The amplitude is the number of waves that pass a given point per unit time.<br> a. True<br> b. False

What separates the terrestrial planets from the jovian or gas planets in space?

Flura [38]

In our solar system, terrestrial planets are separated from the gas giants by the asteroid belt. The asteroid belt is a region in the solar system between Mars and Jupiter where asteroids are located. Gas giants do not have a solid surface and possible a small rocky core. The gas giants are Jupiter, Saturn, Uranus and Neptune. The first four planets, Mercury, Venus, Earth and Mars.

8 0

3 years ago

A 120 resistor a 60 ohm resistor and a 40 ohm resistor are connected in parallel to a 120 volt power source. what is the current

larisa [96]

Answer:

6 A

Explanation:

First of all, we need to calculate the equivalent resistance of the circuit. The three resistors are connected in parallel, so their equivalent resistance is given by:

$\frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}=\frac{1}{120 \Omega}+\frac{1}{60 \Omega}+\frac{1}{40 \Omega}=\frac{3+2+1}{120 \Omega}=\frac{6}{120 \Omega}\\R_T = \frac{120}{6} \Omega$

And now we can use Ohm's law to find the current in the circuit:

$V=R_T II=\frac{V}{R_T}=\frac{120 V}{\frac{120}{6}\Omega}=6 A$

6 0

3 years ago

A horizontal circular curve on a highway is designed for traffic moving at 60 km/h. If the radius of the curve is 150 m, and if

KengaRu [80]

Answer:

The value of the correct angle of banking for the road is $\theta = 67.76$ °

Explanation:

Given data

Velocity (v) = 60 $\frac{m}{s}$

Radius = 150 m

The velocity of the car in this case is given by

$v = \sqrt{r g \tan \theta}$

$v^{2} = r g \tan \theta$

$\tan \theta = \frac{v^{2} }{rg}$

Put all the values in above formula we get

$\tan \theta = \frac{60^{2} }{(150)(9.81)}$

$\tan \theta =$ 2.446

$\theta = 67.76$ °

Therefore the value of the correct angle of banking for the road is $\theta = 67.76$ °

4 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 projectile is launched into the air with the initial speed of vi = 40 m/s at a launch angle of 20 degrees above the horizontal

Sphinxa [80]

The range of the projectile is 188 m

Explanation:

The motion of the arrow in this problem is a projectile motion, so it follows a parabolic path which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction

The path of a projectile is the combination of these two motions: see figure in attachment.

In order to find the horizontal range of the projectile, we just need to calculate the horizontal distance travelled.

We have: