5.Which of the following does not affect rate of evaporation?
O Wind speed
O Surface area
O Temperature
O Insoluble heavy impurities

Answers

Answer 1
Insoluble heavy impurities
Answer 2

Answer:

D

Explanation:

Insoluble impurities would not change the constituent of the substance. Soluble would for example salt water takes longer time for the water to become vapour when subjected to the same temperature that normal water.

Wind would affect, the more windy the tendency for particles of the liquid to be moved into the atmosphere.

With an increase in surface area, the evaporation rate increase . Take a clue from water placed on the ground and exposed to the atmosphere and that same quantity of water is placed in a cup. That on the floor would evaporate faster.

Similarly the higher the temperature a substance is subjected to the easier is it's rate of evaporation. Take for instance water in a cup placed in the sun and that same placed in a room with mild temperatures than that of the sun.With time that in the sun decreases in volume faster than that in the room.


Related Questions

An infinite sheet carries a uniform, positive charge per unit area. The electric field produced by the sheet is represented by parallel lines drawn with a density N lines per m2 that are perpendicular to and away from the sheet. The charge per unit area on the sheet is doubled. How should the density of the electric field lines be changed

Answers

Complete Question

An infinite sheet carries a uniform, positive charge per unit area. The electric field produced by the sheet is represented by parallel lines drawn with a density N lines per m2 that are perpendicular to and away from the sheet. The charge per unit area on the sheet is doubled. How should the density of the electric field lines be changed?

A It should stay the same

B  It should be quadrupled.

C It should be quintupled

D It should be doubled.

E It should be tripled

Answer:

Option D is the correct option

Explanation:

Generally electric field is mathematically represented as

        [tex]E = \frac{\sigma}{\epsilon_o}[/tex]

Where [tex]\sigma[/tex] is the charge per unit area (Charge density )

From the question we are told that [tex]\sigma[/tex] is doubled hence the

     [tex]E = \frac{2 \sigma }{\epsilon_o}[/tex]    

Looking the equation above we see that the value of the electric field will also double given that it is directly proportional to the charge density

ASK YOUR TEACHER A meter stick is found to balance at the 49.7-cm mark when placed on a fulcrum. When a 51.5-gram mass is attached at the 16.0-cm mark, the fulcrum must be moved to the 39.2-cm mark for balance. What is the mass of the meter stick

Answers

Answer:

0.114 kg or 114 g

Explanation:

From the diagram attaches,

Taking the moment about the fulcrum,

sum of clockwise moment = sum of anticlockwise moment.

Wd = W'd'

Where W = weight of the mass, W' = weight of the meter rule, d = distance of the mass from the fulcrum, d' = distance of the meter rule.

make W'  the subject of the equation

W' = Wd/d'................ Equation 1

Given: W = mg = 0.0515(9.8) = 0.5047 N, d = (39.2-16) = 23.2 cm, d' = (49.7-39.2) = 10.5 cm

Substitute these values into equation 1

W' = 0.5047(23.2)/10.5

W' = 1.115 N.

But,

m' = W'/g

m' = 1.115/9.8

m' = 0.114 kg

m' = 114 g

Light bulb 1 operates with a filament temperature of 3000 K, whereas light bulb 2 has a filament temperature of 2000 K. Both filaments have the same emissivity, and both bulbs radiate the same power. Find the ratio A1/A2 of the filament areas of the bulbs.

Answers

Answer:

A₁/A₂ = 0.44

Explanation:

The emissive power of the bulb is given by the formula:

P = σεAT⁴

where,

P = Emissive Power

σ = Stefan-Boltzman constant

ε = Emissivity

A = Surface Area

T = Absolute Temperature of Surface

FOR BULB 1:

Since, emissivity and emissive power are constant.

Therefore,

P = σεA₁T₁⁴   ----------- equation 1

where,

A₁ = Surface Area of Bulb 1

T₁ = Temperature of Bulb 1 = 3000 k

FOR BULB 2:

Since, emissivity and emissive power are constant.

