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3 years ago

The movement of the planets and other celestial bodies in the Solar System is caused by

Physics

2 answers:

katovenus [111]3 years ago

8 0

3. The sun's gravitational force.

Anvisha [2.4K]3 years ago

7 0

The movement of the planets and other celestial bodies in the solar system is actually caused by the sun's gravitational pull or force.

Just like the moon orbits the earth because of the gravitational pull by the earth so does gravitational pull by the sun causes planets' and other celestial bodies' orbit.

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Two loudspeakers emit sound waves along the x-axis. A listener in front of both speakers hears a maximum sound intensity when sp

yKpoI14uk [10]

Answer

given,

difference between the two consecutive maximum

λ = 0.870 - 0.540

λ = 0.33 m

speed of sound = 340 m/s

b) frequency of the sound

v = f x λ

340 = f x 0.33

$f =\dfrac{340}{0.33}$

f = 1030.3 Hz

a) phase difference

the expression of phase difference is given by

$\phi = \dfrac{2\pi}{\lambda}\ \delta$

$\delta = \Delta x - \lambda$

$\delta = 0.540 - 0.33$

$\delta = 0.21\ m$

now,

$\phi = \dfrac{2\pi}{\lambda}\ \times 0.21$

$\phi = \dfrac{2\pi}{0.33}\ \times 0.21$

$\phi = 3.99 rad$

8 0

3 years ago

Two thin concentric spherical shells of radii r1 and r2 (r1 < r2) contain uniform surface charge densities V1 and V2, respect

Lyrx [107]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

So,

a) 0 < r < r1 :

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

Hence, E = 0 for r < r1

b) r1 < r < r2:

Electric field =?

Let, us consider the Gaussian Surface,

E x 4 $\pi$ $r^{2}$ = $\frac{Q1}{E_{0} }$

So,

Rearranging the above equation to get Electric field, we will get:

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$

Multiply and divide by $r1^{2}$

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ x $\frac{r1^{2} }{r1^{2} }$

Rearranging the above equation, we will get Electric Field for r1 < r < r2:

E= (σ1 x $r1^{2}$ ) /( $E_{0}$ x $r^{2}$ )

c) r > r2 :

Electric Field = ?

E x 4 $\pi$ $r^{2}$ = $\frac{Q1 + Q2}{E_{0} }$

Rearranging the above equation for E:

E = $\frac{Q1+Q2}{E_{0} . 4 \pi. r^{2} }$

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ + $\frac{Q2}{E_{0} . 4 \pi. r^{2} }$

As we know from above, that:

$\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ = (σ1 x $r1^{2}$ ) /( $E_{0}$ x $r^{2}$ )

Then, Similarly,

$\frac{Q2}{E_{0} . 4 \pi. r^{2} }$ = (σ2 x $r2^{2}$ ) /( $E_{0}$ x $r^{2}$ )

So,

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ + $\frac{Q2}{E_{0} . 4 \pi. r^{2} }$

Replacing the above equations to get E:

E = (σ1 x $r1^{2}$ ) /( $E_{0}$ x $r^{2}$ ) + (σ2 x $r2^{2}$ ) /( $E_{0}$ x $r^{2}$ )

Now, for

d) Under what conditions, E = 0, for r > r2?

For r > r2, E =0 if

σ1 x $r1^{2}$ = - σ2 x $r2^{2}$

4 0

3 years ago

Change in Energy Quick Check

nika2105 [10]

Answer:

The force of gravity on the object decreases. (FALSE)

The potential energy of the object decreases. (TRUE)

The acceleration due to gravity decreases. (FALSE)

The kinetic energy of the object decrease (FALSE)

Explanation:

FORCE OF GRAVITY:

Force of gravity on the object is the weight of object, which depends upon the mass and value of acceleration due to gravity (W = mg). Since, the value mass is constant on Earth and acceleration due to gravity is also constant on earth.

Therefore, force of gravity remains same on the object. The statement is false.

POTENTIAL ENERGY:

Potential energy of object depends upon the mass, value of acceleration due to gravity, and change in height of object (P.E = mgΔh). Since, the value mass is constant on Earth and acceleration due to gravity increases as the object moves towards the surface of earth. But, the height of object is decreasing.

Therefore, potential energy of object decreases. The statement is true.

ACCELERATION DUE TO GRAVITY:

Acceleration due to gravity depends upon the altitude (gh = g[1 - 2h/Re]). Since, the height of object is decreasing.

Therefore, acceleration due to gravity increases. The statement is false.

But, this change is not significant.

KINETIC ENERGY:

Kinetic Energy of object, which depends upon the mass and velocity of the object (K.E = mv²/2). Since, the value mass is constant on Earth and velocity increases as the object moves towards the surface of earth.

Therefore, kinetic energy of the object also increases. The statement is false.

4 0

3 years ago

angela uses a force of 25 newtons to lift her grocery bag while doing 50 joules of work. how far did she lift the grocery bags

Evgen [1.6K]

We know that:
W=Fs
50J=25N*s
s=50J/25N
s=2m

5 0

3 years ago

Convertir 1200 ms a cs Convertir 0,3 mm a cm.

Vaselesa [24]

Answer:

You can do the reverse unit conversion from cm/s to m/s, or enter any two units below: Metre per second (U.S. spelling: meter per second) is an SI derived unit of both speed (scalar) and velocity (vector quantity which specifies both magnitude and a specific direction), defined by distance in metres divided by time in seconds.

Explanation:

3 0

2 years ago