All categories

3 years ago

What’s is a object that moves through the air space acted on only by gravity

Physics

2 answers:

inessss [21]3 years ago

7 0

This describes projectile motion while neglecting any outside forces other than gravity

Arada [10]3 years ago

4 0

I believe the answer is projectile
As the object reach its maximum height after it thrown, its speed will falls into 0. But due to the gravity force, the speed of the object will start to accelerate again after it came down until it hit another medium.
Example of projectiles: a thrown baseball, a catapult shot.

You might be interested in

Is there any machine that is 100% efficient? why?why not

denis-greek [22]

Answer:

No, it's not there.

Explanation:

For a machine to be 100% efficient, it has to be with an output which is equal to its input. But machines have an out put less than an input, hence efficiency below 100%.

7 0

3 years ago

How can you make the potential energy as high as possible in a magnetic field between one electromagnet and one piece of iron?

harkovskaia [24]

In step 1, to increase the potential energy, the iron will move towards the electromagnet.

In step 2, to increase the potential energy, the iron will move towards the electromagnet.

<h3>Potential energy of a system of magnetic dipole</h3>

The potential energy of a system of dipole depends on the orientation of the dipole in the magnetic field.

$U = \mu B$

where;

$\mu$ is the dipole moment
B is the magnetic field

$B = \frac{\mu_0 I}{2\pi r}$

$U = \mu\times (\frac{\mu_0 I}{2\pi r} )$

Increase in the distance (r) reduces the potential energy. Thus, we can conclude the following;

In step 1, to increase the potential energy, the iron will move towards the electromagnet.
In step 2, when the iron is rotated 180, it will still maintain the original position, to increase the potential energy, the iron will move towards the electromagnet.

Learn more about potential energy in magnetic field here: brainly.com/question/14383738

7 0

2 years ago

The frequency of a wave is 200 Hz. The wavelength is 0.1 m. What is the period of the wave?

Aleks04 [339]

The formula for the period of wave is: wave period is equals to 1 over the frequency. $waveperiod=\frac{1}{frequency}$
To get the value of period of wave you need to divide 1 by 200 Hz. However, beforehand, you have to convert 200 Hz to cycles per second. So that would be, 200 cyles per second or 200/s.
By then, you can start the computation by dividing 1 by 200/s. Since 200/s is in fractional form, you have to find its reciprocal form and multiply it to one which would give you 1 (one) second over 200. This would then lead us to the value 0.005 seconds as the wave period.

wave period= 1/200 Hz
Convert Hz to cycles per second first
200 Hz x 1/s= 200/second
Make 200/second as your divisor, so:

wave period= 1/ 200/s

get the reciprocal form of 200/s which is s/200

then you can start the actual computation:

wave period= 1 x s divided by 200

this would give us an answer of 0.005 s.

6 0

3 years ago

One of the harmonics on a string 1.30m long has a frequency of 15.60 Hz. The next higher harmonic has a frequency of 23.40 Hz. F

Alja [10]

Answer:

$\large \boxed{\text{(a) 7.800 Hz; (b) 20.3 m/s; 40.6 m/s; 60.8 m/s}}$

Explanation:

a) Fundamental frequency

A harmonic is an integral multiple of the fundamental frequency.

$\dfrac{\text{23.40 Hz}}{\text{15.60 Hz}} = \dfrac{1.500}{1} \approx \dfrac{3}{2}$

$f = \dfrac{\text{24.30 Hz}}{3} = \textbf{7.800 Hz}$

b) Wave speed

(i) Calculate the wavelength

In a fundamental vibration, the length of the string is half the wavelength.

$\begin{array}{rcl}L & = & \dfrac{\lambda}{2}\\\\\text{1.30 m} & = & \dfrac{\lambda}{2}\\\\\lambda & = & \text{2.60 m}\\\end{array}$

(b) Calculate the speed s

$\begin{array}{rcl}v_{1}& = & f_{1}\lambda\\& = & \text{7.800 s}^{-1} \times \text{2.60 m}\\& = & \textbf{20.3 m/s}\\\end{array}$

$\begin{array}{rcl}v_{2}& = & f_{2}\lambda\\& = & \text{15.60 s}^{-1} \times \text{2.60 m}\\& = & \textbf{40.6 m/s}\\\end{array}$

$\begin{array}{rcl}v_{3}& = & f_{3}\lambda\\& = & \text{23.40 s}^{-1} \times \text{2.60 m}\\& = & \textbf{60.8 m/s}\\\end{array}$

4 0

3 years ago

A river flows due south at 5 mi/h. A swimmer attempting to cross the river heads due east swimming at 3 mi/h relative to the wat

Dafna11 [192]

Answer:

<u>velocity of swimmer relative to ground = 3 i -5 j</u>

Explanation:

To cross a river the swimmer swims relative to river in perpendicular direction.

Velocity of river = -5 j (south)

Velocity of swimmer relative to river = 3 i(north)

<h2>Velocity of swimmer relative to ground = Velocity of swimmer relative to river + Velocity of river</h2>

Velocity of swimmer relative to ground = 3 i -5 j

So magnitude of total velocity is $\sqrt{3^2+(-5)^2}$ = $\sqrt{9+25}$ = $\sqrt{34}$

3 0

3 years ago