Magnetism at the centre of a bar magnet is

Please answer, do today

Ksenya-84 [330]

Answer:

hshawi hdsdk

done and my name is fricking bella your gonna die

3 0

3 years ago

t is the estimated distance traveled through the air by a marshmallow that is launched from a 15- cm long launcher when it is pl

DerKrebs [107]

Answer:

10

Explanation:

15 cm minus 5 cm

5 0

2 years ago

Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m

netineya [11]

Answer: $m \frac{d}{dt}v_{(t)}$

Explanation:

In the image attached with this answer are shown the given options from which only one is correct.

The correct expression is:

$m \frac{d}{dt}v_{(t)}$

Because, if we derive velocity $v_{t}$ with respect to time $t$ we will have acceleration $a$ , hence:

$m \frac{d}{dt}v_{(t)}=m.a$

Where $m$ is the mass with units of kilograms ( $kg$ ) and $a$ with units of meter per square seconds $\frac{m}{s}^{2}$ , having as a result $kg\frac{m}{s}^{2}$

The other expressions are incorrect, let’s prove it:

$\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)}$ This result has units of $kg\frac{m}{s}$

$m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m$ This result has units of $kg$

$m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C$ This result has units of $kgm^{2}$ and $C$ is a constant

$m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m$ This result has units of $kg$

$m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m$ This result has units of $kg$

$\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C$ This result has units of $kg \frac{m^{3}}{s^{3}}$ and $C$ is a constant

$m\int a_{(t)} dt= \frac{m {a_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{4}}$ and $C$ is a constant

$\frac{m}{2} \frac{d}{dt}{(v_{(x)})}^{2}=0$ because $v_{(x)}$ is a constant in this derivation respect to $t$

$m\int v_{(t)} dt= \frac{m {v_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{2}}$ and $C$ is a constant

6 0

3 years ago

What would happen if i put in different colors in the washer with hot water?

xxTIMURxx [149]

Probably it whould separate becase hot water departs things! Hope this helps

4 0

3 years ago

I will mark you brainlist. How can you use a tuning fork to tune a piano?

Phoenix [80]

A tuning fork's job is to establish a single note that everybody can tune to.

Most tuning forks are made to vibrate at 440 Hz, a tone known to musicians as "concert A." To tune a piano, you would start by playing the piano's "A" key while ringing an "A" tuning fork. If the piano is out of tune, you'll hear a distinct warble between the note you're playing and the note played by the tuning fork; the further apart the warbles, the more out-of-tune the piano. By either tightening or loosening the piano's strings, you reduce the warble until it's in line with the tuning fork. Once the "A" key is in tune, you would then adjust all of the instrument's 87 other keys to match. The method is much the same for most other instruments. Whether you're tuning a clarinet or guitar, simply play a concert A and adjust your instrument accordingly

Explanation:

It can be a bit tricky to hold a tuning fork while manipulating an instrument, which is why some musicians decide to clench the base of a ringing tuning fork in their teeth. This has the unique effect of transmitting sound through your bones, allowing your brain to "hear" the tone through your jaw. According to some urban legends, touching your teeth with a vibrating tuning fork is enough to make them explode. It's a myth, obviously, but if you have a cavity or a chipped tooth, you'll quickly find this method to be unbelievably painful.

Luckily, you can also buy tuning forks that come mounted on top of a resonator, a hollow wooden box designed to amplify a tuning fork's vibrations. In 1860, a pair of German inventors even devised a battery-powered tuning fork that musicians didn't need to ring again and again

6 0

2 years ago