Look at the diagram shown below. The reading on ammeter 1 is 0.15 A. What will the reading on ammeter 2 be

Look At The Diagram Shown Below. The Reading On Ammeter 1 Is 0.15 A. What Will The Reading On Ammeter

Answers

Answer 1

There's no way to tell, without knowing the resistance of either X-1 or X-2, or the reading on A3.


Related Questions

More potential energy can be stored by moving against the magnetic force closer to a magnet?

Answers

Answer:

if your saying can it? then yes or if you are asking what type of magnetic force  then its fr its actual self magnetic force

Explanation:

Find the gravitational potential energy of a body of mass 25kg,kept at a height of 4m If g=10m/s'.

Answers

Answer:

1000 J

Explanation:

The gravitational potential energy of a body can be found by using the formula

GPE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 10 m/s²

From the question we have

GPE = 25 × 10 × 4 = 1000

We have the final answer as

1000 J

Hope this helps you

Kichelle enjoys balloon animals from the carnival. She just received a balloon giraffe that has an initial temperature of 39.0°C and a volume of 1.28 L. If Preston plays a trick on Kichelle, and puts her balloon giraffe into the freezer, what would be the new volume of the balloon if the temperature drops down to 8.0°C?

Answers

39°C=1.28L

8°C=? Less

=8°C/39°C×1.28L

=8/39×1.28L

=10.24/39

=512/195

=2.6L

1. A stone is thrown vertically upward with velocity of 15 m/s at the same time, 10 m vertically above a second stone is allowed to fall. After what time and at what height do they collide (take g = 10 ms^-2)​

Answers

Let the ball at the ground be A and that at the top be B.

Assume that you see the motion of A while sitting at B.

You(B) are obviously at rest w.r.t. yourselves. However you(B) and A have the same acceleration in the same direction. Thus acceleration of A w.r.t. you(B) is 0.

Thus, to you, it will appear as if A is travelling towards you with an uniform speed of 20m/s.

Dividing the distance 20m with that speed, we get that A reaches you(B) in 1s.

Now, you have considered these motions and time by assuming you were stationed at B. However, the time taken for A to meet B shouldn't be dependent on which reference frame you assume. Thus, time taken for A and B to collide is 1s.

Since B falls freely, he covers a distance of  1/2(g)(1²)=5  (assuming g = 10m/s²)

Thus, they meet at a height of (20–5)m = 15m from the ground.

Who is the author "talking” to in the third stanza of Flanders field

Answers

I’m not sure if it’s correct but here u go

In Flanders Fields” is written in the voice of a group of soldiers who have recently died in a World War I battle. By speaking as a group and asking the reader to join in their struggle, these speakers suggest that war is a shared responsibility that affects everyone.

Answer:

One of Canada's best-known poems, “In Flanders Fields” was written on May 2, 1915, when Canadian serviceman John McCrae was stationed at an army hospital in Flanders, Belgium, during World War I.

Explanation:

a soccer ball is kicked at an angle of 35° and it lands on even ground.
A) What angle will produce the same range?
B) Which angle of the two will produce the highest ball?
C) Which angle of the two will produce the ball that is in the air the longest?
D) What angle in this situation would produce the furthest range?

Answers

Hi there!

A)

The angle that will produce the same range is the compliment of 35°.

Thus, kicking the ball at 55° will result in the same range.

We can prove this by using the derived range equation:

[tex]R = \frac{v^2sin2\theta}{g}[/tex]

An angle of 35° yields:

[tex]R = \frac{v^2sin(2*35)}{g} = .939R[/tex]

An angle of 55° yields:

[tex]R = \frac{v^2sin(2*55)}{g} = .939R[/tex]

Both are the same, thus indicating that 55° produces the same range.

B)

The angle of 55° will produce the higher ball because the VERTICAL component of the ball's velocity is greater compared to kicking the ball at 35° degree.

sin(55) > sin(35)

C)

The angle of 55° will result in the ball being in the air the longest because when a ball is in the air (assuming no air resistance), the ball experiences an acceleration due to gravity of -9.8 m/s², causing the vertical velocity to decrease until it eventually reaches 0 m/s at the top of its path. A greater initial vertical velocity means that it will take longer for the ball to fall.

We can prove this using:

vf = vi + at

0 = vy - 9.8t

vy/9.8 = t

Greater vy (vertical component of velocity) ⇒ greater time taken.

D)

The angle that would result in the furthest range is 45°.

