to a diametrically opposite point on the circular path. Therefore, the distance covered by the car must be the half value of the circumference.

Distance = Circumference/2

Distance = 2πr/2

Distance = πr

Since, displacement is the straight line distance between two points. So, the displacement in moving from a point to its diametrically opposite point must be equal to the diameter of circle:

Displacement = Diameter

Displacement = 2r

Answer:

organizing

Explanation:

thinking about where things should go, to make it so that everybody has an easy experience when inside of my store.

Chun's spring exercise would be an anaerobic process which is the correct answer would be an option (A).

The rate-limiting phases in anaerobic processes are the hydrolysis conversion rate and the rate at which methanogenesis uses soluble substrates. The overall amount of solids converted is affected by the hydrolysis of colloidal and solid particles, although neither the process operation nor stability is affected.

Anaerobic means "without oxygen," so this sort of exercise causes the body to break down glucose without the use of oxygen.

At the end of his 100-meter sprint, Chun's muscles are burning.

So Chun's spring workout would involve anaerobic activity.

Therefore, Chun's spring exercise would be an anaerobic process.

Hence, the correct answer would be an option (A).

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Answer:

Option A

Explanation:

Answer:

C

Explanation:

gravity is a pulling force according to Newton

To determine the image characteristics using the 3 principal rays and SALTS (Size, Attitude, Location, Type), we'll consider both lenses and mirrors separately. Here's how you can analyze each case:

Lenses:

Place an object at different distances to the left of a lens with a focal length (f).

a) Object placed beyond 2f:

In this case, the object is placed far beyond twice the focal length of the lens.

Principal ray 1: A ray parallel to the principal axis will pass through the focal point on the opposite side.

Principal ray 2: A ray passing through the optical center will continue in a straight line without any deviation.

Principal ray 3: A ray passing through the focal point on the object side will emerge parallel to the principal axis.

The image will be formed on the opposite side of the lens, between the focal point and twice the focal length.

SALTS:

Size: The image will be smaller than the object.

Attitude: The image will be inverted.

Location: The image will be located between the focal point and twice the focal length.

Type: The image will be real.

b) Object placed at 2f:

In this case, the object is placed at twice the focal length of the lens.

Principal ray 1: A ray parallel to the principal axis will pass through the focal point on the opposite side.

Principal ray 2: A ray passing through the optical center will continue in a straight line without any deviation.

Principal ray 3: A ray passing through the focal point on the object side will emerge parallel to the principal axis.

The image will be formed on the opposite side of the lens at twice the focal length.

SALTS:

Size: The image will be the same size as the object.

Attitude: The image will be inverted.

Location: The image will be located at twice the focal length.

Type: The image will be real.

c) Object placed between f and 2f:

In this case, the object is placed between the focal point and twice the focal length of the lens.

In this case, the object is placed far beyond twice the focal length of the mirror.

Principal ray 1: A ray parallel to the principal axis will reflect through the focal point on the same side.

Principal ray 2: A ray passing through the focal point on the object side will reflect parallel to the principal axis.

Principal ray 3: A ray passing through the center of curvature will reflect back along the same path.

The image will be formed on the opposite side of the mirror, between the focal point and twice the focal length.

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The angle at which the light refracts when it moves from air into the crown glass is 23.3°.

Refraction is the bending of light as it passes through a medium with a different density. It occurs due to a change in the speed of light as it enters a different medium.

We can use Snell's law to solve this problem, which states that the ratio of the sines of the angles of incidence and refraction is equal to the ratio of the refractive indices of the two media:

n1 sin θ1 = n2 sin θ2

where n1 and n2 are the refractive indices of the initial and final media, and θ1 and θ2 are the angles of incidence and refraction, respectively.

From the problem, we have:

n1 = 1.000293 (air)

n2 = 1.520 (crown glass)

θ1 = 25°

We can rearrange Snell's law to solve for θ2:

sin θ2 = (n1/n2) sin θ1

sin θ2 = (1.000293/1.520) sin 25°

sin θ2 = 0.385

Taking the inverse sine of both sides, we get:

θ2 = 23.3°

Therefore, the angle at which the light refracts when it moves from air into the crown glass is 23.3°.

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The Free Body Diagram must include all forces acting on an object, not just weight and normal force. For example,The friction force between a box and a ramp must be considered, directed up the ramp and opposite to the weight force, to accurately analyze equilibrium and predict behavior.

