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Assoli18 [71]
2 years ago
15

I need to write an acrostic poem on Globalisation and i need help. Does anyone have any ideas?

Physics
1 answer:
Aleonysh [2.5K]2 years ago
5 0

Answer:

you could talk about the people and say something like change

Explanation:

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a 64kg skateboarder on a 2.0kg skateboard is on top of a ramp with a vertical height of 5.0 m what is the skateboarders maximum
WITCHER [35]
The formula for a kinetic energy KE of a falling body is
KE = mgh
where m = mass, g = acceleration due to gravity (9.8 m/s^2, constant), h = height.

The total mass of a skateboader and a skateboard is 64 + 2.0 = 66 kg.

Finally,
KE = 66*9.8*5.0 = 32340 J
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A sled is accelerating down a hill at a rate of 1 m s2 . If the mass of the sled is suddenly cut in half and the net force on th
mihalych1998 [28]
We have that F=ma from the 2nd Newton law where F is the force, m is the mass and a is the acceleration. Suppose we have that F' is the new force and m' is the new mass. Then, we have that a'=F'/m' still, by rearranging Newton's law. We are given that F'=2F and m'=m/2. Hence,
a'= \frac{2F}{ \frac{m}{2} } = \frac{4F}{m} = 4\frac{F}{m}
But now, we have from F=ma, that a=F/m and we are given that a=1m/s^2.
We can substitute thus, a'=4a=4*1m/s^2=4m/s^2.
4 0
2 years ago
What results from splitting an atomic nucleus
Burka [1]
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7 0
3 years ago
Read 2 more answers
Find the force necessary to pull a 6 kg object 3 m/s2
mars1129 [50]

Answer:

<h2>18 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 6 × 3

We have the final answer as

<h3>18 N</h3>

Hope this helps you

6 0
2 years ago
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
2 years ago
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