Which circuits are parallel circuits?
0
WA
AHE

Which Circuits Are Parallel Circuits?0WAAHE

Answers

Answer 1

Answer:

the 5 is because u have to select the currents and volts which gives us 2 and 1 plus 3 is 6 but if we rest 1 is 5, thats the answer


Related Questions

What is energy?
Select one:

A force that must be exerted in order to accelerate an object.

A property that must be transferred to perform work.

A particle that can be absorbed to change momentum.

A system that can be rearranged in order to change its state.

Answers

Answer:

Energy is:

A property that must be transferred to perform work.

Answer:

Energy

Explanation:

A property that must be transferred to perform work .

Hope it is helpful to you Stay safe healthy and happy ☺️

What is the magnitude of vector X?​

Answers

Explanation:

by using Pythagoras theorem

h²=p²+b²

z²=Y²+x²

x²=75²-21²

x=√5184

x=72cm is the magnitude of vector X

hope it helps

stay safe healthy and happy.

Two identical satellites orbit the earth in stable orbits. Onesatellite orbits with a speed vat a distance rfrom the center of the earth. The second satellite travels at aspeed that is less than v.At what distance from the center of the earth does the secondsatellite orbit?At a distance that is less than r.At a distance equal to r.At a distance greater than r.Now assume that a satellite of mass m is orbiting the earth at a distance r from the center of the earth with speed v_e. An identical satellite is orbiting the moon at thesame distance with a speed v_m. How does the time T_m it takes the satellite circling the moon to make onerevolution compare to the time T_e it takes the satellite orbiting the earth to make onerevolution?T_m is less than T_e.T_m is equal to T_e.T_m is greater than T_e.

Answers

Answer:

a. At a distance greater than r

b. T_m is greater than T_e.

Explanation:

a. Two identical satellites orbit the earth in stable orbits. One satellite orbits with a speed vat a distance r from the center of the earth. The second satellite travels at a speed that is less than v. At what distance from the center of the earth does the second satellite orbit?

Since the centripetal force on any satellite, F equals the gravitational force F' at r,

and F = mv²/r and F' = GMm/r² where m = mass of satellite, v = speed of satellite, G = universal gravitational constant, M = mass of earth and r = distance of satellite from center of earth.

Now, F = F'

mv²/r = GMm/r²

v² = GM/r

v = √GM/r

Since G and M are constant,

v ∝ 1/√r

So, if the speed decreases, the radius of the orbit increases.

Since the second satellite travels at a speed less than v, its radius, r increases since v ∝ 1/√r.

So, the distance the second satellite orbits is at a distance greater than r

b. An identical satellite is orbiting the moon at the same distance with a speed v_m. How does the time T_m it takes the satellite circling the moon to make one revolution compare to the time T_e it takes the satellite orbiting the earth to make one revolution?

Since the speed of the satellite, v = √GM/r where M = mass of planet

Since the satellite is orbiting at the same distance, r is constant

So, v ∝ √M

Since mass of earth M' is greater than mass of moon, M", the speed of satellite circling moon, v_m is less than v the speed of satellite circling earth at the same distance, r

Now, period T = 2πr/v where r = radius of orbit and v = speed of satellite

Since r is constant for both orbits, T ∝ 1/v

Now, since the speed of the speed of the satellite on earth orbit v  is greater than the speed of the satellite orbiting the moon, v_m, and T ∝ 1/v, it implies that the period of the satellite orbiting the earth, T_e is less than the period of the satellite orbiting the moon, T_m since there is an inverse relationship between T and v. T_e is less T_m implies T_m is greater than T_e

So, T_m is greater than T_e.

What is the equivalent resistance of a circuit that contains two 50.0 0
resistors connected in series to a 12.0V battery?

Answers

Answer:

the correct answer is A :)

instrument used in measurement Amount of substance

Answers

Answer:

For liquids: A measuring cylinder is used.

