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3 years ago

How many sets of planets would you need to create the mass of the Sun?

Physics

1 answer:

kow [346]3 years ago

6 0

Answer:

volume about 1.3 million Earths could fit inside the Sun the mass of the Sun is 1.989 X 10 to the 30 kg about 333000 times the mass of the Earth

Explanation:

step-by-step explanation hope this answer your question

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A pirate fires his cannon parallel to the water but 3.5 m above the water. The cannonball leaves the cannon with a velocity of 1

Bumek [7]

Answer:

202.8m

Explanation:

Given that A pirate fires his cannon parallel to the water but 3.5 m above the water. The cannonball leaves the cannon with a velocity of 120 m/s. He misses his target and the cannonball splashes into the briny deep.

First calculate the total time travelled by using the second equation of motion

h = Ut + 1/2gt^2

Let assume that u = 0

And h = 3.5

Substitute all the parameters into the formula

3.5 = 1/2 × 9.8 × t^2

3.5 = 4.9t^2

t^2 = 3.5/4.9

t^2 = 0.7

t = 0.845s

To know how far the cannonball travel, let's use the equation

S = UT + 1/2at^2

But acceleration a = 0

T = 2t

T = 1.69s

S = 120 × 1.69

S = 202.834 m

Therefore, the distance travelled by the cannon ball is approximately 202.8m.

4 0

2 years ago

A 0.40 kg mass hangs on a spring with a spring constant of 12 N/m. The system oscillated with a constant amplitude of 12 cm. Wha

Vaselesa [24]

Answer:

The maximum acceleration of the system is 359.970 centimeters per square second.

Explanation:

The motion of the mass-spring system is represented by the following formula:

$x(t) = A\cdot \cos (\omega \cdot t + \phi)$

Where:

$x(t)$ - Position of the mass with respect to the equilibrium position, measured in centimeters.

$A$ - Amplitude of the mass-spring system, measured in centimeters.

$\omega$ - Angular frequency, measured in radians per second.

$t$ - Time, measured in seconds.

$\phi$ - Phase, measured in radians.

The acceleration experimented by the mass is obtained by deriving the position equation twice:

$a (t) = -\omega^{2}\cdot A \cdot \cos (\omega\cdot t + \phi)$

Where the maximum acceleration of the system is represented by $\omega^{2}\cdot A$ .

The natural frequency of the mass-spring system is:

$\omega = \sqrt{\frac{k}{m} }$

Where:

$k$ - Spring constant, measured in newtons per meter.

$m$ - Mass, measured in kilograms.

If $k = 12\,\frac{N}{m}$ and $m = 0.40\,kg$ , the natural frequency is:

$\omega = \sqrt{\frac{12\,\frac{N}{m} }{0.40\,kg} }$

$\omega \approx 5.477\,\frac{rad}{s}$

Lastly, the maximum acceleration of the system is:

$a_{max} = \left(5.477\,\frac{rad}{s})^{2}\cdot (12\,cm)$

$a_{max} = 359.970\,\frac{cm}{s^{2}}$

The maximum acceleration of the system is 359.970 centimeters per square second.

7 0

3 years ago

What does compressed mean?

jeka57 [31]

Answer:

flattened by pressure; squeezed or pressed together.

Explanation:

7 0

3 years ago

Read 2 more answers

A horizontal force of 400 N is exerted on a 2.0-kg ball as it rotates (at

frutty [35]

Answer:

the speed of the ball is 10 m/s

Explanation:

Given;

magnitude of exerted force, F = 400 N

mass of the ball, m = 2 kg

radius of the circle, r = 0.5

The speed of the ball is calculated by applying centripetal force formula;

$F = \frac{mv^2}{r} \\\\v^2 = \frac{Fr}{m}\\\\v = \sqrt{\frac{Fr}{m}}\\\\ v = \sqrt{\frac{400*0.5}{2}}\\\\v = 10 \ m/s$

Therefore, the speed of the ball is 10 m/s

6 0

3 years ago

Read 2 more answers

The force diagram represents a girl pulling a sled with a mass of 6.0 kg to the left with a force of 10.0 N at a 30.0 degree ang

Vilka [71]

Normal Force = 54 N
acceleration = 1.2 m/s^2

For Normal Force:
According to the force diagram, we can come up with the equation (all up and down forces):

10 sin 30 + Normal Force - 58.8 = 0
Normal Force = 53.8 N = 54 N

For acceleration:
According to the force diagram, we can come up with the equation (all left and right forces):

10 cos 30 - 1.5 = (6.0) (Acceleration)
Acceleration = 1.19 m/s^2 = 1.2 m/s^2

3 0

3 years ago

Read 2 more answers