1. The Free Body Diagram only shows two forces acting on the box - the weight force and the normal force from the ramp.

2. However, there are other forces that should be included in the diagram, such as the force of friction between the box and the ramp.

3. The force of friction would be directed up the ramp, opposite to the direction the box would slide if it were not stationary.

4. Therefore, the Free Body Diagram should include an arrow representing the force of friction, pointing up the ramp and opposite to the direction of the weight force.

5. Including all the forces in the Free Body Diagram is important to accurately analyze the equilibrium of the box and predict its behavior.

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Answer:

24km/h

Explanation:

go it right on ingenuity 2020

Answer:

The number of electrons entering the thin wire every second is 1.75 x 10⁻³ mobile electrons / second

Explanation:

Given;

emf of the battery, V = 1.5 V

electron density, = 9 × 10²⁸ mobile electrons per m³

mobility of electron, u = 7 × 10⁻⁵ (m/s)/(N/C)

length of thin wire, L = 6 cm = 0.06 m

cross sectional area of the thin wire, A = 1.3 x 10⁻⁸ m²

The magnitude of the electric field in the thin wire is given by;

E = V/L

E = (1.5) / (0.06)

E = 25 N/C

the number of electrons entering the thin wire every second is given by;

\(e/s = mobility \ x \ Electric \ field\\\\number \ of \ electrons \ per \ second =\frac{7*10^{-5} (m/s)}{N/C} *25 (N/C)\\\\number \ of \ electrons \ per \ second = 1.75*10^{-3} \ m/s\)

Therefore, the number of electrons entering the thin wire every second is 1.75 x 10⁻³ mobile electrons / second

The number of electrons entering the thin wire every second is 1.75 x 10⁻³ mobile electrons / second

Since

emf of the battery, V = 1.5 V

electron density, = 9 × 10²⁸ mobile electrons per m³

mobility of electron, u = 7 × 10⁻⁵ (m/s)/(N/C)

length of thin wire, L = 6 cm = 0.06 m

cross sectional area of the thin wire, A = 1.3 x 10⁻⁸ m²

So here the magnitude should be

E = V/L

E = (1.5) / (0.06)

E = 25 N/C

Now the number of electrons should be

= 7 × 10⁻⁵  *25

= 1.75 x 10⁻³ mobile

hence, The number of electrons entering the thin wire every second is 1.75 x 10⁻³ mobile electrons / second

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Answer:

1. some ways I might first observe gravity are... when I first arrive on earth I will see everything staying in one place on the ground and nothing will be floating

everywhere.

2. I might test gravitional force by jumping and seeing how fast I come down or by taking something and dropping it from above my head to the ground to see that it falls and doesn't float.

3.I woukd describe gravity to other aliens back at my home planet by #1 telling them that nothing floats, everything touches the ground and if u drop something from high above you will instantly/immediately fall back down to the ground.

You can say exactly this if you want bc I answered like how if u were gonna answer the question so your welcome

Answer:

1. some ways I might first observe gravity are... when I first arrive on earth I will see everything staying in one place on the ground and nothing will be floating

everywhere.

2. I might test gravitional force by jumping and seeing how fast I come down or by taking something and dropping it from above my head to the ground to see that it falls and doesn't float.

3.I woukd describe gravity to other aliens back at my home planet by #1 telling them that nothing floats, everything touches the ground and if u drop something from high above you will instantly/immediately fall back down to the ground.

Explanation:

You are riding your bike at 5 m/s. In order to avoid hitting a child that has run into the street, you slam on your brakes and slow to a stop. It takes you  1.5 seconds to fully stop. Acceleration will be  3.33 \(m/s^{2}\)

The branch of physics that defines motion with respect to space and time, ignoring the cause of that motion, is known as kinematics. Kinematics equations are a set of equations that can derive an unknown aspect of a body’s motion if the other aspects are provided.

given

speed = 5 m/s

time = 1.5 seconds

using

v = u + at

0 = 5 + a ( 1.5 )

a = -5 / 1.5

 = 3.33 \(m/s^{2}\)

Acceleration will be  3.33 \(m/s^{2}\)

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Answer:

In the given circuit, R

2

,R

6

and R

4

are in series. So,

R

1

′

=7+5+12=24Ω

Now R

1

′

and R

5

′

are in parallel. So,

R

2

′

1

=

8

1

+

24

1

=

24

3+1

=

24

4

=

6

1

R

2

′

=6ohm.

Now R

2

′

,R

1

and R

3

are in series. So,

R=R

2

′

+R

1

+R

3

=6+3+2=11ohm.