For solid: Over flow can is used

Answer:

i think a measuring cylinder

A boy throws a ball straight up with a speed of 21.5 m/s. The ball has a mass of 0.19 kg. How much gravitational potential energy will the ball have at the top of its flight? (Assume there is no air resistance.) A. 43.9 J B. 37.5 J C. 48.5 J D. 41.2 J​

Answers

Answer:

Explanation:

The equation fo potential energy is PE = mgh, where m is the mass of the ball, g is the pull of gravity (constant at 9.8), and h is the max height of the ball. What we do not have here is that height. We need to first solve for it using one-dimensional equations. What we have to know above all else, is that the final velocity of an object at its max height is always 0. That allows us to use the equation

[tex]v_f=v_0+at[/tex] where vf is the final velocity and v0 is the initial velocity. We will find out how long it takes for the object to reach that max height first and then use that time to find out what that max height is. Baby steps here...

0 = 21.5 + (-9.8)t and

-21.5 = -9.8t so

t = 2.19 seconds (Keep in mind that if I used the rules correctly for sig fig's, the answer you SHOULD get is not one shown, so I had to adjust the sig fig's and break the rules. But you know what they say about rules...)

Now we will use that time to find out the max height of the object in the equation

Δx = [tex]v_0t+\frac{1}{2}at^2[/tex] and filling in:

Δx = [tex]21.5(2.19)+\frac{1}{2}(-9.8)(2.19)^2[/tex] which simplifies down a bit to

Δx = 47.1 - 23.5 so

Δx = 23.6 meters.

Now we can plug that in to the PE equation to find the PE of the object:

PE = (.19)(9.8)(23.6) so

PE = 43.9 J

Which one will it be

Answers

Answer:

none

Explanation:

it's to high up to be affected by the gravity

a)Zero
that is the right answer

A car moving in a straight line uniformly accelerated speed increased from 3 m / s to 9 m / s in 6 seconds. With what acceleration did the car move?


a.
2 m/s2


b.
1 m/s2


c.
0 m/s2


d.
3 m/s2

Answers

I think it’s 1 . The answer is B

Answer:

b) 1 m/s

I am sure...........

The image shows the right-hand rule being used for a current-carrying wire.

An illustration with a right hand with fingers curled and thumb pointed up.

Which statement describes what the hand shows?

When the current flows down the wire, the magnetic field flows out on the left side of the wire and in on the right side of the wire.
When the current flows up the wire, the magnetic field flows out on the left side of the wire and in on the right side of the wire.
When the current flows down the wire, the magnetic field flows in on the left side of the wire and out on the right side of the wire.
When the current flows up the wire, the magnetic field flows in on the left side of the wire and out on the right side of the wire.

Answers

D PLS MARK ME AS BRAINLY

Answer:

The answer is (D): When the current flows up the wire, the magnetic field flows in on the left side of the wire and out on the right side of the wire.

Explanation:

If the pressure of a gas is really due to the random collisions of molecules with the walls of the container, why do pressure gauges – even very sensitive ones – give perfectly steady readings? Shouldn't the gauge be continually jiggling and fluctuating? Explain.

Answers

Answer:

there is no fluctuation in the measurement because the quantity of molecule is too large and a quantity of some molecules is imperceptible.

Explanation:

The pressure measurement is carried out by calibrating the force exerted by the air on a surface of known area, suppose a small area 1 mm² = 0.01 cm²

To find out if the random movement of air molecules affects the pressure reading, let's calculate the number of molecules that reaches the pressure gauge.

In a system at atmospheric pressure and in a volume of 1 m³ (walls of 1 m each) there is one mole of air molecules, this mole is evenly distributed, so how many molecules fall on our surface

           # _molecule = 6.02 10²³ 0.01 10⁻⁴ / 1

           #_molecular = 6.02 10¹⁷ molecules per second

therefore the variation of the number of molecules is not very important

Consequently there is no fluctuation in the measurement because the quantity of molecule is too large and a quantity of some molecules is imperceptible.

