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Bess [88]
3 years ago
15

Does someone know how to do math with that equation

Physics
1 answer:
mestny [16]3 years ago
8 0

Answer:

f = 5 cm

Explanation:

using the thin lens equation, given as follows:

\frac{1}{f} = \frac{1}{d_{o}}+\frac{1}{d_{i}}\\

where,

f = focal length = ?

do = the distance of object from lens = 20 cm

di = the distance of image from lens = 6.6667 cm

Therefore,

\frac{1}{f} = \frac{1}{20\ cm}+\frac{1}{6.6667\ cm}\\\\\frac{1}{f} =  0.199999\ cm^{-1}\\\\f = \frac{1}{0.199999\ cm^{-1}}\\\\

<u>f = 5 cm</u>

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Using Excel, or some other graphing software, plot the values of y as a function of x. (You will not submit this spreadsheet. Ho
Evgesh-ka [11]

Answer:

a) > x<-c(1,2,3,4,5)

> y<-c(1.9,3.5,3.7,5.1,6)

> linearmodel<-lm(y~x)

And the output is given by:

> linearmodel

Call:

lm(formula = y ~ x)

Coefficients:

(Intercept)            x  

      1.10         0.98  

b) y = 0.98 x +1.10

And if we compare this with the general model y = mx +b

We see that the slope is m= 0.98 and the intercept b = 1.10

Explanation:

Part a

For this case we have the following data:

x: 1,2,3,4,5

y: 1.9,3.5,3.7,5.1, 6

For this case we can use the following R code:

> x<-c(1,2,3,4,5)

> y<-c(1.9,3.5,3.7,5.1,6)

> linearmodel<-lm(y~x)

And the output is given by:

> linearmodel

Call:

lm(formula = y ~ x)

Coefficients:

(Intercept)            x  

      1.10         0.98  

Part b

For this case we have the following trend equation given:

y = 0.98 x +1.10

And if we compare this with the general model y = mx +b

We see that the slope is m= 0.98 and the intercept b = 1.10

7 0
3 years ago
You are in the forest with some of your friends. You’re being chased by a very angry and hungry bear.
melomori [17]

Answer:

2m head start or else you done for

Explanation:

you cant even out run a bear they run at 35mph the fastest human is 25

8 0
2 years ago
a ship travels a port p and travels 30 km due north. then it changes course and travels 20 km in a direction  30° east of north
liq [111]

When we represent what is given to us on a coordinate plane, we have a figure as shown in the attachment.

To find the distance between P and R, we have to find the Net Displacement of the ship (brown arrow in the figure).

For that, we use the rules for Vector addition.

We see that the first displacement D_{1} = 30 km (blue arrow) is along the y-axis, but the second part of the ship's journey D_{2} = 20 km (red arrow) is at an angle with reference to y-axis.

So, we first find the components of the red arrow along X and Y.

Component of D_{2} along X-axis is given by  D_{2x}  = D_{2} Sin 30 = 10 km

Component of D_{2} along Y-axis is given by  D_{2y}  = D_{2} Cos 30 = 17.32 km

We now add all the vectors along X and along Y separately.

Net Displacement along X  D_{netX} = 10 km

Net Displacement along Y D_{netY} = 30 + 17.32 = 47.32 km

Now that we have the components of the net displacement along X and Y, we make use of Pythagorean Theorem to calculate the D_{net}

D_{net}  = \sqrt{D_{netX} ^{2} + D_{netY} ^{2}}

Therefore, [tex]D_{net} = 48.37 km.

Hence, the distance between the ports P and R is 48 km.

6 0
3 years ago
A ball is kicked horizontally at 4.6 m/s off of a cliff 13.4 m high. How far from the cliff will it land.
Mekhanik [1.2K]
<h3>Answer:</h3>

7.53 m

<h3>Explanation:</h3>

<u>We are given:</u>

Initial Horizontal Velocity of the Ball = 4.6 m/s

Initial Vertical Velocity of the Ball = 0 m/s

Height from which ball is kicked = 13.4 m

<u>Time taken by the ball to reach the ground:</u>

The ball has an initial vertical velocity of 0 m/s

it also has a downward acceleration of 10 m/s² due to gravity

<u>Solving for the time taken:</u>

s = ut + 1/2(at²)                 [second equation of motion]

replacing the values

13.4 = (0)(t) + 1/2 (10)(t²)

13.4 = 5t²

t² = 13.4/5                  [dividing both sides by 5]

t² = 2.68

t = 1.637 seconds     [taking the square root of both sides]

<u>Horizontal distance covered by the ball:</u>

Since there are no horizontal opposing forces on the ball,

the ball will more horizontally at a velocity of 4.6 m/s until it hits the ground

We calculated that the ball will hit the ground in 1.637 seconds

<u>Distance covered:</u>

s = ut + 1/2 (at²)                            [seconds equation of motion]

s = ut                                            [since a = 0m/s² in the horizontal plane]

replacing the values

s = 4.6 * 1.637

s = 7.53 m

Hence, the ball landed 7.53 m from the cliff

5 0
2 years ago
If the mass of a material is 120 grams and the volume of the material is 23 cm3, what would the density of the material be?
oee [108]
The density is 5.22 g/cm³.

Density=\frac{mass}{volume}
Density=\frac{120 g}{23 cm {3} }
Density= 5.22 g/cm³
4 0
3 years ago
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