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Advocard [28]
3 years ago
14

How far will you travel if you walk for 10.0 minutes at a speed of 1.50 m/s?

Physics
1 answer:
fredd [130]3 years ago
7 0

Answer:

15

Explanation:

distance = speed × time.

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what happens when light travels from one medium to another with a different index of refraction at a zero degree angle of incide
zaharov [31]

Answer:

No Refraction takes places

Explanation:

When light travels from one medium to another at an angle of incidence 0 degree then no refraction takes place .

It is because according to Refraction law

                    n_{i}sini = n_{r}sinr

where n_{i} ,n_{r} are the refractive index of medium of incidence and medium of refraction .

When i = 0 , left side of the equation becomes 0 and thus r (angle of refraction) must be zero. So No refraction takes place

4 0
3 years ago
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F = 50 N<br> m = 72 kg<br> m/s2
lyudmila [28]

Answer:

Explanation:

F=ma

a=F/m

a=50/72=

a=0.694

3 0
3 years ago
A block of mass 20 kg sits on a ramp with an angle of 31 degrees above the horizontal. Assuming no friction, how fast will the b
Artyom0805 [142]

Answer:

2.98 m/s^2

Explanation:

I have done this before and it was a question on my physics test

4 0
3 years ago
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Can someone please help me? Will give brainliest.
lora16 [44]
Needs are less than resources, and the population increases.
5 0
3 years ago
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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
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