What are the messing forces that would make the object be in equilibrium?

Answers

Answer:

A) 20 N, B) 20 N, & C) 8 N

Explanation:

For the object to be in equilibrium, the upward forces must be equal to the downward forces and the forward forces must be equal to the backward forces.

1. Determination of A and B.

Forward forces = Backward forces

A + 10 + B = 25 + 25

A + 10 + B = 50

Collect like terms

A + B = 50 – 10

A + B = 40

Assume A and B to be equal. Thus, A is 20 N and B is 20 N.

2. Determination of C

Upward forces = Downward forces

C + 112 = 20 + 100

C + 112 = 120

Collect like terms

C = 120 – 112

C = 8 N

Thus, for the object to be in equilibrium, A must be 20 N, B must be 20 N and C must be 8N.

Mixed powders may be categorized as​

Answers

Answer:

on  flow properties and free-flowing and cohesive.

Explanation:

the power Free flowing powders do not cling together, as cohesive powders stick to each other and form that do not disperse well during mixing

a. A horse pulls a cart along a flat road. Consider the following four forces that arise in this situation.

1. the force of the horse pulling on the cart
2. the force of the cart pulling on the horse
3. the force of the horse pushing on the road
4. the force of the road pushing on the horse

b. Suppose that the horse and cart have started from rest; and as time goes on, their speed increases in the same direction. Which one of the following conclusions is correct concerning the magnitudes of the forces mentioned above?

1. Force 1 exceeds Force 2.
2. Force 2 is less than Force 3.
3. Force 2 exceeds Force 4.
4. Force 3 exceeds Force 4.
5. Forces 1 and 2 cannot have equal magnitudes.

Answers

Answer:

a) F₁ = F₂,  F₃ = F₄,  b)  the correct answer is 3

Explanation:

a) In this exercise we have several action and reaction forces, which are characterized by having the same magnitude, but different direction and being applied to different bodies

Forces 1 and 2 are action and reaction forces F₁ = F₂

Forces 3 and 4 are action and reaction forces F₃ = F₄

as it indicates that the

b) how the car increases if speed implies that force 1> force3

      F₁ > F₃

therefore the correct answer is 3

When landing after a spectacular somersault, a 40.0-kg gymnast decelerates by pushing straight down on the mat. Calculate the force she must exert if her deceleration is 7.00 times the acceleration due to gravity. Explicitly show how you follow the steps in the Problem-Solving Strategy for Newton’s laws of motion.

Answers

Answer:

[tex]F=3139.2N[/tex]

Explanation:

Mass [tex]m=40.0kg[/tex]

Acceleration [tex]a=7g=68.67m/s^2[/tex]

With g as 9.81

Generally the equation for Force is mathematically given by

  [tex]F_net=F-w\\\\F-w=ma\\\\F=ma-w\\\\[/tex]

 [tex]F=ma-mg\\\\F=m(a+g)[/tex]

Therefore

 [tex]F=m*8g[/tex]

 [tex]F=40*8*9.81[/tex]

 [tex]F=3139.2N[/tex]

Select the correct answer from each drop-down menu.
Balloon B is negatively charged.
Balloon A is ( )
charged, and balloon C is( ) charged. If balloon A approaches balloon C, there'll be a force of ( )
between them.

Answers

Answer:

it would look like what I did in the picture

Explanation:

Calculate the equivalent resistance

Answers

Not enough information

15. A car travelling towards the right has a mass of 1332 kg and has a speed of 25 m/s. A truck is
travelling towards the left with a mass of 3000 kg and a speed of 15 m/s. They collide head
on with each other. What is the total momentum after the crash? In which direction will the
vehicles travel after the collision?

