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ss7ja [257]
3 years ago
9

A dog accelerates at 1.50 m/s2 to reach a velocity of 13.5 m/s while covering a distance of 49.3 m. What was his initial velocit

y?
Physics
1 answer:
vovangra [49]3 years ago
7 0

Let <em>v</em> be the dog's initial velocity. Then

(13.5 m/s)^2 - <em>v</em> ^2 = 2 (1.50 m/s^2) (49.3 m)

==>   <em>v</em> ^2 = (13.5 m/s)^2 - 2 (1.50 m/s^2) (49.3 m)

==>   <em>v</em> = √((13.5 m/s)^2 - 2 (1.50 m/s^2) (49.3 m))

==>   <em>v</em> ≈ 5.86 m/s

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Two converging lenses are placed 30 cm apart. The focal length of the lens on the right is 20 cm while the focal length of the l
Masja [62]

Answer:

a)   I2 = 3 (o-10) / (o- 30) , b)   h ’/h=  3 (o-10) / o (o-30)

Explanation:

The builder's equation is

          1 / f = 1 / o + 1 / i

Where f is the focal length, or e i are the distance to the object and image, respectively

As the separation between the lenses is greater than the focal distances, we must work them individually and separately. Let's start with the leftmost lens with focal length f = 15 cm

Let's calculate the position of the image of this lens

         1 / i1 = 1 / f - 1 / o

         1 / i1 = 1/15 - 1 / o

         i1 = o 15 / (o-15)

Let's calculate the distance to the image of the second lens, for this the image of the first is the distance to the object of the second

        o2 = d-i1

We write the builder equation

       1 / f2 = 1 / o2 + 1 / i2

       1 / i2 = 1 / f2 -1 / o2

       1 / i2 = 1 / f2 - 1 / (d-i1)

       1 / i2 = 1/20 - 1 / (d-i1)            (1)

Let's evaluate the last term

      d-i1 = d - 15 o / (o-15)

      d-i1 = (d (o-15) - 15 o) / (o-15)

      d- i1 = (30 or -30 15 -15 o) / (o-15)

      d-i1 = (15 or - 450) / (o- 15)

      d-i1 = = (15 or -450) / (o-15)

replace in 1

      1 / i2 = 1/20 - (or - 15) / (15 or -450)

      1 / i2 = [(15 o-450) - (o-15) 20] / (15 or -150)

      1 / i2 = (15 or - 450 - 20 or + 300) / (15 or - 150)

      1 / i2 = (-5 or -150) / (15 or -150)

      1 / i2 = (or -30) / (3 or - 30)

      I2 = 3 (o-10) / (o- 30)

Part B

The height of the image, we use the magnification equation

     m = h ’/ h = - i / o

     h ’= - h i / o

In our case

     h ’= h i2 / o

     h ’= h 3 (o-10) / o (o-30)

If they give the distance to the object it is easier

5 0
3 years ago
For adults, the RDA of the amino acid lysine is 12 mg per kg of body weight. How many grams per day should a 78kg adult receive?
kari74 [83]
RDA stands for Recommended Daily Allowance. To determine the amount needed of a certain adult per day, we simply multiply the mass of the adult to the value of RDA. For this case, we do as follows:

Daily requirement = 12 mg/kg ( 78 kg ) = 936 mg of lysine = 936000 g
6 0
3 years ago
two masses are separated by 1 m. suppose the masses are moved so they are 2 m apart how will the gravitational force change
Verdich [7]
The gravitational force would get stronger because the farther the two masses are separated the more gravitational force will be used to pull them together the closer they are the less gravitational pull is used to pull them together
5 0
3 years ago
Samantha is checking the weather for her upcoming trip to Mexico City. The weather forecast predicts a high-pressure system for
Zanzabum

Calm, sunny days with wind moving away from the center.

5 0
3 years ago
Read 2 more answers
Four blocks of weights are required using which any body whose weight is between 1kg and 40 kg can be weighed. Find the four wei
Vikentia [17]

Answer:

The weights are 1 kg, 3kg, 9kg and 27kg.

Explanation:

The weights are 1 kg, 3kg, 9kg and 27kg.

1+3+9+27= 40

27+9+3= 39

27+9+3-1=38

27+9+1=37

27+9=36

27+9-1=35

27+9+1-3=34

27+9-3=33

27+9-3-1=32

27+3+1=31

27+3=30

27+3-1=29

27+1=28

27

27-1=26

27+1-3=25

27-3=24

27-3-1=23

27+3+1-9=22

27+3-9=21

27+3-9-1=20

Like this all the weights from 1 to 40 kg can be made using 1,3,9 and 27 kg.

6 0
3 years ago
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