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3 years ago

SC.912.P.12.10

Physics

1 answer:

Korolek [52]3 years ago

8 0

Answer:

Explanation:

You can see the 1st Law thermodynamatic, that energy flow of thermo source.

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The loudness of a sound will be determined by its

adoni [48]

Answer:

By its amplitude.

Explanation:

loudness is sound intensity & intensity depends on square of amplitude. for example higher the amplitude higher the intensity which means higher the loudness.

4 0

3 years ago

Two children, Ferdinand and Isabella, are playing with a waterhose on a sunny summer day. Isabella is holding the hose in herhan

sweet [91]

Answer:

84.196%

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

$y_0=1\ m$

x = 10 m

t = Time taken

$v_0$ = 3.5 m/s (assumed, as it is not given)

$A_0$ = $\pi 1.5^2$

We have the equation

$y=y_0+ut+\dfrac{1}{2}gt^2\\\Rightarrow 0=y_0-\dfrac{1}{2}gt^2\\\Rightarrow t=\sqrt{\dfrac{2y_0}{g}}\\\Rightarrow t=\sqrt{\dfrac{2\times 1}{9.81}}\\\Rightarrow t=0.45152\ s$

$x=x_0+vt\\\Rightarrow 10=0+v0.45152\\\Rightarrow v=\dfrac{10}{0.45152}\\\Rightarrow v=22.14741\ m/s$

From continuity equation we have

$Av=A_0v_0\\\Rightarrow A=\dfrac{A_0v_0}{v}\\\Rightarrow A=\dfrac{\pi 1.5^2\times 3.5}{22.14741}\\\Rightarrow A=1.11706\ cm^2$

Fraction is given by

$f=\dfrac{A_0-A}{A_0}\times 100\\\Rightarrow f=\dfrac{\pi 1.5^2-1.11706}{\pi 1.5^2}\times 100\\\Rightarrow f=84.196\ \%$

The fraction is 84.196%

4 0

3 years ago

Parallaxes of stars are sometimes measured relative to the positions of galaxies or distant objects called quasars. Why is this

Elenna [48]

Answer:

quasars are massive celestrial bodies that are far or remotely placed emitting large energy radiations.

parallax is the best way of determining distances between astronomical bodies and parallaxes of stars are measured relative to a quasar to minimise the measurement error. each observations is stimulated with the basic angle variations leading to minimum uncertainty in the results as the assumptions in the measurements are as minimal as possible.

7 0

3 years ago

a 2.6 kg block is attached to a horizontal spring that has a spring constant of 126 N/m. At the instant when the displacement of

NeTakaya

Answer:

The acceleration of the object is $5.57\ m/s^2$ .

Explanation:

Given that,

Mass of the block, m = 2.6 kg

Spring constant of the spring, k = 126 N/m

At the instant when the displacement of the spring from its unstained length is 0.115 m. We need to find the acceleration of the object.

When the block is displaced, the force acting on the spring is equal to the force due to its motion. Such as :

$kx=ma$

a is acceleration of the object

$a=\dfrac{kx}{m}\\\\a=\dfrac{126\times 0.115}{2.6}\\\\a=5.57\ m/s^2$

So, the acceleration of the object is $5.57\ m/s^2$ .

8 0

3 years ago

Two Earth satellites, A and B, each of mass m, are to be launched into circular orbits about Earth's center. Satellite A is to o

Juliette [100K]

(a) 0.473

The potential energy of a satellite orbiting around Earth is given by

$U=-\frac{GMm}{R+h}$

where

G is the gravitational constant

M is the Earth's mass

m is the satellite's mass

R is the Earth's radius

h is the altitude of the satellite above the Earth's surface

So the potential energy of satellite A is

$U_A=-\frac{GMm}{R+h_A}$

while potential energy of satellite B is

$U_B=-\frac{GMm}{R+h_B}$

Therefore the ratio of the potential energy of satellite B to that of satellite A is

$\frac{U_B}{U_A}=\frac{R+h_A}{R+h_B}$

and using

hA = 5920 km

hB = 19600 km

R = 6370 km

we find

$\frac{U_B}{U_A}=\frac{6370+5920}{6370+19600}=0.473$

(b) 0.473

The kinetic energy of a satellite orbiting around Earth instead is given by

$K=\frac{GMm}{2(R+h)}$

So the kinetic energy of satellite A is

$K_A=\frac{GMm}{2(R+h_A)}$

while kinetic energy of satellite B is

$K_B=\frac{GMm}{2(R+h_B)}$

Therefore the ratio of the kinetic energy of satellite B to that of satellite A is

$\frac{K_B}{K_A}=\frac{R+h_A}{R+h_B}$

which is identical to before, so it gives

$\frac{K_B}{K_A}=\frac{6370+5920}{6370+19600}=0.473$

(c) Satellite B

The total energy of a satellite in orbit is given by

$E=U+K = -\frac{GMm}{R+h}+\frac{GMm}{2(R+h)}=-\frac{GMm}{2(R+h)}$

We see that the total energy is:

1) negative (because the satellite is on a bound orbit)

2) inversely proportional to the distance of the satellite from the Earth's center, R+h

So the magnitude of the fraction in the equation is larger for the satellite which is closer to the Earth's surface (satellite A), but since the energy is negative, this means that the total energy of this satellite is smaller than that of satellite B. So, satellite B has a greater total energy.

(d) $1.03\cdot 10^7 J$