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3 years ago

What are the two principle advantages of telescopes over eyes?

Physics

1 answer:

Stells [14]3 years ago

8 0

Answer:

i) Telescopes can be used to view far distant objects but the human eye can't view far distant objects.

ii) Telescopes uses two convex lenses producing a magnified image while the human eye only possesses one convex lens (image seen are smaller than that viewed under telescopes)

Explanation:

The telescopes can be used to view far distant objects due to their presence of two convex lenses. The two convex lenses are the objective lens (lens closer to object) and the eye piece lens (lens closer to eye). The object to be viewed forms an intermediate image first before the final image is seen using the eye piece lens.

The human eye only possess one convex lens and as such cannot view far ranged objects.

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What effect does the vertical acceleration have on the horizontal velocity of the projectile?

KengaRu [80]

Answer:
None, if air resistance is ignored.

Explanation:
At any instant, the projectile has vertical and horizontal components of velocity.
Vertical acceleration due to gravity affects the vertical velocity by accelerating the object toward the center of the earth, and by decreasing the upward vertical velocity..
The horizontal component of velocity makes the object travel horizontally as long as the projectile is airborne.
Thsi discussion assumes that air resistance is ignored.

3 0

3 years ago

______________ are chemicals in the body affecting how you feel.

7nadin3 [17]

Answer:

emotions

Explanation:

emotions are how you feel and can happen any time

Hope it helps <333

3 0

2 years ago

Read 2 more answers

A top-fuel dragster starts from rest and has a constant acceleration of 42.0 m/s2. What are (a) the final velocity of the dragst

disa [49]

Answer:

a) Final velocity of the dragster at the end of 1.8 s = 75.6 m/s

b) Final velocity of the dragster at the end of 3.6 s = 151.2 m/s

c) The displacement of the dragster at the end of 1.8 s = 68.04 m

d) The displacement of the dragster at the end of 3.6 s = 272.16 m

Explanation:

a) We have equation of motion v = u + at

Initial velocity, u = 0 m/s

Acceleration , a = 42 m/s²

Time = 1.8 s

Substituting

v = u + at

v = 0 + 42 x 1.8 = 75.6 m/s

Final velocity of the dragster at the end of 1.8 s = 75.6 m/s

b) We have equation of motion v = u + at

Initial velocity, u = 0 m/s

Acceleration , a = 42 m/s²

Time = 3.6 s

Substituting

v = u + at

v = 0 + 42 x 3.6 = 75.6 m/s

Final velocity of the dragster at the end of 3.6 s = 151.2 m/s

c) We have equation of motion s= ut + 0.5 at²

Initial velocity, u = 0 m/s

Acceleration , a = 42 m/s²

Time = 1.8 s

Substituting

s= ut + 0.5 at²

s = 0 x 1.8 + 0.5 x 42 x 1.8²

s = 68.04 m

The displacement of the dragster at the end of 1.8 s = 68.04 m

d) We have equation of motion s= ut + 0.5 at²

Initial velocity, u = 0 m/s

Acceleration , a = 42 m/s²

Time = 3.6 s

Substituting

s= ut + 0.5 at²

s = 0 x 3.6 + 0.5 x 42 x 3.6²

s = 272.16 m

The displacement of the dragster at the end of 3.6 s = 272.16 m

3 0

3 years ago

During lightning strikes from a cloud to the ground, currents as high as 2.50×10^4 Amps can occur and last for about 40.0 micros

dangina [55]

Answer:

1 C

Explanation:

The intensity of electric current is defined as

$I=\frac{q}{t}$

where

I is the current

q is the amount of charge transferred

t is the time interval during which the charge is transferred

For the lightning in this problem, we have

$I=2.50\cdot 10^4 A$ is the current

$t=40.0 \mu s = 40.0\cdot 10^{-6} s$ is the time interval

Solving the formula for q, we find the amount of charge transferred:

$q=I t = (2.50\cdot 10^4 A)(40.0\cdot 10^{-6}s)=1 C$

6 0

3 years ago

I'm not really sure how to go about creating the equation, can anyone help me?

AlexFokin [52]

The displacement vector (SI units) is
$\vec{r} =At\hat{i}+A[t^{3}-6t^{2}]\hat{j}$

The speed is a scalar quantity. Its magnitude is
$v= \sqrt{A^{2}t^{2}+A^{2}(t^{3}-6t^{2})^{2}} \\ v=A \sqrt{t^{2}+t^{6}-12t^{5}+36t^{4}} \\ v=At \sqrt{t^{4}-12t^{3}+36t^{2}+1}$

Answer: At√(t⁴ - 12t³ + 36t² + 1)

3 0

3 years ago