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3 years ago

10

A pool can be filled by a pipe in 3 hours. it takes 3 times as long for another pipe to empty the pool. how long will it take to

fill the pool if both pipes are open?

1 answer:

Mila [183]3 years ago

7 0

There is need 4.5 hours to fill the pool in case if both pipes are open

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What is the wavelength of an earthquake wave if it has a speed of 6 km/s and a frequency of 13 Hz?

PtichkaEL [24]

Velocity can be calculated using the following rule:
velocity = lambda * frequency where:
velocity is given = 6 km/sec
lambda is the wavelength that we want to calculate
frequency is given = 13 Hz

Substitute with the givens in the above equation to get the wavelength (lambda) as follows:
6 = 13*lambda
wavelength = 6/13 = 0.461538 Km = 461.538 meters

5 0

3 years ago

Compare the gravitational acceleration on the following objects compared to the Sun using:

arsen [322]

The gravitational acceleration of White dwarf compared to Sun is 13,675.86.

The gravitational acceleration of Neutron star compared to Sun is 6.79 x 10⁻²⁴.

The gravitational acceleration of Star Betelgeuse compared to Sun is 8.5 x 10¹⁰.

<h3>Mass of the planets</h3>

Mass of sun = 2 x 10³⁰ kg

Mass of white dwarf = 2.765 x 10³⁰ kg

Mass of Neutron star = 5.5 x 10¹² kg

Mass of star Betelgeuse = 2.188 x 10³¹ kg

<h3>Radius of the planets</h3>

Radius of sun = 696,340 km

Radius of white dwarf = 7000 km

Radius of Neutron star = 11 km

Radius of star Betelgeuse = 617.1 x 10⁶ km

<h3>Gravitational acceleration of White dwarf compared to Sun</h3>

$\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{2.765 \times 10^{30}}{2\times 10^{30}} \times [\frac{696,340,000}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 13,675.86$

<h3>Gravitational acceleration of Neutron star compared to Sun</h3>

$\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{5.5 \times 10^{12}}{2\times 10^{30}} \times [\frac{11,000}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 6.79\times 10^{-24}$

<h3>Gravitational acceleration of Star Betelgeuse compared to Sun</h3>

$\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{2.188 \times 10^{31}}{2\times 10^{30}} \times [\frac{617.1 \times 10^9}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 8.5\times 10 ^{10}$

Learn more about acceleration due to gravity here: brainly.com/question/88039

3 0

3 years ago

How to find the magnitude of a centripetal force?<br>(formula)

frozen [14]

Answer:

F=mv^2÷r

Explanation:

i know every thing

the magnitude f of the centripetal force is equal to the mass m of the body times it veloctiy squared v^2 divided by the radius r of its path

4 0

3 years ago

A car of mass 650 kg slows from a speed of 20.0 m/s to a speed of 10.0 m/s. How much kinetic energy has it lost. How much force

Natalija [7]

Formulas you need for this problem:
F= mass•acceleration
KE= mass•velocity^/2
Acceleration= final velocity-intial velocity/time
time= distance/speed

t= 97.5\10= 9.75 seconds
Acc= 10-20/9.75= -1.03 m/s/s

These are the answers :)
F=650•-1.03= -669.5N
KE= 650•100/2= 32,500J or 32.5KJ

7 0

4 years ago

On a distance-versus-time graph, _____.

babymother [125]

Answer:

A

Explanation:

The slope of the graph has the units of vertical axis divided by horizontal axis. This means that the slope of a distance vs time graph is distance/time, or velocity.

Slope is calculated by "rise over run" so C is incorrect.

Since the slope represents velocity, a constant slope equates to a constant velocity, hence B is incorrect. Same reasoning for D being incorrect: if the slope is zero, the object is not moving.

3 0

3 years ago