Therefore,

P = σεA₂T₂⁴   ----------- equation 2

where,

A₂ = Surface Area of Bulb 2

T₂ = Temperature of Bulb 1 = 2000 k

Dividing equation 1 by equation 2, we get:

P/P = σεA₁T₁⁴/σεA₂T₂⁴

1 = A₁(3000)²/A₂(2000)²

A₁/A₂ = (2000)²/(3000)²

A₁/A₂ = 0.44

the distance between 2 station is 5400 m find the time taken by a train to cover this distance, if the train travels with speed 60m/s

Answers

Answer:

I dont know bro

Explanation:

Ask an expert

Answer:

Time=90s

Explanation:

Speed=distance /time

[tex]60 = \frac{5400}{t} where \: t \: is \: time \\60t = 5400 \\ t = \frac{5400}{60} \\ t =90 \\ hope \: this \: helps..good \: luck [/tex]

A small rock with mass 0.12 kg is fastened to a massless string with length 0.80 m to form a pendulum. The pendulum is swinging so as to make a maximum angle of 45 ∘ with the vertical. Air resistance is negligible. Part A What is the speed of the rock when the string passes through the vertical position

Answers

Answer:

v = 3.33 m/s

Explanation:

In the position of 45 degrees, all the energy of the rock is gravitational, then we have:

E = m*g*L*cos(angle)

and in the vertical position of the string, all the energy is kinetic, so we have:

E = m*v^2/2

If there is no dissipation, both energies are equal, so we have:

m*g*L*cos(45) = m*v^2/2

9.81 * 0.8 * 0.7071 * 2 = v^2

v^2 = 11.0986

v = 3.33 m/s

Work out the velocity v at the end of a rollercoaster ride (0). (rearrange the equation for KE to make velocity v the subject)
KE=1/2mv^2

Answers

Explanation:

If the kinetic energy of an object is given and we need to find its velocity of motion, then we can find it by using the formula of kinetic energy as :

[tex]K=\dfrac{1}{2}mv^2[/tex]

m is mass of the object

We can rearrange the above equation such that,

[tex]v=\sqrt{\dfrac{2K}{m}}[/tex]

Hence, this is the velocity at the end of a rollercoaster ride.

A low C (f = 65Hz) is sounded on a piano. If the length of the piano wire is 2.0 m and its mass density is 5.0 g/m2, determine the tension of the wire.

Answers

Answer:

Tension of the wire(T) = 169 N

Explanation:

Given:

f = 65Hz

Length of the piano wire (L) = 2 m

Mass density = 5.0 g/m² = 0.005 kg/m²

Find:

Tension of the wire(T)

Computation:

f = v / λ

65 = v / 2L

65 = v /(2)(2)

v = 260 m/s

T = v² (m/l)

T = (260)²(0.005/2)

T = 169 N

Tension of the wire(T) = 169 N

A rod of mass M = 154 g and length L = 35 cm can rotate about a hinge at its left end and is initially at rest. A putty ball of mass m = 11 g, moving with speed V = 9 m/s, strikes the rod at angle θ = 29° a distance D = L/3 from the end and sticks to the rod after the collision.Calculate the rotational kinetic energy, in joules, of the system after the collision.

Answers

Answer:

Explanation:

moment of inertia of the rod = 1/3 mL² , m is mass and L is length of rod.

1/3 x .154 x .35²

= .00629

moment of inertia  of putty about the axis of rotation

= m d² , m is mass of putty and d is distance fro axis

= .011 x( .35 / 3 )²

= .00015

Total moment of inertia I = .00644 kgm²

angular momentum of putty about the axis of rotation

= mvRsinθ

m is mass , v is velocity , R is distance where it strikes the rod and θ is angle  with the rod at which putty strikes

= .011 x 9 x .35 / 3 x sin 29

= .0056

Applying conservation of angular momentum

angular momentum of putty = angular momentum of system after of collision

.0056 =  .00644 ω where ω is angular velocity of the rod after collision

ω = .87 rad /s .

Rotational energy

= 1/2 I ω²

I is total moment of inertia

=  .5 x .00644 x .87²

= 2.44 x 10⁻³ J .

