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2 years ago

Answer these questions about a light year.?

Physics

2 answers:

aliina [53]2 years ago

4 0

These are five questions and five answers:

a. State the speed in m/sec using scientific notation

i) Find the place value of the digit with the highest place value (the digit most to the left) in the number 300,000,000: hundred millions.

ii) Determine the number of zeros corresponding to that place value: 9

ii) Write the digit with the highest place value, mulitplied by the power of ten with exponent equal to the number of zeros corresponding to the place of such digit less 1 (9 - 1 = 8): 3 × 10⁸

iii) Complete with the units: 3 × 10⁸ m/s ← answer

b. State the speed in km/sec using scientific notation

i) State the conversion factor: 1 km = 1000 m = 10³ m

⇒ 1 = 1 km / 10³ m

ii) Multiply by the conversion factor:

3 × 10⁸ m/s × 1 km / (10³ m)

iii) Simplify (cancel the units that are common in the numerator and denominator)

3 × (10⁸ / 10³) km/s = 3 × 10⁵ km/s ← answer

c. State the number of seconds in one year using scientific notation

i) Find the place value of the digit with the highest place value (the digit most to the left) in the number 32,000,000: ten millions.

ii) Determine the number of zeros corresponding to that place value: 8

ii) Write the digit with the highest place value, as integer, add the other significant figures as decimals, mulitplied by the power of ten with exponent equal to the number of zeros corresponding to the place of the digit with highest place value less 1 (8 - 1 = 7): 3.2 × 10⁷

iii) Complete with the units: 3.2 × 10⁷ s ← answer

d. Calculate the distance in meters of one light year

i) distance formula: distance = speed × time

ii) Replace with the numbers in scientific notation:

distance = 3 × 10⁸ m/s × 3.2 × 10⁷ s

iii) Simplify (due the operations and cancel the units that are common in the numerator and denominator)

3 × 10⁸ m/s × 3.2 × 10⁷ s = 9.6 × 10 ¹⁵ m ← answer

e. State this distance in centimeters

i) State the conversion factor: 1m = 100 cm = 10² cm

⇒ 1 = 10² cm / 1 m

ii) Multiply by the conversion factor: 9.6 × 10 ¹⁵ m × 10² cm / 1 m

iii) Simplily: 9.6 × 10⁷ cm ← answer

FrozenT [24]2 years ago

4 0

A. state the speed in m/sec using scientific notation

3.0 x 10^8 m/sec

b. state the speed in km/sec using scientific notation

3.0 x 10^5 km/sec

c. state the number of seconds in one year using scientific notation

<span> 3.2 x 10^7 sec
</span>
d. calculate the distance in meters of one light year

1 light year = <span>9.4605284 x 10^15 meters
</span>
e. state this distance in centimeters

9.4605284 x 10^12<span> centimeters</span>

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The average velocity of an object over 6.0 seconds interval is 2 m/s what is the total distance traveled and M by the object doi

lina2011 [118]

Answer:

<h2>The answer is 12 m</h2>

Explanation:

The distance covered by an object given it's velocity and time taken can be found by using the formula

distance = velocity × time

From the question we have

distance = 2 × 6

We have the final answer as

Hope this helps you

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2 years ago

A projectile is launched at an angle of 30 and lands 20 s later at the same height as it was launched. (a) What is the initial s

Pavlova-9 [17]

Answer:

(a) 196 m/s

(b) 490 m

(c) 3394.82 m

(d) 2572.5 m

Explanation:

First of all, let us know one thing. When an object is thrown in the air, it experiences two forces acting in two different directions, one in the horizontal direction called air resistance and the second in the vertically downward direction due to its weight. In most of the cases, while solving numerical problems, air resistance is neglected unless stated in the numerical problem. This means we can assume zero acceleration along the horizontal direction.

Now, while solving our numerical problem, we will discuss motion along two axes according to our convenience in the course of solving this problem.

<u>Given:</u>

Time of flight = t = 20 s
Angle of the initial velocity of projectile with the horizontal = $\theta = 30^\circ$

<u>Assume:</u>

Initial velocity of the projectile = u
R = Range of the projectile during the time of flight
H = maximum height of the projectile
D = displacement of the projectile from the initial position at t = 15 s

Let us assume that the position from where the projectile was projected lies at origin.

Initial horizontal velocity of the projectile = $u\cos \theta$
Initial horizontal velocity of the projectile = $u\sin \theta$

Part (a):

During the time of flight the displacement of the projectile along the vertical is zero as it comes to the same vertical height from where it was projected.

$\therefore u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow u\sin \theta t=\dfrac{1}{2}(g)t^2\\\Rightarrow u=\dfrac{gt^2}{2\sin \theta t}\\\Rightarrow u=\dfrac{9.8\times 20^2}{2\sin 30^\circ \times 20}\\\Rightarrow u=196\ m/s$

Hence, the initial speed of the projectile is 196 m/s.

Part (b):

For a projectile, the time take by it to reach its maximum height is equal to return from the maximum height to its initial height is the same.

So, time taken to reach its maximum height will be equal to 10 s.

And during the upward motion of this time interval, the distance travel along the vertical will give us maximum height.

$\therefore H = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow H = 196\times \sin 30^\circ \times 10 + \dfrac{1}{2}\times(-9.8)\times 10^2\\ \Rightarrow H =490\ m$

Hence, the maximum altitude is 490 m.

Part (c):

Range is the horizontal displacement of the projectile from the initial position. As acceleration is zero along the horizontal, the projectile is in uniform motion along the horizontal direction.

So, the range is given by:

$R = u\cos \theta t\\\Rightarrow R = 196\times \cos 30^\circ \times 20\\\Rightarrow R =3394.82\ m$

Hence, the range of the projectile is 3394.82 m.

Part (d):

In order to calculate the displacement of the projectile from its initial position, we first will have to find out the height of the projectile and its range during 15 s.

$\therefore h = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow h = 196\times \sin 30^\circ \times 15 + \dfrac{1}{2}\times(-9.8)\times 15^2\\ \Rightarrow h =367.5\ m\\r = u\cos \theta t\\\Rightarrow r = 196\times \cos 30^\circ \times 15\\\Rightarrow r =2546.11\ m\\\therefore D = \sqrt{r^2+h^2}\\\Rightarrow D = \sqrt{2546.11^2+367.5^2}\\\Rightarrow D =2572.5\ m$

Hence, the displacement from the point of launch to the position on its trajectory at 15 s is 2572.5 m.

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2 years ago

You have a string with a mass of 0.0133 kg. You stretch the string with a force of 8.89 N, giving it a length of 1.97 m. Then, y

Kazeer [188]

Answer:

(i) The wavelength is 0.985 m

(ii) The frequency of the wave is 36.84 Hz

Explanation:

Given;

mass of the string, m = 0.0133 kg

tensional force on the string, T = 8.89 N

length of the string, L = 1.97 m

Velocity of the wave is:

$V = \sqrt{\frac{F_T}{M/L} } \\\\V = \sqrt{\frac{8.89}{0.0133/1.97} } \ = 36.29 \ m/s$

(i) The wavelength:

Fourth harmonic of a string with two nodes, the wavelength is given as,

L = 2λ

λ = L/2

λ = 1.97 / 2

λ = 0.985 m

(ii) Frequency of the wave is:

v = fλ

f = v / λ

f = 36.29 / 0.985

f = 36.84 Hz

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