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2 years ago

Blank is the the decrease in a materail's volume when it's blank decrease

Physics

1 answer:

natulia [17]2 years ago

7 0

Answer:

Explanation:

becAUSE IT POOOPYES

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What would be your estimate of the age of the universe if you measured a value for Hubble's constant of H0 = 30 km/s/Mly ? You c

Maksim231197 [3]

Answer:

The age of the universe would be 9.9 billion years

Explanation:

We can calculate an estimate for the age of the Universe from Hubble's Law. Let's suppose the distance between two galaxies is D and the apparent velocity with which they are separating from each other is v. At some point, the galaxies were touching, and we can consider that time the moment of the Big Bang.

Thus, the time it has taken for the galaxies to reach their current separations is:

$\displaystyle{t=D/v}$

and from Hubble's Law:

$v =H_0D$

Therefore:

$\displaystyle{t=D/v=D/(H_0\times D)=1/H_0}$

With the given value for the Hubble's constant we have:

$H_0=(30\ km/s/Mly) \times (1 Mly/ 9.461 \times 10^{18} km) = 3.17\times 10^{-18}\ 1/s$

and thus,

$t=1/H_0 = 1/(3.17\times 10^{-18} 1/s) = 0.315 \times 10^{18}\ s \approx 9988584474.8858\ years \approx 9.9\ billion\ years$

6 0

3 years ago

2.(Ramp section) Suppose the height of the ramp is h1= 0.40m, and the foot of the ramp is horizontal, and is h2= 1.5m above the

frozen [14]

Answer:

a) the distance that the solid steel sphere sliding down the ramp without friction is 1.55 m

b) the distance that a solid steel sphere rolling down the ramp without slipping is 1.31 m

c) the distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping is 1.2 m

d) the distance that a solid aluminum sphere rolling down the ramp without slipping is 1.31 m

Explanation:

Given that;

height of the ramp h1 = 0.40 m

foot of the ramp above the floor h2 = 1.50 m

assuming R = 15 mm = 0.015 m

density of steel = 7.8 g/cm³

density of aluminum = 2.7 g/cm³

a) distance that the solid steel sphere sliding down the ramp without friction;

we know that

distance = speed × time

d = vt --------let this be equ 1

according to the law of conservation of energy

mgh₁ = $\frac{1}{2}$ mv²

v² = 2gh₁

v = √(2gh₁)

from the second equation; s = ut + $\frac{1}{2}$ at²

that is; t = √(2h₂/g)

so we substitute for equations into equation 1

d = √(2gh₁) × √(2h₂/g)

d = 2√( h₁h₂ )

we plug in our values

d = 2√( 0.40 × 1.5 )

d = 1.55 m

Therefore, the distance that the solid steel sphere sliding down the ramp without friction is 1.55 m

distance that a solid steel sphere rolling down the ramp without slipping;

we know that;

mgh₁ = $\frac{1}{2}$ mv² + $\frac{1}{2}$ $I_{}$ ω²

mgh₁ = $\frac{1}{2}$ mv² + $\frac{1}{2}$ ( $\frac{2}{5}$ mR²) ω²

v = √( $\frac{10}{7}$ gh₁ )

so we substitute √( $\frac{10}{7}$ gh₁ ) for v and t = √(2h₂/g) in equation 1;

d = vt

d = √( $\frac{10}{7}$ gh₁ ) × √(2h₂/g)

d = 1.69√( h₁h₂ )

we substitute our values

d = 1.69√( 0.4 × 1.5 )

d = 1.31 m

Therefore, the distance that a solid steel sphere rolling down the ramp without slipping is 1.31 m

distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping;

we know that;

mgh₁ = $\frac{1}{2}$ mv² + $\frac{1}{2}$ $I_{}$ ω²

mgh₁ = $\frac{1}{2}$ mv² + $\frac{1}{2}$ ( $\frac{2}{3}$ mR²) ω²

v = √( $\frac{6}{5}$ gh₁ )

so we substitute √( $\frac{6}{5}$ gh₁ ) for v and t = √(2h₂/g) in equation 1 again

d = vt

d = √( $\frac{6}{5}$ gh₁ ) × √(2h₂/g)

d = 1.549√( h₁h₂ )

d = 1.549√( 0.4 × 1.5 )

d = 1.2 m

Therefore, the distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping is 1.2 m

d) distance that a solid aluminum sphere rolling down the ramp without slipping.