A ball rolls over the edge of a platform with only a horizontal velocity. The height of the platform is 1.60m and the horizontal range of the ball from the base of the platform is 20.0m. What is the velocity of the ball just before it touches the ground

Answers

Answer:

v = 46.99 m/s

Explanation:

The velocity of the ball just before it touches the ground, is given by the following formula:

[tex]v=\sqrt{v_x^2+v_y^2}[/tex]           (1)

vx: horizontal component of the velocity

vy: vertical component of the velocity

The vertical component vy is calculated by using the following formula:

[tex]v_y^2=v_{oy}^2+2gh[/tex]   (2)

vy: final velocity

voy: initial vertilal velocity = 0m/s  (because it is a semi parabolic motion)

g: gravitational acceleration = 9.8 m/s^2

h: height = 1.60m

You replace the values of the parameters in the equation (2):

[tex]v_y=2(9.8m/s^2)(1.60m)=31.36\frac{m}{s}[/tex]

vx is calculated by using the information about the horizontal range of the ball:

[tex]R=v_o\sqrt{\frac{2h}{g}}[/tex]    (3)

R: horizontal range of the ball = 20.0 m

You solve the previous equation for vo, the initial horizontal velocity:

[tex]v_o=R\sqrt{\frac{g}{2h}}=(20.0m)\sqrt{\frac{9.8m/s^2}{2(1.60m)}}\\\\v_o=35\frac{m}{s}[/tex]

The horizontal component of the velocity is constant in the complete trajectory, hence, you have that

vx = vo = 35 m/s

Finally, you replace the values of vx and vy in the equation (1):

[tex]v=\sqrt{(35m/s)^2+(31.36m/s)^2}=46.99\frac{m}{s}[/tex]

The velocity of the ball just before it touches the ground is 46.99 m/s

a research submarine what is the maximum depth it can go

Answers

Answer: 36, 200 feet deep according to information on google

Explanation:

A small submarine, the bathyscape Trieste, made it to 10,916 meters (35,813 feet) below sea level in the deepest point in the ocean, the Challenger Deep in the Marianas Trench, a few hundred miles east of the Philippines. This part of the ocean is 11,034 m (36,200 ft) deep, so it seems that a submarine can make it as deep as it's theoretically possible to go

A force of 640 newtons stretches a spring 4 meters. A mass of 40 kilograms is attached to the end of the spring and is initially released from the equilibrium position with an upward velocity of 6 m/s. Find the equation of motion.

Answers

Bhbbv h Gucci Janice)6225

A car moving in a straight line starts at X=0 at t=0. It passesthe point x=25.0 m with a speed of 11.0 m/s at t=3.0 s. It passes the point x=385 with a speed of 45.0 m/s at t=20.0 s. Find the average velocity and the average acceleration between t=3.0 s and 20.0 s.

Answers

Answer:

Average velocity v = 21.18 m/s

Average acceleration a = 2 m/s^2

Explanation:

Average speed equals the total distance travelled divided by the total time taken.

Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)

Average acceleration equals the change in velocity divided by change in time.

Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)

Where;

v1 and v2 are velocities at time t1 and t2 respectively.

And x1 and x2 are positions at time t1 and t2 respectively.

Given;

t1 = 3.0s

t2 = 20.0s

v1 = 11 m/s

v2 = 45 m/s

x1 = 25 m

x2 = 385 m

Substituting the values;

Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)

v = (385-25)/(20-3)

v = 21.18 m/s

Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)

a = (45-11)/(20-3)

a = 2 m/s^2

Question 10
Air with a density of 1.20 kg/m3 flows through a 75.0 cm diameter pipe with a velocity of 2.00 m/s. What is the mass flow rate?

