All categories

3 years ago

PLS HELP FIND THE AVREGE KENETIC ENERGY!!

Physics

1 answer:

ANEK [815]3 years ago

4 0

Answer: if you gogle it it will tell you the awenser

lanation:

You might be interested in

The magnitude of a component of a vector must be

den301095 [7]

Less than or equal to the magnitude of the vector

6 0

3 years ago

Q. A mass of 300g is lifted to a<br> height of 10m<br> 205 by a person. Calculate his work done

sergiy2304 [10]

A=mgh
m=300g=0.3kg
g=9,81 m/s^2
h=10m
A=29.43J

3 0

2 years ago

A force can exist between two charged particles or objects even if they're as small as subatomic particles. Between which of the

ASHA 777 [7]

If both particles have the SAME electrical charge, then they repel.
If they have DIFFERENT electrical charge, then they attract.

Protons have + charge .
Electrons have - charge .

So two protons (A) or two electrons (D) push apart.

One proton and one electron (C) pull together.

3 0

3 years ago

Read 2 more answers

For work to be accomplished we must have

Nat2105 [25]

The answer they are looking for is the last one. However the last two are technically correct but the third one would result in negative work.

8 0

2 years ago

Read 2 more answers

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