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lesantik [10]
3 years ago
14

From because when held inhands stmah 3.in an experiment to determine the density of Sand using a density bottle, the following m

easurements were recorded: Mass of empty of density bottle=43.2g Mass of density bottle full of water=66.4g Mass of density bottle with some sand =67.5g Mass of density bottle with sand, filled up with water=82.3g Use the above data to determine the: a) mass of the water that filled the bottle 1mks b) Volume of the water that filled the bottle. 2mk c) Volume of the density bottle Imk. d) Mass of sandlmk e) Mass of water that filled the space above the sand lmk • • • f) Volume of the sand 3mk Density of the sand​

Physics
1 answer:
BARSIC [14]3 years ago
4 0

What argument does Minow make in his speech?

A. Parents should demand better programs for their children to watch in the evenings.

B. People watch too much television, which leads to boredom and violence.

C. Television executives have a responsibility to provide better programming.

D. The nation's children depend on television to entertain and educate them.

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You are the science officer on a visit to a distant solar system. Prior to landing on a planet you measure its diameter to be 1.
babymother [125]

Answer:

Mass of the planet = 1.48 × 10²⁵ Kg

Mass of the star = 5.09 × 10³⁰ kg

Explanation:

Given;

Diameter = 1.8 × 10⁷ m

Therefore,

Radius = \frac{\textup{Diameter}}{\textup{2}}  = \frac{\textup{1.8}\times10^7}{\textup{2}}

or

Radius of the planet = 0.9 × 10⁷ m

Rotation period = 22.3 hours

Radius of star = 2.2 × 10¹¹ m

Orbit period = 407 earth days = 407 × 24 × 60 × 60 seconds = 35164800 s

free-fall acceleration = 12.2 m/s²

Now,

we have the relation

g = \frac{\textup{GM}}{\textup{R}^2}

g is the free fall acceleration

G is the gravitational force constant

M is the mass of the planet

on substituting the respective values, we get

12.2 = \frac{6.67\times10^{-11}\times M}{(0.9\times10^7)^2}

or

M = 1.48 × 10²⁵ Kg

From the Kepler's law we have

T² = \frac{\textup{4}\pi^2}{\textup{G}M_{star}}(R_{star})^3

on substituting the respective values, we get

35164800² = \frac{\textup{4}\pi^2}{6.67\times10^{-11}\timesM_{star}}(2.2\times10^{11})^3

or

M_{star} = 5.09 × 10³⁰ kg

6 0
3 years ago
A bee beats its wings approximately 184 times per second. what is its wave period?
sergejj [24]

Answer: It's 0.00543

4 0
3 years ago
Read 2 more answers
After creating a question and forming a hypothesis, what is the next step in the scientific method
Keith_Richards [23]
The next step is to create research
3 0
3 years ago
A long-distance runner is running at a constant speed of 5 m/s.
vivado [14]

Answer:

3.33 minutes (3 minutes and 20 seconds)

Explanation:

Speed of the runner = s = 5 m/s

We need to calculate how will it take for runner to complete 1 km. We have the speed, the distance and we need to find the time. Before performing any calculations, we must convert the values to same units.

Speed is in m/s and distance is in kilometers. So we have to either convert speed to km/s or distance into meters. In this case, converting distance into meters would be a convenient option.

1 kilo meters = 1000 meters

The distance, speed and time are related by the equation:

Distance = Speed x Time

So,

Time = Distance/Speed

Using the values, we get:

t = 1000/5

t = 200 seconds

This means, the runner can complete 1 kilometers in 200 seconds. Since, there are 60 seconds in a minute, we can convert this time to minutes, by dividing it by 60. i.e.

200 \text{ sec} = \frac{200}{60} \text{ min} = 3.33 \text{ min}

Thus, it will take the runner 3.33 minutes (3 minutes and 20 seconds) to travel 1 km.

3 0
3 years ago
During a baseball game, a batter hits a pop-up to a fielder 93 m away.The acceleration of gravity is 9.8 m/s2.If the ball remain
velikii [3]

Explanation:

It is given that, a batter hits a pop-up to a fielder 93 m away, range of the projectile, R = 93 m

The ball remains in the air for 5.4 s, the time of flight is 5.4 s

Time of flight : T=\dfrac{2v\sin\theta}{g}

5.4=\dfrac{2v\sin\theta}{g}\\\\v\sin\theta=\dfrac{5.4\times 9.8}{2}\\\\v\sin\theta=26.46

Maximum height of the projectile : H=\dfrac{v^2\sin^2\theta}{2g}

We need to find H.

So,

H=\dfrac{(v\sin\theta)^2}{2g}\\\\H=\dfrac{(26.46)^2}{2\times 9.8}\\\\H=35.72\ m

So, it will rise to a height of 35.72 m.

6 0
3 years ago
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