Answers

Answer:

75.0 cm

Explanation:

becouse i don,t no the right answer

A hydraulic lift is made by sealing an ideal fluid inside a container with an input piston of cross-sectional area 0.004 m2 , and an output piston of cross-sectional area 1.2 m2 . The pistons can slide up or down without friction while keeping the fluid sealed inside. What is the maximum weight that can be lifted when a force of 60 N is applied to the input piston

Answers

Answer:

Maximum weight that can be lifted = 18,000 N

Explanation:

Given:

Cross-sectional area of input (A1) = 0.004 m²

Cross-sectional area of the output (A2) = 1.2 m ²

Force (F) = 60 N

Computation:

Pressure on input piston (P1) = F / A1

Assume,

Maximum weight lifted by piston = W

Pressure on output piston (P2) = W / A2

We, know that

P1 = P2

[F / A1]  = [W / A2]

[60 / 0.004] = [W / 1.2]

150,00 = W / 1.2

Weight = 18,000 N

Maximum weight that can be lifted = 18,000 N

A truck has a bed that is 4.50 metres long,and 2.50 metres wide, and 1.50 metres high. What is maximum volume of sand can the truck carry within this dimensions?​

Answers

Answer:

since the bed is a cuboid, we find the volume by L×W×H

4.50 × 2.50 × 1.50 = 16.875m³

HOPE THIS HELPS

A 2500 kg truck moving at 10.00 m/s strikes a car waiting at the light. Assume there is no friction on the road. The hook bumpers continue to move at 7.00 m/s. What is the mass of the struck car

Answers

M2=(M1Vi/Vf)-M1=[2500*(10/7)]-2500
M2=(3/7)*2500=1070kg

A sphere of diameter 6.0cm is moulded into a thin uniform wire of diameter 0.2mm. Calculate the length of the wire in metres (Take π = 22/7) *​

Answers

Answer:

2025m

Explanation:

Since all materials of the sphere is made to a cylindrical wire, it implies the volume of the sphere material is same as that of the cylinder. This is expressed mathematically thus.

Volume of Sphere= volume of cylinder

4/3 ×π×R^3= π× r2× L

4/3 ×R^3= r^2×L

Hence

L = 3/4 × R^3/ r^2

But R = 6.0/2 = 3.0cm{ Diameter is twice raduis}

r= 0.2/2 = 0.1mm=>0.01cm{ Diameter is twice raduis and unit converted by dividing by 10 since 10mm = 1cm}

Substituting R and r into the expression for L, we have :

L = 3/4 × 3^3/ 0.01^2= 0.75 ×27/0.0001 = 202500cm

202500/100= 2025m{ we divide by 100 because 100cm=1m}

A bus travelling at a speed of 40 kmph reaches its destination in 8 minutes and 15 seconds. How far is the destination? a. 5.43 km         b. 5.44 km        c. 5.50 km         d. 9.06 km

Answers

Answer:

c. 5.50 km

Explanation:

8 min * 1h/(60 min) = 8/60 = 2/15 h

15 sec* 1 min/60 sec = 1/4 min * 1h/(60 min) =  1/240 h

8 min 15 sec = (2/15+1/240)h

40 km/h *(2/15 +1/240)h =5.50 km

Answer: 5.50 km

Explanation:

A certain metal with work function of Φ = 1.7 eV is illuminated in vacuum by 1.4 x 10-6 W of light with a wavelength of λ = 600 nm. 1)How many photons per second, N, are incident on the metal? N = photons per second Submit 2)What is KEmax, the maximum kinetic energy of the electron that is emitted from the metal? KEmax = eV

Answers

Answer:

1) n = 4.47*10^12 photons

2) K = 0.25 eV

Explanation:

This is a problem about the photoelectric effect.

1) In order to calculate the number of photons that impact the metal, you take into account the power of the light, which is given by:

[tex]P=\frac{E}{t}=1.4*10^{-6}\frac{J}{s}[/tex]     (1)

Furthermore you calculate the energy of a photon with a wavelength of 600nm, by using the following formula:

[tex]E_p=h\frac{c}{\lambda}[/tex]         (2)

c: speed of light = 3*10^8 m/s

h: Planck's constant = 6.626*10^-34 Js

λ: wavelength = 600*10^-9 m

You replace the values of the parameters in the equation (2):

[tex]E_p=(6.626*10^{-34}Js)\frac{3*10^8m/s}{600*10^{-9}m}=3.131*10^{-19}J[/tex]

Next, you calculate the quotient between the power of the light (equation (1)) and the energy of the photon:

[tex]n=\frac{P}{E_p}=\frac{1.4*10^{-6}J/s}{3.131*10^{-19}J}=4.47*10^{12}photons[/tex]

