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BartSMP [9]
3 years ago
12

a new movie earn 114000000 in its opening weekend.express this amount in(a) gigadollars and (b) teradollars

Physics
1 answer:
forsale [732]3 years ago
3 0

The earnings of movie in its opening weekend = 114000000$

We have

       1 kilo = 10^3

       1 Mega = 10^6

       1 Giga = 10^9

       1 Tera = 10^{12}

       1 Peta = 10^{15}

       1 Exa = 10^{18}

       1 Zetta = 10^{21}

       1 Yotta = 10^{24}

a) We have 1 Giga = 10^9 and earnings of movie in its opening weekend = 114000000$

Earnings = 0.114*10^9 $ =0.114 Giga dollars

b) We have 1 Tera = 10^{12} and earnings of movie in its opening weekend = 114000000$

Earnings = 0.000114*10^{12}$ =0.000114 Tera dollars

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zheka24 [161]

Answer:

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6 0
3 years ago
wo ships, one 200200 metres in length and the other 100100 metres in length, travel at constant but different speeds. When trave
shutvik [7]

Answer:

The speed of the faster ship is 22.5 m/s

Explanation:

The length of the ships are;

Ship 1 = 200 m

Ship 2 = 100 m

The time it takes for the ships to completely cross each other when travelling in opposite directions = 10 seconds

The time it takes both ships to cross each other when travelling in the same direction = 25 seconds

Let x represent the speed of the first ship, ship 1, and y represent the speed of the second ship, ship, 2, we have;

(x + y) × 10 = 200 + 100 = 300

10·x + 10·y = 300...(1)

(x - y) × 20 = 200 + 100 = 300

20·x - 20·y = 300...(2)

Multiply equation (1) by 2, to get;

(x + y) × 10 × 2 = 300 × 2

20·x + 20·y = 600...(3)

Adding equation (1) to equation (3) gives;

20·x + 20·y + 20·x - 20·y = 600 + 300

40·x = 900

x = 900/40 = 22.5

x = 22.5

The speed of the first ship, ship 1 = x = 22.5 m/s

From equation (1), we have;

10·x + 10·y = 300

y = (300 - 10·x)/10 = (300 - 10×22.5)/10 = 7.5

y = 7.5

The speed of the second ship, ship 2 = y = 7.5 m/s

Therefore, the faster ship is ship 1 with a speed of 22.5 m/s

7 0
3 years ago
Alfredo leaves camp and, using a compass, walks 4 km E, then 6 km S, 3 km E, 5 km N, 10 km W, 8 km N, and, finally, 3 km S. At t
kari74 [83]

Answer: 3km E, 1km N

Explanation:

4E + 6S + 3E + 5N + 10W + 8N + 3S= 3W, 1S

3W, 1S= 3E, 1N

8 0
2 years ago
Vector A has a magnitude of 16 m and makes an angle of 44° with the positive x axis. Vector B also has a magnitude of 13 m and i
marshall27 [118]

Answer with explanation:

The given vectors in are reduced to their componednt form as shown

For vector A it can be written as

\overrightarrow{v}_{a}=16cos(44^{o})\widehat{i}+16sin(44^{o})\widehat{j}

Similarly vector B can be written as

\overrightarrow{v}_{b}=-13\widehat{i}

Hence The sum and difference is calculated as

\overrightarrow{v}_{a}+\overrightarrow{v}_{b}=16cos(44^{o})\widehat{i}+16sin(44^{o})\widehat{j}+(-13\widehat{i})\\\\\overrightarrow{v}_{a}+\overrightarrow{v}_{b}=(16cos(44^{o})-13)\widehat{i}+16sin(44^{o})\widehat{j}\\\\\therefore \overrightarrow{v}_{a}+\overrightarrow{v}_{b}=-1.49\widehat{i}+11.11\widehat{j}\\\\\therefore |\overrightarrow{v}_{a}+\overrightarrow{v}_{b}|=\sqrt{(-1.49)^{2}+11.11^{2}}=11.21m

The direction is given by

\theta =tan^{-1}\frac{r_{y}}{r_{x}}\\\\\theta =tan^{-1}\frac{11.11}{-1.49}=97.64^{o}with positive x axis.

Similarly

\overrightarrow{v}_{a}-\overrightarrow{v}_{b}=16cos(44^{o})\widehat{i}+16sin(44^{o})\widehat{j}-(-13\widehat{i})\\\\\overrightarrow{v}_{a}-\overrightarrow{v}_{b}=(16cos(44^{o})+13)\widehat{i}+16sin(44^{o})\widehat{j}\\\\\therefore \overrightarrow{v}_{a}-\overrightarrow{v}_{b}=24.51\widehat{i}+11.11\widehat{j}\\\\\therefore |\overrightarrow{v}_{a}-\overrightarrow{v}_{b}|=\sqrt{(24.51)^{2}+11.11^{2}}=26.91m

The direction is given by

\theta =tan^{-1}\frac{r_{y}}{r_{x}}\\\\\theta =tan^{-1}\frac{11.11}{24.51}=24.38^{o}with positive x axis.

3 0
3 years ago
A bullet is shot vertically upward with an initial velocity of 128 ft/s. The bullet's height after t seconds is y(t) = 128t - 16
saul85 [17]

The height of the bullet when the velocity is zero is 256 ft.

<h3>Height of the bullet when the velocity is zero </h3>

The height of the bullet when the velocity is zero is determined by taking derivative of the function as shown below;

v = \frac{dy}{dt} = 128 -32t \\\\when \ v \ is \ zero\\\\v = 0\\\\128 - 32t = 0\\\\32t = 128\\\\t = \frac{128}{32} \\\\t = 4 \  s

The height of the bullet at this time is calculated as follows;

y(4) = 128(4) - 16(4)^2\\\\y(4) = 256 \ ft

Learn more about height of projectiles here: brainly.com/question/10008919

6 0
2 years ago
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