Answers

Explanation:

Given that,

The mass of a car, m₁ = 1332 kg

The speed of the car, u₁ = 25 m/s (right)

The mass of a truck, m₂ = 3000 kg

The speed of the truck, u₂ = -15 m/s

The total momentum after the crash is given by :

p=m₁u₁ + m₂u₂

Put all the values,

P = 1332(25) + 3000(-15)

= −11700 kg-m/s

So, the total momentum after the crash is equal to 11700 kg-m/s and it is in the left direction.

The lever of a car lift has an area of 0.2 meters squared, and the area of the lift under the car is 8
meters squared. If you push with a force of 3 newtons, how much force will be applied to the
car?

Answers

Answer:

THE ANSWER IS SOMETHING LIKE 55

On page 14 of the call of the wild jack London, writes,in vague ways he remembered back to the youth of the breed ,this statement is example of

Answers

Answer:

On page 14 of the call of the wild jack London, writes,in vague ways he remembered back to the youth of the breed ,this statement is example of the racial unconscious. As a general rule, really great novels contain universal truths. 

Explanation:

Hey mate dont worry! My answer is correct!!

Answer:

racial unconscious

Explanation:

took the test and it was correct

A block with length of 1.5m,width 1=1m,height=0.5m and mass 300kg.what is the pressure at the bottom surface of the block

Answers

Answer:

Find The area and calculate pressure.

Explanation:

Using only astronomical data from the Appendix E in the textbook, calculate the speed of the planet Venus in its essentially circular orbit around the sun.
Venus = 4.87x10^24

Answers

Answer:

[tex]v=3.49\times 10^4\ m/s[/tex]

Explanation:

Given that,

Mass of Venus, [tex]M_V=4.87\times 10^{24}\ kg[/tex]

We know that,

Mass of Sun, [tex]M_s=1.98\times 10^{30}\ kg[/tex]

The distance between the center of Sun and the center of Venus is [tex]1.08\times 10^{11}\ m[/tex]

We need to find the peed of the planet Venus in its essentially circular orbit around the sun. using the formula,

[tex]v=\sqrt{\dfrac{GM_s}{r}}[/tex]

Put all the values,

[tex]v=\sqrt{\dfrac{6.67\times 10^{-11}\times 1.98\times 10^{30}}{1.08\times 10^{11}}}\\\\v=3.49\times 10^4\ m/s[/tex]

So, the speed of the planet venus is [tex]3.49\times 10^4\ m/s[/tex].

A block of mass m is moved over a distance d. An applied force F is directed perpendicularly to the block’s displacement. How much work is done on the block by the force F?​

Answers

zero

Explanation:

Work W is defined as

W = Fd = Fdcos(theta)

and it is a dot product of the force and displacement and theta is angle between F and d Since the force is perpendicular to d, angle is 90° thus cos90 = 0. Hence work is zero.

A 20-g bullet moving at 1000 m/s is fired through a 1kg block of wood emerging at a speed of 100m/s.what is the change in kinetic energy of the bullet

Answers

Answer:

D. 0.16 KJ

Explanation:

Given the following data;

Initial velocity, U = 1000 m/s

Mass of bullet, M1 = 20 g to kilograms = 20/1000 = 0.02 kg

Mass of block, M2 = 1 kg

Final velocity, V = 100 m/s

To find the change in kinetic energy;

First of all, we would determine the velocity after the collision by applying the law of conservation of momentum.

[tex] M_{1}U - M_{1}V = M_{2}V_{f} [/tex]

Substituting into the above formula, we have;

[tex] 0.02*1000 - 0.02*100 = 1*V_{f} [/tex]

[tex] 20 - 2 = V_{f} [/tex]

Vf = 18 m/s

Next, we would find the kinetic energy of block-bullet;

[tex] K.E = \frac{1}{2}MV^{2}[/tex]

Substituting into the formula, we have;

[tex] K.E = \frac{1}{2}*1*18^{2}[/tex]

[tex] K.E = \frac{1}{2}*324 [/tex]

K.E = 162 Joules.