The number of photons is 4.47*10^12

2) The kinetic energy of the electrons emitted by the metal is given by the following formula:

[tex]K=E_p-\Phi[/tex]     (3)

Ep: energy of the photons

Φ: work function of the metal = 1.7 eV

You first convert the energy of the photons to eV:

[tex]E_p=3.131*10^{-19}J*\frac{6.242*10^{18}eV}{1J}=1.954eV[/tex]

You replace in the equation (3):

[tex]K=1.95eV-1.7eV=0.25eV[/tex]

The kinetic energy of the electrons emitted by the metal is 0.25 eV

(1). The Number of photons per second is,[tex]4.23*10^{12}[/tex]

(2). The maximum kinetic energy of the electron is 0.37eV.

(1). The power of light is given as,

            [tex]P=1.4*10^{-6}W[/tex]

    Energy is given as,

             [tex]E=\frac{hc}{\lambda} =\frac{6.626*10^{-34}*3*10^{8} }{600*10^{-9} } \\\\E=3.313*10^{-19} Joule\\\\E=\frac{3.313*10^{-19}}{1.6*10^{-19} }=2.07eV[/tex]

Number of photons per second is,

                    [tex]N=\frac{P}{E}=\frac{1.4*10^{-6} }{3.313*10^{-19} } =4.23*10^{12}[/tex]

(2). the maximum kinetic energy of the electron is,

              [tex]K.E=E-\phi[/tex]

Where [tex]\phi[/tex] is work function.

         [tex]K.E=2.07-1.7=0.37eV[/tex]

Learn more:

https://brainly.com/question/12337396

Determined to test the law of gravity for himself, a student walks off a skyscraper 180 m high, stopwatch in hand, and starts his free fall (zero initial velocity). Five seconds later, Superman arrives at the scene and dives off the roof to save the student.
a) Superman leaves the roof with an initial velocity that he produces by pushing himself downward from the edge of the roof with his legs of steel. He then falls with the same acceleration as any freely falling body. What must the value of the initial velocity be so that Superman catches the student just before they reach the ground?
b) On the same graph, sketch the positions of the student and of Superman as functions of time. Take Superman's initial speed to have the value calculated in part (a).
c) If the height of the skyscraper is less than some minimum value, even Superman can't reach the student before he hits the ground. What is this minimum height?

Answers

Answer:

a)  v₀ = - 164.62 m / s , c) y = 122.5 m

Explanation:

We can solve this exercise using the free fall kinematic relations.

We put our reference system on the floor, so the height of the skyscraper is y₀ = 180m and the floor level is y = 0 m

 

For the boy

         y = y₀ + v₀ t - ½ g t²

with free fall its initial speed is zero

        y = ½ g t2

For superman

        y = y₀ + v₀ (t-5) - ½ g (t-5)²

how superman grabs the lot just before hitting the ground

we look for the time it takes the boy down

         t = √ (2 y₀ / g)

         t = √ (2 180 / 9,8)

         t = 6.06 s

in the equation for superman, we clear the volume and calculate

         v₀ (t-5) = -y₀ + ½ g (t-5)²

         v₀ (6.06 -5) = -180 + ½ 9.8 (6.06 -5)²

         v₀ 1.06 = -174.49

         v₀ = - 174.49 / 1.06

         v₀ = - 164.62 m / s

the negative sign indicates that the initial speed is down

b) to graph the position of the two we use the table

  t (s)      Y_boy (m)   Y_superman (m)

    0             180                 180

   1              175.1               180

   5              57.5              180

   6                3.6                10.18

see attachment for the two curves

c) calculate the height that falls a lot in the 5 seconds (t = 5)

           y = -1/2 g t²

           y = ½ 9.8 5²

           y = 122.5 m

for this height superman has not yet left the skyscraper, so the boy hits the ground

This problem explores the behavior of charge on conductors. We take as an example a long conducting rod suspended by insulating strings. Assume that the rod is initially electrically neutral. For convenience we will refer to the left end of the rod as end A, and the right end of the rod as end B. In the answer options for this problem, "strongly attracted/repelled" means "attracted/repelled with a force of magnitude similar to that which would exist between two charged balls.A small metal ball is given a negative charge, then brought near (i.e., within about 1/10 the length of the rod) to end A of the rod. What happens to end A of the rod when the ball approaches it closely this first time?