We can prove this using calculus. Recall the above range equation:

[tex]R = \frac{v^2sin2\theta}{g}[/tex]

We can take the derivative and use the first-derivative test to find its critical point:

[tex]\frac{dR}{d\theta} = \frac{v^22cos2\theta}{g} = 0[/tex]

Evaluate:

[tex]v^22cos2\theta = 0 \\\\cos2\theta = 0 \\\\2\theta = 90^o\\\\\boxed{\theta = 45^o}[/tex]

A. The angle that will produce the same range as a 35° kick is 53°.

B. The angle that produces the highest ball is 45°.

C. The angle that produces the ball that is in the air the longest is 45°.

D. There is no single angle that will produce the furthest range for all initial velocities and accelerations due to gravity.

A. The range of a projectile is the horizontal distance it travels before it hits the ground. The range of a projectile is determined by the initial velocity, the angle of projection, and the acceleration due to gravity.

For a given initial velocity, the range of a projectile is maximized when the angle of projection is 45°. However, if the ground is not level, the range of a projectile can be maximized at other angles.

In the case of a soccer ball kicked at an angle of 35°, the range will be maximized at an angle of 53°. This is because the range of a projectile is maximized when the vertical component of the initial velocity is equal to the horizontal component of the initial velocity.

The angle of 53° is the angle that produces a vertical component of the initial velocity that is equal to the horizontal component of the initial velocity when the ball is kicked at an angle of 35°.

Therefore, the angle that will produce the same range as a 35° kick is 53°.

B. The angle that produces the highest ball is 45°.

C. The angle that produces the ball that is in the air the longest is 45°.

D. The angle that produces the furthest range depends on the initial velocity and the acceleration due to gravity. There is no single angle that will produce the furthest range for all initial velocities and accelerations due to gravity.

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The length of a road is 35.5cm what will be the length in meters?

Answers

안녕하세요

answer is 355 meters ...

Answer:

0.355m

Explanation:

To convert from cm to m, divide by 100

35.5÷100=0.355m

A 60 kg gorilla named Anthony Falcon is standing on his skateboard. This is on planet Erf,

where a = g = -10m/s2

If Anthony Falcon has a weight of 600 N, with what amount of force is their body pushing

down on the skateboard with?

Answers

F= ma
F= (600/-10) -10
F= 580n

At least I think that’s the answer

2. Un niño hace girar con la mano una pelota de hule que se encuentra sujeta mediante un cordón de
0.75 m de longitud. Si la pelota da 0.7 vueltas a cada segundo, entonces,
a) ¿Cuál es el periodo de la pelota?
b) ¿Cuál es la velocidad lineal de la pelota?

Answers

Cuando el niño hace girar 0.7 vueltas por segundo una pelota de hule que se encuentra sujeta mediante un cordón de 0.75 m de longitud, tenemos que:

a) El periodo de la pelota es 1.43 segundos.

b) La velocidad lineal de la pelota es 3.3 m/s.

a) El periodo de la pelota está dado por:

[tex] T = \frac{2\pi}{\omega} [/tex]

En donde:

ω: es la velocidad angular

Dado que la pelota da 0.7 vueltas (revoluciones) cada segundo, la velocidad angular es:

[tex] \omega = \frac{0.7 \:rev}{s}*\frac{2\pi rad}{1 \:rev} = 4.40 rad/s [/tex]

Entonces, el periodo es:

[tex]T = \frac{2\pi}{\omega} = \frac{2\pi}{4.40 rad/s} = 1.43 s[/tex]

b) La velocidad lineal de la pelota se puede calcular usando la siguiente ecuación:

[tex] v = \omega r [/tex]

En donde:

r: es el radio de la circunferencia = longitud del cordón = 0.75 m

[tex]v = \omega r =4.40 rad/s*0.75 m = 3.3 m/s[/tex]

Por lo tanto, la velocidad lineal de la pelota es 3.3 m/s.

Puedes encontar mas aquí:

https://brainly.com/question/19380743?referrer=searchResultshttps://brainly.com/question/17689540?referrer=searchResults

Espero que te sea de utilidad!

If the ball has a positive electric charge, what should the charge of the coilgun be to push the ball away?

A)positive
B) negative
C)neutral

Answers

Answer:

A

Explanation:if the things have the same charge they will repel makeing the ball move away form its positively charged guy = that the coil gun works.

explain why the moon orbits the earth while the earth orbits the sun

Answers

As the Earth rotates, it also moves, or revolves, around the Sun. ... As the Earth orbits the Sun, the Moon orbits the Earth. The Moon's orbit lasts 27 1/2 days, but because the Earth keeps moving, it takes the Moon two extra days, 29 1/2, to come back to the same place in our sky .hope it help

Explain the relationship between the frequency, energy, and wavelength of a wave.​

Answers

Answer:

Explanation:

The energy equation is E=hv

Which explains why scientists often prefer to measure mass rather than weight?

It is always the same.

It is affected by gravity.

