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2 years ago

How many stars are in a galaxy? hundreds thousands millions billions.

Physics

1 answer:

faust18 [17]2 years ago

7 0

Answer:

Billions :)

hope this helps! <3

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How much heat is released when 35kg of water freezes?

inn [45]

You need an additional point of data here: the enthalpy of fusion, or conversely the enthalpy of melting (they differ only by their sign). For water (or ice) that value is gotten from sources such as the internet

ΔH°(fus) = 6.01 kJ/mole 

Since you have 35 000g, how many moles do you have? 
Moles H2O = 35000 g/(18.015 g/mole) = 1942.8 moles

So, take that ΔH°(fus) in kJ/mole, multiply by the number of moles, and there ya go!

6.01 x 1942.8 = 11,676 kJ of energy is released

Hope I helped!! xx

3 0

2 years ago

Your grandmother in Ireland sends you her favorite cookie recipe. Her instructions say to bake the cookies at

Lapatulllka [165]

$F=\frac{9}{5}C+32$

$F=\frac{9}{5}190.5C+32$

$F=342.9+32$

$F=374.9$

$374.9F=190.5C$

3 0

3 years ago

Read 2 more answers

An ideal spring hangs from the ceiling. A 2.15 kg mass is hung from the spring, stretching the spring a distance d = 0.0895 m fr

Igoryamba

Answer:

The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

Explanation:

Given that,

Mass = 2.15 kg

Distance = 0.0895 m

Amplitude = 0.0235 m

We need to calculate the spring constant

Using newton's second law

$F= mg$

Where, f = restoring force

$kx=mg$

$k=\dfrac{mg}{x}$

Put the value into the formula

$k=\dfrac{2.15\times9.8}{0.0895}$

$k=235.41\ N/m$

We need to calculate the kinetic energy of the mass

Using formula of kinetic energy

$K.E=\dfrac{1}{2}mv^2$

Here, $v = A\omega$

$K.E=\dfrac{1}{2}m\times(A\omega)^2$

Here, $\omega=\sqrt{\dfrac{k}{m}}^2$

$K.E=\dfrac{1}{2}m\times A^2\sqrt{\dfrac{k}{m}}^2$

$K.E=\dfrac{1}{2}kA^2$

Put the value into the formula

$K.E=\dfrac{1}{2}\times235.41\times(0.0235)^2$

$K.E=0.06500\ J$

Hence, The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

8 0

3 years ago

If the frog lands with a velocity equal to its average velocity and comes to a full stop 0.25s later, what is the frog’s average

Nina [5.8K]

Answer:

Average accelation = -4V

Explanation:

$a=\frac{V-V0}{t}$

V=0 m/s (because the frog stopped)

V0 = V (average velocity)

t= 0,25 s

So;

$a=\frac{V-V0}{t}=\frac{0-V}{0.25}=-4V$

4 0

2 years ago

What is the kinetic energy of a 1.40 kg discus with a speed of 22.5 m/s?

Oksana_A [137]

Kinetic energy = (1/2) (mass) (speed)²

   = (1/2) (1.4 kg) (22.5 m/s)²

   = (0.7 kg) (506.25 m²/s² )

   = 354.375 kg-m²/s² = 354.375 joules .

This is just the kinetic energy associated with a 1.4-kg glob of
mass sailing through space at 22.5 m/s. In the case of a frisbee,
it's also spinning, and there's some additional kinetic energy stored
in the spin.

8 0

3 years ago