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2 years ago

The synodic period for a planet is different from its sidereal period because

Physics

1 answer:

IrinaK [193]2 years ago

5 0

<span>Earth (and hence the observer) moves.</span>

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A book is thrown downward from the library window with a speed of 2.0\,\dfrac{\text m}{\text s}2.0 s m 2, point, 0, start frac

dem82 [27]

Answer: final Velocity v = 10.2m/s

Explanation:

Final speed v(t) is given as

v(t) = u + at .......1

Where; u = the initial speed

a = acceleration

t = time taken

The total distance travelled d is given as

d = ut + 1/2(at^2)

Given

d = 5.0m

u = 2.0m

a = g = 10m/s2 (acceleration due to gravity)

Substituting into the equation above we have

5 = 2t + 5t^2

5t^2 +2t -5 = 0

Applying the quadratic formula. We have;

t = 0.82s & t = -1.22s

t cannot be negative

t = 0.82s

From equation 1 above

v = 2.0m/s + 10(0.82)m/s

v = 10.2m/s

7 0

2 years ago

Read 2 more answers

A 9.0 volt battery is connected to four resistors in a parallel circuit. The resistors have 7.0 Ω, 5.0 Ω, 4.0 Ω, and 2.0 Ω respe

lidiya [134]

Series,effective resistance =R₁+R₂+R₃...
parallel,effective resistance 1/R=1/R₁ +1/R₂ +1/R₃...

Here,effective resistance 1/R =1/7 +1/5+ 1/4+1/2
=1.092
R = 1/1.092 =0.915Ω
voltage V=9 V
current I=V/R
I=9 / 0.915
=9.83 A

5 0

3 years ago

Read 2 more answers

The proper time between two events is measured by clocks at rest in a reference frame in which the two events: The proper time b

alekssr [168]

Answer:

a - As long as the time between 2 events is reconcilable with a light signal, the time between the events, in that frame, can be determined.

8 0

2 years ago

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Which of Newton's laws

Virty [35]

Answer:

Newton's 2nd law think soo

8 0

2 years ago

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A thin double convex glass lens with an index of 1.56 while surrounded by air has a 10 cm focal length. If it is placed under wa

bearhunter [10]

Explanation:

Formula which holds true for a leans with radii $R_{1}$ and $R_{2}$ and index refraction n is given as follows.

$\frac{1}{f} = (n - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

Since, the lens is immersed in liquid with index of refraction $n_{1}$ . Therefore, focal length obeys the following.

$\frac{1}{f_{1}} = \frac{n - n_{1}}{n_{1}} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

$\frac{1}{f(n - 1)} = [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

and, $\frac{n_{1}}{f(n - n_{1})} = \frac{1}{R_{1}} - \frac{1}{R_{2}}$

or, $f_{1} = \frac{fn_{1}(n - 1)}{(n - n_{1})}$

$f_{w} = \frac{10 \times 1.33 \times (1.56 - 1)}{(1.56 - 1.33)}$

= 32.4 cm

Using thin lens equation, we will find the focal length as follows.

$\frac{1}{f} = \frac{1}{s_{o}} + \frac{1}{s_{i}}$

Hence, image distance can be calculated as follows.

$\frac{1}{s_{i}} = \frac{1}{f} - \frac{1}{s_{o}} = \frac{s_{o} - f}{fs_{o}}$

$s_{i} = \frac{fs_{o}}{s_{o} - f}$

$s_{i} = \frac{32.4 \times 100}{100 - 32.4}$

= 47.9 cm

Therefore, we can conclude that the focal length of the lens in water is 47.9 cm.

4 0

2 years ago