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Zielflug [23.3K]

2 years ago

13

Recent findings in astrophysics suggest that the observable universe can be modeled as a sphere of radius R = 13.7 × 109 light-y

ears = 13.0 × 1025 m with an average total mass density of about 1 × 10-26 kg/m3. Only about 4% of total mass is due to "ordinary" matter (such as protons, neutrons, and electrons). Part A Estimate how much ordinary matter (in kg) there is in the observable universe.

1 answer:

-BARSIC- [3]2 years ago

6 0

Answer:

$3.7\times 10^{51})$ kg

Explanation:

$R$ = radius of the sphere modeled as universe = $13\times 10^{25}$ m

Volume of sphere is given as

$V = \frac{4\pi R^{3}}{3}$

$V = \frac{4(3.14) (13\times 10^{25})^{3}}{3}$

$V = 9.2\times 10^{78}$ m³

$\rho$ = average total mass density of universe = $1\times 10^{-26}$ kg/m³

$m$ = Total mass of the universe = ?

We know that mass is the product of volume and density, hence

$m = \rho V$

$m = (1\times 10^{-26}) (9.2\times 10^{78})$

$m = 9.2\times 10^{52}$ kg

$M$ = mass of "ordinary" matter = ?

mass of "ordinary" matter is only about 4% of total mass, hence

$M = (0.04) m$

$M = (0.04)(9.2\times 10^{52})$

$M = 3.7\times 10^{51}$ kg

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A plane traveling north at 100.0 km/h through the air gets caught in a 40.0 km/h crosswind blowing west. This turbulence caused

mojhsa [17]

Answer:

Explanation:

If I assume that the wind did not cause the plane to chage its velocity.

The plane will have a velocity of vp = (0*i + 100*j) km/h relative to ground

The cart has a velocity of vc = (0*i - 20*j) km/h relative to the plane

vc' = vc + vp

vc' = (0*i + 100*j) + (0*i - 20*j) = (0*i + 80*j) km/h relative to the ground.

If I assume that the wind move the plane:

The plane will have a velocity of vp = (-40*i + 100*j) km/h relative to ground

The cart has a velocity of vc = (0*i - 20*j) km/h relative to the plane

vc' = vc + vp

vc' = (-40*i + 100*j) + (0*i - 20*j) = (-40*i + 80*j) km/h relative to the ground.

In reality the wind would move the plane a little, not to the full speed of the wind, somewhere between these two values, but without more data it cannot be calculated.

6 0

2 years ago

A wave travels through a medium at 251 m/s and has a wavelength of 5.10 cm. What is its frequency? What is its angular frequency

allochka39001 [22]

Explanation:

It is given that,

Speed of a wave, v = 251 m/s

Wavelength of the wave, λ = 5.1 cm = 0.051 m

(1) The frequency of the wave is given by :

$\nu=\dfrac{v}{\lambda}$

$\nu=\dfrac{251\ m/s}{0.051\ m}$

$\nu=4921.56\ Hz$

(2) Angular frequency of the wave is given by :

$\omega=2\pi\nu$

$\omega=2\pi\times 4921.56\ Hz$

$\omega=30923.07\ rad/s$

(3) The period of oscillation is given by T as :

$T=\dfrac{1}{\nu}$

$T=\dfrac{1}{4921.56}$

T = 0.000203 seconds

or

T = 0.203 milliseconds

Hence, this is the required solution.

5 0

2 years ago

 A bartender slides a beer mug at 1.50 m/s toward a customer at the end of a frictionless bar that is 1.20 m tall. The customer

Andrew [12]

Answer:

a) the mug hits the floor 0.7425m away from the end of the bar. b) |V|=5.08m/s θ= -72.82°

Explanation:

In order to solve this problem, we must first start by doing a drawing of the situation. (see attached picture).

a)

From the drawing we can see that we are dealing with a two dimensions movement problem. So in order to find out how far away from the bar the mug will fall, we need to start by finding how long it will take the mug to be in the air, so we analyze the vertical movement of the mug.