Answers

Answer:

scientists often prefer to measure mass rather than weight because weight is affected by gravity while mass is not. Mass is always the same.

Explanation:

Weight is a force rather than a measure of the amount of matter something contains. The same amount of matter can have different weights depending on where it is. Mass directly measures the amount of matter present and remains the same no matter where it's measured.

A car goes from 60 m/s to 75 m/s in 10 seconds.calculate the car’s acceleration?

Answers

Explanation:

Uhhhh here is an example

a=vt. =27 m/s10 s. =2.7 m/s2.

[tex]▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪[/tex]

Acceleration of the car is ~

[tex] \boxed{1.5 \: \: m/s}[/tex]

[tex] \large \boxed{ \mathfrak{Step\:\: By\:\:Step\:\:Explanation}}[/tex]

We know that,

Acceleration = rate of change in velocity ~

that is ~

[tex]a = \dfrac{v - u}{t} [/tex]

where,

v = final velocity

u = initial velocity

t = time taken ~

now, let's solve ~

[tex]a = \dfrac{75 - 60}{10} [/tex]

[tex]a = \dfrac{15}{10} [/tex]

[tex]a = 1.5 \: \: m/s[/tex]

cecily is inflating her bicyble tyre with the pump below. when she pushes the plunger down, it is moving against a force appliefd by the air inside the cynlinder. this means that the plunger is doing ___

Answers

Answer:

"work against the force of the air in the tire"

The air in the tire provides a force opposing the force of the air provided by the plunger.

1 point
What is the potential energy of a bird flying, if the bird has 1000 J of total
energy and 450 J of kinetic energy? (only put the number, no units or
commas)

Answers

Answer:

the bird has 550J of potential energy

Explanation:

PEtotal=PE+KE

1000J=PE+450J

subtracting the kinetic energy from the total we get:

1000J-450J=PE

550J=PE

What is the total amount of potential and kinetic energy in a system? *
1 point
A: Electrical energy
B: Heat energy
C: Mechanical energy
D: Nuclear energy

Answers

Answer:

C. Mechanical Energy

Explanation:

The total amount of mechanical energy is merely the sum of the potential energy and the kinetic energy.

When there is a temperature inversion, you would expect to experience Group of answer choices clouds with extensive vertical development above an inversion aloft. good visibility in the lower levels of the atmosphere and poor visibility above an inversion aloft. an increase in temperature as altitude increases.

Answers

Temperature inversion leads to an increase in temperature as altitude increases.

The term temperature inversion refers to a situation in which a layer of warm air lies over a layer of cool air. This is also referred to as thermal inversion. This occurs when the air below to loose heat rapidly.

One of the effects of temperature inversion is reduction in visibility. So, thermal inversion leads to an increase in temperature as altitude increases.

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A small ball is attached to one end of a spring that has an unstrained length of 0.201 m. The spring is held by the other end, and the ball is whirled around in a horizontal circle at a speed of 3.41 m/s. The spring remains nearly parallel to the ground during the motion and is observed to stretch by 0.0176 m.

Required:
By how much would the spring stretch if it were attached to the ceiling and the ball allowed to hang straight down, motionless?

Answers

The extension of the spring when the ball is allowed to hang straight down, motionless is 0.0032 m.

The given parameters;

unstrained length, l₁ = 0.201 mextension of the string, x = 0.0176 mspeed of the ball, v = 3.41 m/s

The radius of the circular path when spring is stretched is calculated as;

R = l₁ + x

R = 0.201 + 0.0176

R = 0.2186 m

The spring constant is calculated as follows;

[tex]F = ma\\\\Kx = \frac{mv^2}{R} \\\\K = \frac{mv^2}{Rx} \\\\K = \frac{(3.41)^2 m}{0.2186 \times 0.0176} \\\\K = 3,022.4 m \ N/m[/tex]

The extension of the spring when the ball is allowed to hang straight down, motionless;

[tex]F = mg\\\\Kx=mg\\\\x = \frac{mg}{K} \\\\x = \frac{9.8 m}{3022.4 m} \\\\x = 0.0032 \ m[/tex]

Thus, the extension of the spring when the ball is allowed to hang straight down, motionless is 0.0032 m.

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Which phrase is best described as the energy of motion?

Answers

Answer:

The energy of motion is known as kinetic energy.

Explanation:

The diagram shows what happened in the Rutherford and Marsden scattering experiment. Complete this sentence: The particle shown in red will come straight back from the foil because it is __________ by the charge in the gold nucleus.

Answers

From Rutherford  and Marsden scattering experiment, we can conclude that the particle shown in red will come straight back from the foil because it is deflected by the charge in the gold nucleus.

Rutherford proposed Planetary model atom, which visualized an atom to consists of a positively charged heavy core called the nucleus around which negatively charged electrons circle in orbits much as planets move round the sun

Thus, from Rutherford  and Marsden scattering experiment, we can conclude that the particle shown in red will come straight back from the foil because it is deflected by the charge in the gold nucleus.

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the answer is reflected or u can say repelled

a
(2) A 800 g block is pushed up an inclined plane (angled at 18°) with a velocity of 11.8 m/s. The first block slides up the
incline a distance of 2.2 m and strikes a second block with a mass of 300 g also moving at 3.4 m/s up the incline.
The two blocks hit and stick together. Determine the following:
(i) The maximum vertical height of the two blocks when they stop.
(ii) The time needed for the blocks to reach the bottom of the incline after the moment of impact.
(u = 0.19)

Answers

this is my attachment answer hope it's helpful to you

Solve this anyone please fast ..!​

Answers

Answer:

Explanation:

Ax= 20*sin37

Ay =20*cos37

Bx=15*cos40

By =  -15*sin40

Cx = -6*cos60

Cy = -6*sin60

Answer:

Ax = 12 cos 53

Ay = 12 sin 53

Bx = 15 cos 320 = 15 cos 40

By = 15 sin 320 = -15 sin 40

Cx = 6 cos 240 = -6 cos 60

Cy = 6 sin 240 = -6 sin 60

please help me please help​

Answers

Answer:

Answer Distance is 5 × 10⁴ km

Explanation:

Gravitational formular

[tex]{ \tt{F = \frac{GMm}{ {r}^{2} } }} \\ [/tex]

F is the gravitational forceG is the universal gravitational constantr is the separation distanceM & m are the masses

For the first case;

[tex]{ \tt{F _{1} = \frac{GMm}{2.5 \times {10}^{4} } }} \\ [/tex]

For the second case;

[tex]{ \tt{F _{2} = \frac{GMm}{r _{2}} }} \\ [/tex]

but F2 = ½F1

Therefore, F1 = 2F2

Hence:

[tex]{ \tt{ \frac{GMm}{2.5 \times {10}^{4} } =2 \times \frac{GMm}{r _{2} } }} \\ \\ { \tt{ \frac{1}{2.5 \times {10}^{4} } = \frac{2}{r _{2} } }} \\ \\ { \tt{r _{2} = 2 \times 2.5 \times {10}^{4} }} \\ \\ { \underline{ \tt{ \: \: r _{2} = 5 \times {10}^{4} \: km \: \: }}}[/tex]

Answer: Distance is 5 × 10⁴ km

Explanation:

ignore this i put the wrong grade sorry

Answers

Answer:

Oh its fine i am not in college but its still fine

Explanation:

what is the purpose of delivering medical aid through drones

Answers

Drones can help overcome transport challenges and delays in the delivery of small, low weight supplies, through the re-supply of essential medicines and delivery of medical diagnostic kits and return samples.

A long-distance runner runs at a constant speed of 4.8 m/s. How long does it take them to run 1.5 km?

Answers

Convert km to meters:

1km = 1000 m

1.5 km x 1000 = 1500 m

1500m / 4.8 m/s = 312.5 seconds

312.5 seconds / 60 seconds per minute =5.2 minutes = 5 minutes 12.5 seconds

What is the potential energy of a 50kg car on top of a 600m hill?

Answers

Answer:

294,000 J

Explanation:

The potential energy of a body can be found by using the formula

PE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 9.8 m/s²

From the question we have

PE = 50 × 9.8 × 600 = 294,000

We have the final answer as

294,000 J

Hope this helps you

A runner slows down after completing a race. Her deceleration is 0.25 m/s2. After 5 s she is travelling at 4 m/s, determine her initial velocity.​

Answers

Please find attached photograph for your answer.

Hope it helps.

Do comment if you have any query.

The initial velocity of the runner is 5.25 m/s.

What is deceleration?

You've probably seen that when there's a lot of traffic and there are more bikes blocking us, we tend to slow down the speed of our bikes.

Therefore, deceleration is defined as a reduction in speed as the body moves away from the beginning location. The opposite of acceleration is deceleration.

Given parameters:

Deceleration of the runner: a = 0.25 m/s².

Time interval: t = 5 second.

Final velocity: v = 4 m/s.

We have to find initial velocity of the runner: u = ?

From definition of deceleration, it can be written that:

deceleration = (initial velocity - final velocity) time interval

0.25 = (u -4)/5

u =  5.25 m/s.

Hence, initial velocity of the runner is 5.25 m/s.

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A car starts from rest and for 15.0 s its wheels has a constant linear acceleration of 0.800 m/s^2 to the right. What is the angle through which each wheel rotated in 20.0 s interval if the radius of the tires is 0.330 m?

Answers

237.375 rad. …………………
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