In Kilojoules;

K.E = 162/1000

K.E = 0.162 ≈ 0.16 KJ

Which two factors affect the electric force between two particles?
A. The distance between them
B. Their charges
C. The strong nuclear force
D. Gravitational force
E. Their mass

Answers

Answer:

a the distance between the

For a standard production car, the highest road-tested acceleration ever reported occurred in 1993, when a Ford RS200 Evolution went from zero to 26.8 m/s (60 mi/h) in 3.323 s. Find the magnitude of the car's acceleration.

Answers

Answer:

a = 8.06 m/s²

Explanation:

The acceleration of this car can be found using the first equation of motion:

[tex]v_f = v_i + at\\\\a = \frac{v_f-v_i}{t}[/tex]

where,

a = acceleration = ?

vf = final speed = 26.8 m/s

vi = initial speed = 0 m/s

t = time = 3.323 s

Therefore,

[tex]a = \frac{26.8\ m/s-0\ m/s}{3.323\ s}[/tex]

a = 8.06 m/s²

According to the model, when was the universe at its most dense?

A) During the Dark Ages where matter increased in mass.

B) Just before the Big Bang where all matter existed in a singularity.

C) During the nuclear fusion events, as the atoms become more massive.

D) Current day, as the number of galaxies, solar systems, and planets have increased.

Answers

it d cus you look at the gragh

Answer:

The Answer is D

Explanation:

Hope this helps!!!!

A surfactant with a Hydrophile-
Lipophile Balance (HLB) value of 18 is expected to function as a solubilizing agent
O False
O True​

Answers

Answer:

Explanation:

True

The Best answer from the options is TRUE

A surfactant with a Hydrophile-lipophile Balance value of 18 is a solubilizing agent because a hydrophilic-lipophilic balance is used to measure the degree /level of hydrophilicness or liphophilicness of a Surfactant is at . and the

method used by the Hydrophile-lipophile balance to achieve this is by calculating the values for the different regions of the surfactant.

Hence a Surfactant with a hydrophile-lipophile Balance value of 18 is expected to function as a solubilizing agent.

Attached below is the HLB

Learn more : https://brainly.com/question/11338364

Which option is the best blackbody radiator?

A.
The Sun

B.
A red laser pointer

C.
A tennis ball

D.
Boiling water

Answers

Answer:

A. The Sun

Explanation:

The Sun is to be considered a perfect black body.

can Denel be regarded as monopoly in south africa​

Answers

Answer:

Denel (Pty) Ltd was established as a private company, incorporated in terms of the Companies Act on 1 April 1992 with the State as the sole shareholder. ... Denel can at present, without doubt, be regarded as a public monopoly.

A string that is under 50.0N of tension has linear density 5.0g/m. A sinusoidal wave with amplitude 3.0cm and wavelength 2.0m travels along the string. What is the maximum speed of a particle on the string

Answers

Answer:

9.42 m/s

Explanation:

Applying,

V' = Aω.............. Equation 1

Where V' = maximum speed of the string, A = Amplitude of the wave, ω = angular velocity.

But,

ω = 2πf................. Equation 2

Where f = frequency, π = pie

And,

f = v/λ................ Equation 3

Where, λ = wave length, v = velocity

Also,

v = √(T/μ)................. Equation 4

Where T = Tension, μ = linear density.

From the question,

Given: T = 50.0 N, μ = 5.0 g/m = 0.005 kg/m

Substitute into equation 4

v = √(50/0.005)

v = √(10000)

v = 100 m/s

Also Given: λ = 2.0 m

Substitute into equation 3

f = 100/2

f = 50 Hz.

Substitute the value of f into equation 2

Where π = constant = 3.14

ω = 2(3.14)(50)

ω = 314 rad/s

Finally,

Given: A = 3.0 cm = 0.03 m

Substitute into equation 1

V' = 0.03(314)

V' = 9.42 m/s

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