Did I do these questions correctly?

If something is traveling at 20 m/s constant velocity, is it in equilibrium? If a projectile is launched at a velocity of 20 m/s

Irina18 [472]

If something is traveling at 20 m/s constant speed AND its direction isn't changing, then its velocity is constant. Another way to say that is: Its acceleration is zero. Zero acceleration means zero NET force acting on the object, or a group of BALANCED forces acting on it, also called EQUILIBRIUM. The required answer is: YES.

If a real projectile is launched, the force of gravity acts on it vertically downward. There's no upward force acting on it to balance gravity. Therefore, the forces on the projectile are NOT balanced, there IS a net vertical force on it, and it's NOT in equilibrium. Too bad.

3 0

2 years ago

A testing instrument that's used to measure electrical signals in a circuit and display them as waveforms on a screen is called

disa [49]

A testing instrument that's used to measure electrical signals
in a circuit and display them as waveforms on a screen is called
an oscilloscope.

8 0

3 years ago

the speed of travel of the moon around the earth, using the formula for the speed of a moving object in a circular path

Svetach [21]

Answer: 1018.26 m/s

Explanation:

Approaching the orbit of the Moon around the Earth to a circular orbit (or circular path), we can use the equation of the speed of an object with uniform circular motion:

$V=\sqrt{G\frac{M}{r}}$

Where:

$V$ is the speed of travel of the Moon around the Earth

$G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$ is the Gravitational Constant

$M=5.972(10)^{24} kg$ is the mass of the Earth

$r=384400(10)^{3} m$ is the distance from the center of the Earth to the center of the Moon

Solving:

$V=\sqrt{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{5.972(10)^{24} kg}{384400(10)^{3} m}}$

$V=1018.26 m/s$ This is the speed of travel of the Moon around the Earth

5 0

2 years ago

A

malfutka [58]

Answer:

<em>The penny will hit the ground at 6.39 seconds</em>

Explanation:

The penny is dropped from a height of y=200 m. The equation of the height on a free-fall motion is given by:

$\displaystyle y=\frac{gt^2}{2}$

Where $g=9.8\ m/s^2$ , and t is the time.

Solving for t:

$\displaystyle t=\sqrt{\frac{2y}{g}}$

Using the value y=200:

$\displaystyle t=\sqrt{\frac{2*200}{9.8}}$

t=6.39 sec

The penny will hit the ground at 6.39 seconds

4 0

3 years ago

In an electricity experiment, a 1.20 g plastic ball is suspended on a 59.0 cm long string and given an electric charge. A charge

stich3 [128]

Answer:

a.) 5.24 10⁻³ N . b) 0.013 N

Explanation:

a) In absence of other forces, the plastic ball is only subject to the force of gravity (downward) , and to the tension in the string, which are equal each other.

We are told that there exists an horizontal force , of an electric origin, that causes the ball to swing out to a 24º angle (respect the normal) and remain there, so there exists a new equilibrium condition.

In this situation, both the vertical and horizontal components of the external forces acting on the ball (gravity, tension and the electrical force) must be equal to 0.

The only force that has horizontal and vertical components, is the tension in the string.

We can apply Newton's 2nd Law to both directions, as follows:

T cos 24º - mg = 0

-T sin 24º + Fe = 0

where T= Tension in the string.

Fe = Electrical Force

mg = Fg = gravity force

⇒ T = mg/ cos 24º

Replacing in the horizontal forces equation:

-mg/cos 24º . sin 24º = -Fe

∴ Fe = mg. tg 24º = 0.0012 kg. 9.8 m/s². tg 24º = 5.24 10⁻³ N

b) In order to get the value of T, we can simply solve for T the vertical forces component equation , as follows:

T = mg/ cos 24º = 0.0012 kg. 9.8 m/s² / 0.914 = 0.013 N

6 0

2 years ago