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2 years ago

3. What is a manometer used to determine?

Physics

2 answers:

kolbaska11 [484]2 years ago

4 0

Used to indicate pressure!

Gemiola [76]2 years ago

3 0

Answer:

It's used to indicate pressure

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A satellite circles the Earth in an orbit whose radius is twice the Earth’s radius. The Earth’s mass is 5.98 x 1024 kg, and its

gavmur [86]

Hello!

Recall the period of an orbit is how long it takes the satellite to make a complete orbit around the earth. Essentially, this is the same as 'time' in the distance = speed * time equation. For an orbit, we can define these quantities:

$d = 2\pi r$ ← The circumference of the orbit

speed = orbital speed, we will solve for this later

time = period

Therefore:

$T = \frac{2\pi r}{v}$

Where 'r' is the orbital radius of the satellite.

First, let's solve for 'v' assuming a uniform orbit using the equation:
$v = \sqrt{\frac{Gm}{r}}$

G = Gravitational Constant (6.67 × 10⁻¹¹ Nm²/kg²)

m = mass of the earth (5.98 × 10²⁴ kg)

r = radius of orbit (1.276 × 10⁷ m)

Plug in the givens:
$v = \sqrt{\frac{(6.67*10^{-11})(5.98*10^{24})}{(1.276*10^7)}} = 5590.983 m/s$

Now, we can solve for the period:

$T = \frac{2\pi (1.276*10^7)}{5590.983} =\boxed{ 14339.776 s}$

7 0

2 years ago

What happens when a light bulb is disconnected in a parralel circuit?

VARVARA [1.3K]

Answer:

Hope i could help!

Explanation:

so all but one light could be burned out, and the last one will still function.

5 0

3 years ago

A 1980-kg car is traveling with a speed of 15.5 m/s. What is the magnitude of the horizontal net force that is required to bring

Dennis_Churaev [7]

Answer: 6067.5 N

Explanation:

Work = Change in Energy. To start, all of the energy is kinetic energy, so find the total KE using: KE = 1/2(m)(v^2). Plug in 1980 kg for m and 15.5 m/s for v and get KE = 237847.5 J.

Now, plug this in for work: Work = Force * Distance; so, divide work by distance to get 6067.5 N.

5 0

2 years ago

Two ladybugs sit on a rotating disk that is slowing down at a constant rate. The ladybugs are at rest with respect to the surfac

d1i1m1o1n [39]

The two ladybugs have same rotational (angular) speed

Explanation:

The rotational (angular) speed of an object in circular motion is defined as:

$\omega=\frac{\theta}{t}$

where

$\theta$ is the angular displacement

t is the time interval considered

Here we have two ladybugs, which are located at two different distances from the axis. In particular, ladybug 1 is halfway between ladybug 2 and the axis of rotation. However, since they rotate together with the disk, and the disk is a rigid body, every point of the disk cover the same angle $\theta$ in the same time $t$ : this means that every point along the disk has the same angular speed, and therefore the two ladybugs also have the same angular speed.