In the circuit shown in the figure above, suppose that the value of R1 is 100 k ohms and the value of R2 is 470 k ohms. At which of the following locations in the circuit would you measure the highest voltage with your meter? A. Between points A and B
B. Between points A and C
C. Between points B and E
D. Between points B and C

Answers

Answer 1

Answer:

Answer:

Explanation:

voltage between A and C is equal battery's voltage.

Related Questions

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An object essentially at infinity is moved to a distance of 90 cm in front of a thin positive lens. In the process its image distance triples. Determine the focal length of the lens.

Answers

Answer:

67.5 cm

Explanation:

u = - 90 cm, v = 3 x u = 3 x 90 = 270 cm

let f be the focal length

Use lens equation

1 / f = 1 / v - 1 / u

1 / f = 1 / 270 + 1 / 90

1 / f = 4 / 270

f = 67.5 cm

Final answer:

To determine the focal length of the lens, we use the lens formula and set up an equation based on the given information. Solving for the image distance, we find that it is zero, indicating the image is formed at infinity. Therefore, the focal length of the lens is 90 cm.

Explanation:

To determine the focal length of the lens, we can use the lens formula:

1/f = 1/v - 1/u

Where f is the focal length, v is the image distance, and u is the object distance.

Given that the image distance triples when the object is moved from infinity to 90 cm in front of the lens, we can set up the following equation:

1/f = 1/(3v) - 1/(90)

Multiplying through by 90*3v, we get:

90*3v/f = 270v - 90*3v

90*3v/f = 270v - 270v

90*3v/f = 0

Simplifying further, we find that: v = 0

When the image distance is zero, it means the image is formed at infinity, so the lens is focused at the focal point. Therefore, the focal length of the lens is 90 cm.

Light from a sodium lamp (λ = 589 nm) illuminates two narrow slits. The fringe spacing on a screen 150 cm behind the slits is 4.0 mm. What is the spacing (in mm) between the two slits?

Answers

The spacing between the two slits is 0.221mm.

The spacing between the two slits is given as,

$D=(\lambda L)/(y)$

Where $\lambda$ is wavelength, y is fringe spacing and L is length of screen.

Given that, $\lambda=589nm,L=150cm,y=4mm$

Substitute in above equation.

$D=(589*10^(-9)*150*10^(-2) )/(4*10^(-3) )\n \nD=2.21*10^(-4) m\n\nD=0.221mm$

Hence, the spacing between the two slits is 0.221mm.

Learn more about the sodium lamp here:

brainly.com/question/24867424

A quarterback claims that he can throw the football a horizontal distance of 167 m. Furthermore, he claims that he can do this by launching the ball at the relatively low angle of 33.1 ° above the horizontal. To evaluate this claim, determine the speed with which this quarterback must throw the ball. Assume that the ball is launched and caught at the same vertical level and that air resistance can be ignored. For comparison a baseball pitcher who can accurately throw a fastball at 45 m/s (100 mph) would be considered exceptional.

Answers

Answer:u=42.29 m/s

Explanation:

Given

Horizontal distance=167 m

launch angle $=33.1^(\circ)$

Let u be the initial speed of ball

Range $=(u^2\sin 2\theta )/(g)$

$167=(u^2\sin (66.2))/(9.8)$

$u^2=1788.71$

$u=√(1788.71)$

$u=42.29 m/s$

Light with a wavelength of 495 nm is falling on a surface and electrons with a maximum kinetic energy of 0.5 eV are ejected. What could you do to increase the maximum kinetic energy of electrons to 1.5 eV?

Answers

Answer:

To increase the maximum kinetic energy of electrons to 1.5 eV, it is necessary that ultraviolet radiation of 354 nm falls on the surface.

Explanation:

First, we have to calculate the work function of the element. The maximum kinetic energy as a function of the wavelength is given by:

$K_(max)=(hc)/(\lambda)-W$

Here h is the Planck's constant, c is the speed of light, $\lambda$ is the wavelength of the light and W the work function of the element:

$W=(hc)/(\lambda)-K_(max)\nW=((4.14*10^(-15)eV\cdot s)(3*10^8(m)/(s)))/(495*10^(-9)m)-0.5eV\nW=2.01eV$

Now, we calculate the wavelength for the new maximum kinetic energy:

$W+K_(max)=(hc)/(\lambda)\n\lambda=(hc)/(W+K_(max))\n\lambda=((4.14*10^(-15)eV\cdot s)(3*10^8(m)/(s)))/(2.01eV+1.5eV)\n\lambda=3.54*10^(-7)m=354*10^(-9)m=354nm$

This wavelength corresponds to ultraviolet radiation. So, to increase the maximum kinetic energy of electrons to 1.5 eV, it is necessary that ultraviolet radiation of 354 nm falls on the surface.

A cat leaps to catch a bird. If the cat's jump was at 60.0° off the ground and its initial velocity was point of its trajectory? 0.30 m 3.44 m/s, what is the highest O 13.76 m 0.45 m 0.90 m

Answers

Answer:

The highest of its trajectory = 0.45 m

Option C is the correct answer.

Explanation:

Considering vertical motion of cat:-

Initial velocity, u = 3.44 sin60 = 2.98 m/s

Acceleration , a = -9.81 m/s²

Final velocity, v = 0 m/s

We have equation of motion v² = u² + 2as

Substituting

v² = u² + 2as

0² = 2.98² + 2 x -9.81 x s

s = 0.45 m

The highest of its trajectory = 0.45 m

Option C is the correct answer.

A driver drives for 30.0 minutes at 80.0 km/h, then 45.0 minutes at 100 km/h. She then stops 30 minutes for lunch. She then travels for 30 minutes at 80 km/h. (a) Sketch a plot of her displacement versus time and speed versus time. (b) Calculate her average speed.

Answers

Answer:

b) 68,9 km/h a) picture

Explanation:

In this problem, since velocity is expressed in km/h and time in minutes, we have to convert either time to hours or velocity to km/min. It is easier to use hours.

Using this formula we pass time to hours:

$t_(hours)=t_(min)*(1 h)/(60 min)\n30min*(1 h)/(60 min)=0,5h\n45min*(1 h)/(60 min)=0,75h$

Now we can plot speed vs time (image 1). The problem says that the driver uses constant speed, so all lines have to be horizontal.

Using the values of the speed we calculate the distance in each interval

$d=v*t\n80km/h*0.5h=40km\n100km/h*0.75h=75km$

Using these values and the fact that she was having lunch in the third one (therefore stayed in the same position), we plot position vs time, using initial position zero (image 2, distance is in km, not meters).

Finally, we compute the average speed with the distance over time:

$v_(average)=(155km)/(2.25h)=68.9km/h$

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