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3 years ago

I need help on this question

Physics

2 answers:

Mamont248 [21]3 years ago

4 0

Mabye the bottom one?

iVinArrow [24]3 years ago

3 0

Answer:

Explanation:

less dense water rising and more dense water sinking

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A lens is used to make an image of magnification -1.50. The image is 30.0 cm from the lens. What is the object’s distance (unit=

photoshop1234 [79]

Answer:

The object distance is 20cm

Explanation:

Given

Magnification = -1.5

Image distance = 30 cm.

Required

Object Distance

We can calculate the object's distance using magnification formula;

M = -V/U

Where M = Magnification = -1.50

V = Image Distance = 30cm

U = Object Distance

Substitute the above parameters in the formula above.

-1.50 = -30/U

Multiply both sides by -1

-1.50 * -1 = -30/U * -1

1.50 = 30/U

Multiply both sides by U

1.50 * U = 30/U * U

1.50U = 30

Divide through by 1.50

1.50U/1.50 = 30/1.50

U = 30/1.50

U = 20cm

Recall that U represented the object distance.

Hence, the object distance is 20cm

5 0

4 years ago

Can someone help? Thanks.

Lilit [14]

The average speed of Frank's car is 2 m/s and the average speed of Noah's car is 1.8 m/s.

<u>Explanation:</u>

Average speed is the measure of total distance covered at different time period. Since, the formula used for calculating the average speed is the ratio of total distance covered by each car to the time taken to cover that distance.

As there is only one set of data i.e., distance and time, the average speed will be equal to the speed of the car.

So in this case, the total distance covered by franks car is 300 cm = 3 m in 1.5 s. Then, the average speed will be

$Avg. S_{frank} = \frac{Total\ distance\ rolled\ by\ his\ car}{Time\ taken}$

$Avg.S_{frank} =\frac{300 \times 10^{-2} }{1.5}=2\ m/s$

Similarly, the average speed of Noahs car which rolled a distance of 360 cm = 3.6 m in time 2 s, will be

$Avg.S_{Noah} =\frac{360 \times 10^{-2} }{2}=1.8\ m/s$

Thus, the average speed of Frank's car is 2 m/s and the average speed of Noah's car is 1.8 m/s.

4 0

4 years ago

Monochromatic light is incident on a pair of slits that are separated by 0.220 mm. The screen is 2.60 m away from the slits. (As

Naddik [55]

Answer:

$\lambda = 1.667 nm$

$\theta = 0.8681^o$

Explanation:

From the question we are told that

The distance of separation is $d = 0.220 \ mm = 0.00022 \ m$

The is distance of the screen from the slit is $D = 2.60 \ m$

The distance between the central bright fringe and either of the adjacent bright $y = 1.97 cm = 1.97 *10^{-2}\ m$

Generally the condition for constructive interference is

$d sin \tha(\theta ) = n \lambda$

From the question we are told that small-angle approximation is valid here.

So $sin (\theta ) = \theta$

=> $d \theta = n \lambda$

=> $\theta = \frac{n * \lambda }{d }$

Here n is the order of maxima and the value is n = 1 because we are considering the central bright fringe and either of the adjacent bright fringes

Generally the distance between the central bright fringe and either of the adjacent bright is mathematically represented as

$y = D * sin (\theta )$

From the question we are told that small-angle approximation is valid here.

$y = D * \theta$

=> $\theta = \frac{ y}{D}$

$\frac{n * \lambda }{d } = \frac{y}{D}$

$\lambda =\frac{d * y }{n * D}$

substituting values

$\lambda = \frac{0.00022 * 1.97*10^{-2} }{1 * 2.60 }$

$\lambda = 1.667 *10^{-6}$

$\lambda = 1.667 nm$

In the b part of the question we are considering the next set of bright fringe so n= 2

Hence

$dsin (\theta ) = n \lambda$

$\theta = sin^{-1}[\frac{ n * \lambda }{d} ]$

$\theta = sin^{-1}[\frac{ 2 * 1667 *10^{-9}}{ 0.00022} ]$

$\theta = 0.8681^o$

7 0

4 years ago

Average running speed is 8km/h. Convert this to m/s :

rosijanka [135]

Answer:

Im not sure if you mean miles per hour or meters per hour but if it is miles it is 4.97 miles per hour and if it is meters then it is 8000 meters per hour

Explanation:

Sorry if I'm wrong

6 0

3 years ago

Read 2 more answers

Under normal conditions, what color do we expect diamonds to be?

Ann [662]

Under normal lighting conditions, most diamonds appear to be colorless the naked eye. Many consumers think that all diamonds are colorless, however, in reality, they come in a wide range of colors. Actualcolor differences are very subtle.

8 0

3 years ago