At what frequency does the driver of the car hear the ambulance’s siren? The velocity of sound in air is 343 m/s.

While on vacation, a student picks up surface rocks from around the world to add to her rock collection. The composition of her

kykrilka [37]

I want to say that they will be primarily flat but I honestly don't know

7 0

3 years ago

A 20 cm-radius ball is uniformly charged to 71 nC.

artcher [175]

Answer:

Part a)

$\rho = 2.12\mu C/m^3$

Part b)

$q_1 = 1.11 nC$

$q_2 = 8.88 nC$

$q_3 = 71 nC$

Part c)

$E_1 = 3996 N/C$

$E_2 = 7992 N/C$

$E_3 = 15975 N/C$

Explanation:

Part a)

As we know that charge density is the ratio of total charge and total volume

So here the volume of the charge ball is given as

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

$V = \frac{4}{3}\pi(0.20)^3$

$V = 0.0335 m^3$

now the charge density of the ball is given as

$\rho = \frac{71 nC}{0.0335} = 2.12\mu C/m^3$

Part b)

Now the charge enclosed by the surface is given as

$q = \rho V$

at radius of 5 cm

$q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.05)^3$

$q = 1.11 nC$

at radius of 10 cm

$q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.10)^3$

$q = 8.88 nC$

at radius of 20 cm

$q = 71 nC$

Part c)

As we know that electric field is given as

$E = \frac{kq}{r^2}$

so we have electric field at r = 5 cm

$E_1 = \frac{(9\times 10^9)(1.11 nC)}{0.05^2}$

$E_1 = 3996 N/C$

electric field at r = 10 cm

$E_2 = \frac{(9\times 10^9)(8.88 nC)}{0.10^2}$

$E_2 = 7992 N/C$

electric field at r = 20 cm

$E_3 = \frac{(9\times 10^9)(71 nC)}{0.20^2}$

$E_3 = 15975 N/C$

3 0

3 years ago

I dont know how to do this at all please help

worty [1.4K]

Wow ! I understand your shock. I shook and vibrated a little
when I looked at this one too.

The reason for our shock is all the extra junk in the question,
put there just to shock and distract us.

"Neutron star", "5.5 solar masses", "condensed burned-out star".
That's all very picturesque, and it excites cosmic fantasies in
out brains when we read it, but it's just malicious decoration.
It only gets in the way, and doesn't help a bit.

The real question is:

What is the acceleration of gravity 2000 m from
the center of a mass of 1.1 x 10³¹ kg ?

Acceleration of gravity is

   G · M / R²

= (6.67 x 10⁻¹¹ N·m²/kg²) · (1.1 x 10³¹ kg) / (2000 m)²

= (6.67 x 10⁻¹¹ · 1.1 x 10³¹ / 4 x 10⁶) (N) · m² · kg / kg² · m²

=   1.83 x 10¹⁴ (kg · m / s²) · m² · kg / kg² · m²

= 1.83 x 10¹⁴    m / s²

That's about 1.87 x 10¹³ times the acceleration of gravity on
Earth's surface.

In other words, if I were standing on the surface of that neutron star,
I would weigh 1.82 x 10¹² tons, give or take.

3 0

3 years ago

A playground slide is in the form of an arc of a circle with a maximum height of 3.0 m, with a radius of 8.5 m, and with the gro

julia-pushkina [17]

Answer:

a) $s \approx 6.676\,m$

Explanation:

a) Let assume that the ground is not inclined, since the bottom of the playground slide is tangent to ground. Then, the length of given by the definition of a circular arc:

$s = \frac{\pi}{4}\cdot R$

$s=\frac{\pi}{4}\cdot (8.5\,m)$

$s \approx 6.676\,m$

The bottom of the slide has a height of zero. The physical phenomenon around Dr. Ritchey's daughter is modelled after Principle of Energy Conservation. The child begins at rest:

$U_{g,A} = K_{B} + W_{fr}$