In order to find the time we need to use the following formula, which contains the data we know:

$y_{f}=y_{0}+v_{y0}t+\frac{1}{2}at^{2}$

we know that $y_{f}=0$ and that $v_{y0}=0$ as well, so the formula is simplified to:

$0=y_{0}+\frac{1}{2}at^{2}$

we can now solve this for t, so we get:

$-y_{0}=\frac{1}{2}at^{2}$

$-2y_{0}=at^{2}$

$\frac{-2y_{0}}{a}=t^{2}$

$t=\sqrt{\frac{-2y_{0}}{a}}$

we know that $y_{0}=1.20m$ and that $a=g=-9.8m/s^{2}$

the acceleration of gravity is negative because the mug is moving downwards. So we substitute them into the given formula:

$t=\sqrt{\frac{-2(1.20m)}{(-9.8m/s^{2})}}$

which yields:

t=0.495s

we can now use this to find the horizontal distance the mug travels. We know that:

$V_{x}=\frac{x}{t}$

so we can solve this for x, so we get:

$x=V_{x}t$

and we can now substitute the values we know:

x=(1.5m/s)(0.495s)

which yields:

x=0.7425m

b) Now that we know the time it takes the mug to hit the floor, we can use it to find the final velocity in the y-direction by using the following formula:

$a=\frac{v_{f}-v_{0}}{t}$

we know the initial velocity in the vertical direction is zero, so we can simplify the formula:

$a=\frac{v_{f}}{t}$

so we can solve this for the final velocity:

$V_{yf}=at$

in this case the acceleration is the same as the acceleration of gravity (which is negative) so we can substitute that and the time we found on the previous part to get:

$V_{yf}=(-9.8m/s^{2})(0.495s)$

which yields:

$V_{yf}=-4.851m/s$

so now we know the components of the final velocity, which are:

$V_{xf}=1.5m/s$ and $V_{yf]=-4.851m/s$

so now we can find the speed by determining the magnitude of the vector, like this:

$|V|=\sqrt{V_{x}^{2}+V_{y}^{2}}$

so we get:

$|V|=\sqrt{(1.5m/s)^{2}+(-4.851m/s)^{2}$

which yields:

|V|=5.08m/s

now, to find the direction of the impact, we can use the following equation:

$\theta = tan^{-1} (\frac{V_{y}}{V_{x}})$

so we get:

$\theta = tan^{-1} (\frac{-4.851m/s}{(1.5m/s)})$

which yields:

$\theta = -72.82^{o}$

4 0

2 years ago

A plane flying at 70.0 m/s suddenly stalls. If the acceleration during the stall is 9.8 m/s2 directly downward, the stall lasts

tino4ka555 [31]

Answer:

v = 66.4 m/s

Explanation:

As we know that plane is moving initially at speed of

$v = 70 m/s$

now we have

$v_x = 70 cos25$

$v_x = 63.44 m/s$

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$v_y = 29.6 m/s$

now in Y direction we can use kinematics

$v_y = v_i + at$

$v_y = 29.6 - (9.81 \times 5)$

$v_y = -19.5 m/s$

since there is no acceleration in x direction so here in x direction velocity remains the same

so we will have

$v = \sqrt{v_x^2 + v_y^2}$

$v = \sqrt{63.44^2 + 19.5^2}$

$v = 66.4 m/s$

4 0

2 years ago

A flywheel of mass M is rotating about a vertical axis with angular velocity ω0. A second flywheel of mass M/5 is not rotating a

Contact [7]

Answer:

0.83 ω

Explanation:

mass of flywheel, m = M

initial angular velocity of the flywheel, ω = ωo

mass of another flywheel, m' = M/5

radius of both the flywheels = R

let the final angular velocity of the system is ω'

Moment of inertia of the first flywheel , I = 0.5 MR²

Moment of inertia of the second flywheel, I' = 0.5 x M/5 x R² = 0.1 MR²

use the conservation of angular momentum as no external torque is applied on the system.

I x ω = ( I + I') x ω'

0.5 x MR² x ωo = (0.5 MR² + 0.1 MR²) x ω'

0.5 x MR² x ωo = 0.6 MR² x ω'

ω' = 0.83 ω

Thus, the final angular velocity of the system of flywheels is 0.83 ω.

5 0

2 years ago