Answers

Answer:

rod end A is strongly attracted towards the balls

rod end B is weakly repelled by the ball as it is at a greater distance

Explanation:

When the ball with a negative charge approaches the A end of the neutral bar, the charge of the same sign will repel and as they move they move to the left end, leaving the rod with a positive charge at the A end and a negative charge of equal value at end B.

Therefore rod end A is strongly attracted towards the balls and

rod end B is weakly repelled by the ball as it is at a greater distance

Compare the energy consumption of two commonly used items in the household. Calculate the energy used by a 1.40 kW toaster oven, Wtoaster , which is used for 5.40 minutes , and then calculate the amount of energy that an 11.0 W compact fluorescent light (CFL) bulb, Wlight , uses when left on for 10.50 hours .

Answers

Energy = (power) x (time)

-- For the toaster:

Power = 1.4 kW  =  1,400 watts

Time = 5.4 minutes = 324 seconds

Energy = (1,400 W) x (324 s)  =  453,600 Joules

-- For the CFL bulb:

Power = 11 watts

Time = 10.5 hours = 37,800 seconds

Energy = (11 W) x (37,800 s)  =  415,800 Joules

-- The toaster uses energy at 127 times the rate of the CFL bulb.

-- The CFL bulb uses energy at 0.0079 times the rate of the toaster.

-- The toaster is used for 0.0086 times as long as the CFL bulb.

-- The CFL bulb is used for 116.7 times as long as the toaster.    

-- The toaster uses 9.1% more energy than the CFL bulb.

-- The CFL bulb uses 8.3% less energy than the toaster.  

A certain type of laser emits light that has a frequency of 4.6 x 1014 Hz. The light, however, occurs as a series of short pulses, each lasting for a time of 3.1 x 10-11s. The light enters a pool of water. The frequency of the light remains the same, but the speed of light slows down to 2.3 x 108 m/s. In the water, how many wavelengths are in one pulse

Answers

Answer:

14,260

Explanation:

Relevant data provided for computing the wavelengths are in one pulse is here below:-

The number of wavelengths in Ls = [tex]4.6\times 10_1_4[/tex]

Therefore the Number of in time = Δt = [tex]3.1\times 10_-_1_1[/tex]

The number of wavelengths are in one pulse is shown below:-

[tex]Number\ of\ wavelengths = \triangle t\times f[/tex]

[tex]= 3.1\times 10_-_1_1\times 4.6\times 10_1_4[/tex]

= 14,260

Therefore for computing the number of wavelengths are in one pulse we simply applied the above formula.

what is the speed of light in quartz

Answers

Answer:

1.95 x 10^8 m/s.

Explanation:

Answer:

the answer is 1.95 x 10^8 m/s

Explanation:

a) Write the names of the materials used in the ohm law according to the Figure 1?
b) If the voltage of a circuit is 12 V and the resistance is 40 , What is the generated power?

Answers

Answer:

a. i. conducting wire

ii high-pass and low-pass filters

iii. Cobra-4 Xpert-link

iii. voltage source

b. Power generated is 3.6 W.

Explanation:

Ohm's law state that the current passing through a metallic conductor, e.g wire is directly proportional to the potential difference across its ends, provided temperature is constant.

i.e      V = IR

i. conducting wire

ii high-pass and low-pass filters

iii. Cobra-4 Xpert-link

iii. voltage source

b. Given that; V = 12 V and R = 40 Ohm's.

P = IV

From Ohm's law, I = [tex]\frac{V}{R}[/tex]

So that;

P = [tex]\frac{V^{2} }{R}[/tex]

   = [tex]\frac{12^{2} }{40}[/tex]

  = [tex]\frac{144}{40}[/tex]

 = 3.6 W

The power is 3.6 W.

