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3 years ago

How many galaxies are present in the observable universe?

Physics

2 answers:

strojnjashka [21]3 years ago

6 0

Answer:

100 billion i think

Explanation:

marissa [1.9K]3 years ago

5 0

There are at least 2 trillion galaxies present in the observable universe.

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Suppose you hit a 0.058-kg tennis ball so that the ball then moves with an acceleration of 10 m/s2. If you were to hit a basketb

stiks02 [169]

Answer:

1 m/s²

Explanation:

Force = mass × acceleration

F = ma ............ Equation 1

Where F = force, m = mass, a = acceleration.

Given: m = 0.058 kg, a = 10 m/s²

Substitute into equation 1

F = 0.058(10)

F = 0.58 N.

If the same force was used to hit the baseball,

F = m'a

a = F/m'.............. Equation 2

Where M' = mass of the baseball.

Given: F = 0.58 N, m' = 0.58 kg.

Substitute into equation 2

a = 0.58/0.58

a = 1 m/s²

8 0

3 years ago

Read 2 more answers

A mother and father have two daughters. The daughters have some similar and some different physical features. Which statement be

disa [49]

C. The daughters received a random set of genes from both of their parents

4 0

3 years ago

Read 2 more answers

5 What is the maximum speed at which a car round a curve of 25m radius on a level road if the coefficient of static friction bet

pshichka [43]

Hi there!

On a level road:

∑F = Ff (Force due to friction)

The net force is the centripetal force, so:

mv²/r = Ff

Rewrite the force due to friction:

mv²/r = μmg

Cancel out the mass:

v²/r = μg

Solve for v:

v = √rμg

v = √(25)(9.81)(0.8) = 14.01 m/s

8 0

2 years ago

The altitude of the International Space Station ttt minutes after its perigee (closest point), in kilometers, is given by \qquad

Dmitriy789 [7]

Answer:

T = 92.8 min

Explanation:

Given:

The altitude of the International Space Station t minutes after its perigee (closest point), in kilometers, is given by:

$A(t) = 415 - sin(\frac{2*\pi (t+23.2)}{92.8})$

Find:

- How long does the International Space Station take to orbit the earth? Give an exact answer.

Solution:

- Using the the expression given we can extract the angular speed of the International Space Station orbit:

$A(t) = 415 - sin({\frac{2*\pi*t }{92.8} + \frac{23.2*2*\pi }{92.8} )$

- Where the coefficient of t is angular speed of orbit w = 2*p / 92.8

- We know that the relation between angular speed w and time period T of an orbit is related by:

T = 2*p / w

T = 2*p / (2*p / 92.8)

Hence, T = 92.8 min

7 0

3 years ago

The membrane that surrounds a certain type of living cell has a surface area of 5.1 x 10-9 m2 and a thickness of 1.4 x 10-8 m.

ziro4ka [17]

a) The charge on the outer surface is $1.2\cdot 10^{-12} C$

b) The number of ions is $7.5\cdot 10^6$

Explanation:

The membrane behaves as a parallel plate capacitor, whose capacity is given by the equation

$C=\frac{k\epsilon_0 A}{d}$

where

k = 4.3 is the dielectric constant

$\epsilon_0 =8.85\cdot 10^{-12} F/m$ is the vacuum permittivity

$A=5.1\cdot 10^{-9} m^2$ is the surface area

$d=1.4\cdot 10^{-8} m$ is the distance between the two plates

Substituting,

$C=\frac{(4.3)(8.85\cdot 10^{-12})(5.1\cdot 10^{-9})}{1.4\cdot 10^{-8}}=1.4\cdot 10^{-11} F$

The capacity of the membrane is related to the potential difference between the two surfaces by

$C=\frac{Q}{\Delta V}$

where here we have

Q = excess charge on one surface

$\Delta V = 85.5 mV = 0.0855 V$ is the potential difference between the two surfaces

Solving for Q, we find

$Q=C\Delta V=(1.4\cdot 10^{-11})(0.0855)=1.2\cdot 10^{-12} C$

We said that the net charge on the outer surface is

$Q=1.2\cdot 10^{-12} C$

The charge of one K+ ions is equal to the electron charge

$+e=1.6\cdot 10^{-19} C$

Therefore, the number of ions on the outer surface can be found by dividing the total charge by the charge of a single ion:

$N=\frac{Q}{e}=\frac{1.2\cdot 10^{-12}}{1.6\cdot 10^{-19}}=7.5\cdot 10^6$

Learn more about capacity:

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6 0

3 years ago