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Inessa [10]
3 years ago
6

If you could travel with speed of light (3.0e5 km/s) it would take only 2.5 minutes to reach venus. how far is venus from earth

in km?
Physics
1 answer:
motikmotik3 years ago
7 0
D=S/T ,S=3.0e5km/s × 2.5min×60
D=4.5×e6km
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An object initially at rest accelerates at 5 meters per second2 until it attains a speed of 30 meters per second. What distance
Ipatiy [6.2K]

Answer:

Distance s  = 90 meter

Explanation:

Given:

Initial velocity u = 0 m/s

Final velocity v = 30 m/s

Acceleration a = 5 m/s²

Find:

Distance s

Computation:

v² = u² + 2as

30² = 0² + 2(5)(s)

Distance s  = 90 meter

5 0
3 years ago
Which of the following is an example of energy changing from one form to another?
Anastaziya [24]

The correct option is

a fire


In fact, a fire is a conversion of chemical energy (contained in the molecules of the initial substances) into thermal energy (the heat released by the fire itself), therefore this is an example of energy changing from one form to another. All the other objects, instead, do not represent any form of energy conversion.

6 0
3 years ago
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Which are characteristics of fungi? Check all that apply.
Inessa05 [86]

they are eukaryotes that have cell walls, are heterotrophs that feed by absorbing their food, and use spores to reproduce.

3 0
2 years ago
They are called short-period comets. Short-period comets are thought to originate in the Kuiper Belt, which is a belt of icy obj
yaroslaw [1]

Answer:

Do you still need this answer?

Explanation:

3 0
3 years ago
Sometimes a person cannot clearly see objects close up or far away. To correct this type of vision, bifocals are often used. The
Rudik [331]

Answer:

1)   P₁ = -2 D,   2) P₂ = 6 D

Explanation:

for this exercise in geometric optics let's use the equation of the constructor

          \frac{1}{f} = \frac{1}{p} + \frac{1}{q}

where f is the focal length, p and q are the distance to the object and the image, respectively

1) to see a distant object it must be at infinity (p = ∞)

          \frac{1}{f_1} = \frac{1}{q}

           q = f₁

2) for an object located at p = 25 cm

            \frac{1}{f_2} = \frac{1}{25} + \frac{1}{q}

We can that in the two expressions we have the distance to the image, this is the distance where it can be seen clearly in general for a normal person is q = 50 cm

we substitute in the equations

1) f₁ = -50 cm

2)  

        \frac{1}{f_2} = \frac{1}{25} + \frac{1}{50}

        \frac{1}{f_2} = 0.06

         f₂ = 16.67 cm

the expression for the power of the lenses is

          P = \frac{1}{f}

where the focal length is in meters

           

1)       P₁ = 1/0.50

        P₁ = -2 D

2)     P₂ = 1 /0.16667

        P₂ = 6 D

4 0
2 years ago
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