we know that;

mgh₁ = $\frac{1}{2}$ mv² + $\frac{1}{2}$ $I_{}$ ω²

mgh₁ = $\frac{1}{2}$ mv² + $\frac{1}{2}$ ( $\frac{2}{5}$ mR²) ω²

v = √( $\frac{10}{7}$ gh₁ )

so we substitute √( $\frac{10}{7}$ gh₁ ) for v and t = √(2h₂/g) in equation 1;

d = vt

d = √( $\frac{10}{7}$ gh₁ ) × √(2h₂/g)

d = 1.69√( h₁h₂ )

we substitute our values

d = 1.69√( 0.4 × 1.5 )

d = 1.31 m

Therefore, the distance that a solid aluminum sphere rolling down the ramp without slipping is 1.31 m

8 0

3 years ago

In a physics laboratory experiment, a coil with 200 turns enclosing an area of 13.1 cm2 is rotated during the time interval 3.10

sergij07 [2.7K]

Answer:

A) $\Phi=83.84\times 10^{-9}$

B) $\Phi=0 Wb$

C) $emf=5.4090\times 10^{-4}V$

Explanation:

Given that:

no. of turns i the coil, $n=200$

area of the coil, $a=13.1 \times 10^{-4}\,m^2$

time interval of rotation, $t=3.1\times 10^{-2}\,s$

intensity of magnetic field, $B=6.4\times 10^{-5}\,T$

(A)

Initially the coil area is perpendicular to the magnetic field.

So, magnetic flux is given as:

$\Phi=B.a\,cos \theta$ ..................................(1)

$\theta$ is the angle between the area vector and the magnetic field lines. Area vector is always perpendicular to the area given. In this case area vector is parallel to the magnetic field.

$\Phi=6.4\times 10^{-5}\times 13.1 \times 10^{-4}\, cos 0^{\circ}$

$\Phi=83.84\times 10^{-9} Wb$

(B)

In this case the plane area is parallel to the magnetic field i.e. the area vector is perpendicular to the magnetic field.

∴ $\theta=90^{\circ}$

From eq. (1)

$\Phi=6.4\times 10^{-5}\times 13.1 \times 10^{-4}\, cos 90^{\circ}$

$\Phi=0 Wb$

(C)

According to the Faraday's Law we have:

$emf=n\frac{B.a}{t}$

$emf=\frac{200\times 6.4\times 10^{-5}\times 13.1 \times 10^{-4}}{3.1\times 10^{-2}}$

$emf=5.4090\times 10^{-4}V$

7 0

3 years ago

Suppose you were bungee jumping from a bridge while blowing a hand-held air horn. How would someone remaining on the bridge hear

Genrish500 [490]

Answer:

The pitch will progressively lower

Explanation:

If i were bungee jumping from a bridge while blowing a hand-held air horn and someone who remains on the bridge will hear a decreased pitch or frequency as the source is moving away from the stationary listener as per the Dooplers effect. Hence, the pitch will progressively lower as the source is moving away from the observer.

4 0

3 years ago

Read 2 more answers

This is a two part question: What is the force of friction acting on a 100 kg steel slab at rest on a steel floor?

Crazy boy [7]

1) 0N... friction opposes the motion of an object, since the block is at rest there is no motion thus no friction

2) F=ma
= (5.5kg)(30m/s)
=165 N

7 0

2 years ago

Blank is the the decrease in a materail's volume when it's blank decrease ​

Blank is the the decrease in a materail's volume when it's blank decrease