We say that the displacement of a particle is a vector quantity. Our best justification for this assertion is: A. a displacement is obviously not a scalar. B. displacement can be specified by a magnitude and a direction. C. operating with displacements according to the rules for manipulating vectors leads to results in agreement with experiments. D. displacement can be specified by three numbers. E. displacement is associated by motion.

Answers

Answer:

Option B - displacement can be specified by a magnitude and a direction.

Explanation:

A Vector quantity is defined as a physical quantity characterized by the presence of both magnitude as well as direction. Examples include displacement, force, torque, momentum, acceleration, velocity e.t.c

Whereas a scalar quantity is defined as a physical quantity which is specified with the magnitude or size alone. Examples include length, speed, work, mass, density, etc.

Displacement is the difference between the initial position and the final position of a body. Displacement is a vector quantity and not a scalar quantity because it can be described by using both magnitude as well as direction.

Looking at the options, the only one that truly justifies this definition is option B.

state Ohm`s law as applied in electricity

Answers

Answer:

Ohm's Law (E = IR) is as fundamentally important as Einstein's Relativity equation (E = mc²) is to physicists. When spelled out, it means voltage = current x resistance, or volts = amps x ohms, or V = A x Ω.

Observe: Air pressure is equal to the weight of a column of air on a particular location. Airpressure is measured in millibars (mb). Note how the air pressure changes as you move StationB towards the center of the high-pressure system.
A. What do you notice?
B. Why do you think this is called a high-pressure system?

Answers

Answer:

a) When moving towards a high pressure center the pressure values ​​increase in the equipment

b) This area is called high prison since the weight of the atmosphere on top is maximum

Explanation:

A) A high atmospheric pressure system is an area where the pressure is increasing the maximum value is close to 107 Kpa, the other side as low pressure can have small values ​​85.5 kPa.

When moving towards a high pressure center the pressure values ​​increase in the equipment

B) This area is called high prison since the weight of the atmosphere on top is maximum

in general they are areas of good weather

A car of mass 410 kg travels around a flat, circular race track of radius 83.4 m. The coefficient of static friction between the wheels and the track is 0.286. The acceleration of gravity is 9.8 m/s 2 . What is the maximum speed v that the car can go without flying off the track

Answers

Answer:

The maximum speed v that the car can go without flying off the track = 15.29 m/s

Explanation:

let us first lay out the information clearly:

mass of car (m) = 410 kg

radius of race track (r) = 83.4 m

coefficient of friction (μ) = 0.286

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

maximum speed = v m/s

For a body in a constant circular motion, the centripetal for (F) acting on the body is given by:

F = mass × ω

where:

F = maximum centripetal force = mass × μ × g

ω = angular acceleration = [tex]\frac{(velocity)^2}{radius}[/tex]

∴ F = mass × ω

m × μ × g = m × [tex]\frac{v^{2} }{r}[/tex]

410 × 0.286 × 9.8 = 410 × [tex]\frac{v^{2} }{83.4}[/tex]

since 410 is on both sides, they will cancel out:

0.286 × 9.8 = [tex]\frac{v^{2} }{83.4}[/tex]

2.8028 = [tex]\frac{v^{2} }{83.4}[/tex]

now, we cross-multiply the equation

2.8028 × 83.4 = [tex]v^{2}[/tex]

[tex]v^{2}[/tex] = 233.754

∴ v = √(233.754)

v = 15.29 m/s

Therefore, the maximum speed v that the car can go without flying off the track = 15.29 m/s

Astronauts are testing the gravity on a new planet. A rock is dropped between two photogates that are 0.5 meters apart. The first photogate reads a velocity of 1.2 m/s and the the second photogate reads a velocity of 4.3 m/s . What is the acceleration of gravity on this new planet?

Answers

Answer:

a = 17 m / s²

Explanation:

For this experiment that the astronauts are carrying out, the value of the relation of gravity is cosecant, therefore we can use the kinematic relations

         v² = v₀² + 2a y

They indicate the initial speed 1.2 m / s the final speed 4.3 m / s and the distance remembered 0.5 m

we clear

        a = (v² - v₀²) / 2y

we calculate

       a = (4.3² -1.2²) / 2 0.5

       a = 17 m / s²

this is the gravity of the new planet

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