We know i=

R+r

E

=

11+1

6

=

12

6

=

2

1

i=0.5amp.

Answer:

Overall charge still remains zero on conductor until touched by charged rod.

Explanation:

Here, we want to know what has happened to the overall charge on the conductor.

Since the conductor is neutral, the overall charge on the conductor must remain zero because positive charge is induced on close end to rod then equal and negaitve charge is induced on far end to rod.

Thus, overall charge still remains zero on conductor until touched by charged rod.

Answer:

A. 200m/s

Explanation:

Using the law of conservation of momentum expressed as;

m1u1 + m2u2 = (m1+m2)v

m1 and m2 are the masses of the object

u1 and u2 are the respective velocities

v is the common velocity

Given

m1 = 1.2kg

u1 = 0m/s (block is a stationary object)

m2 = 50g= 0.05kg

u2 = ?

v = 8.0m/s

Substitute the values into the formula and get u2 (speed of the bullet before hitting the block)

1.2(0)+0.05u2 = (1.2 + 0.05)(8)

0.05u2 = 1.25(8)

0.05u2 = 10

u2 = 10/0.05

u2 = 200m/s

Hence the speed of the bullet before it hit the block is 200m/s

Given data:

* The distance traveled in east in time 0.5 h is 9.9 km

* The distance traveled in east of north in time 0.77 h is 9 km at 15 degree.

* The distance traveled in east in 0.5 h is 9.9 km.

Solution:

The diagramatic representation of the give system is

The net distance traveled in east is,

where 9 cos( 15 degree ) terms is representing the component of 9 distance along the east direction.

Solving the value of total distance in east is,

The net distance traveled in the north direction is,

The total time taken to complete the trip is,

The average velocity of bicycle along the east is,

Substituting the known values,

The average velocity of the bicycle along the north direction is,

Thus, the magnitude of the average velocity is,

The direction of the average velocity is,

Hence, the magnitude of the average velocity is 16.15 km per hour and direction of average velocity is at 4.69 degree east of north.

The mass of the planet is  5.98 × 10^24 kg.

To calculate the mass of the planet, we can use Kepler's Third Law of Planetary Motion. This law states that the square of the period of revolution of a planet around the sun is directly proportional to the cube of the semi-major axis of its orbit.

First, we need to convert the period of the satellite's orbit to seconds. We know that there are 60 minutes in an hour, so the period can be expressed as (6 × 60 + 12) minutes, which equals 372 minutes. Multiplying this by 60 seconds, we get a period of 22,320 seconds.

Next, we need to find the semi-major axis of the orbit. In a circular orbit, the semi-major axis is equal to the radius of the orbit. Therefore, the semi-major axis is 8.8 × 10^6 m.

Now, we can apply Kepler's Third Law to calculate the mass of the planet. The formula is T^2 = (4π^2/GM) × a^3, where T is the period of revolution, G is the gravitational constant, M is the mass of the planet, and a is the semi-major axis of the orbit.

Rearranging the formula, we can solve for the mass of the planet:
M = (4π^2/G) × a^3 / T^2

Plugging in the values, we get:
M = (4 × π^2 / 6.67430 × 10^-11) × (8.8 × 10^6)^3 / (22,320)^2

Evaluating this expression, we find that the mass of the planet is approximately 5.98 × 10^24 kg.

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1) The starting temperature of the gas is 288 K

b) When the pressure is doubled, the temperature is 576 K

We know that we have to use the ideal gas equation so as to be able to obtain the temperature of the starting of the gas and we know that;

Pressure = 200 kPa or 1.97 atm

Volume =  3000 cm3  or 3 L

Temperature = ?

Number of moles = mass/molar mass = 7 g/28 g/mol = 0.25 moles

Then we have;

PV = nRT

T = PV/nR

T =  1.97 * 3/0.25 * 0.082

T = 5.91/0.0205

T = 288 K

If the pressure doubles and we have;

P1/T1 = P2/T2

P1T2 = P2T1

T2 = 2(1.97) * 288/1.97

T2 = 576 K

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Answer:

neutral

Explanation:

ization

Answer:

A neutralization reaction is when an acid and a base react to form water and a salt and involves the combination of H+ ions and OH- ions to generate water. The neutralization of a strong acid and strong base has a pH equal to 7.

Explanation:

Answer:

43,750 s

Explanation:

Power= work/time

125=350/time

125 x 350 = 43,750 s