Calculate the current flowing if a charge of 36 kilocoulombs flows in 1 hour.

How is it possible to get three precise but inaccurate measurements of the same volume of water

The inaccurate measurements must be similar to the other two measurements (ex; 590, 589, 599), but different from the actual volume of water. (Ex; the actual volume is let say.. 100, but you measured 50, 49, 40)

8 0

2 years ago

A 4-kg cat is resting on a bookshelf that is 3m high. What is the cats gravitational potential energy relative to the floor if t

Kobotan [32]

Potential energy = mass x gravity x height

P.E = 4 x 9.8 x 3
P.E = 117.6 J

8 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

2 years ago

What happens to a falling object when the force of air resistance equals the force of gravity?

12345 [234]

Answer:

A falling object will continue to accelerate to higher speeds until they encounter an amount of air resistance that is equal to their weight.

Explanation:

7 0

2 years ago

Read 2 more answers

Discuss the relationship between electric and magnetic fields. How are they related?

evablogger [386]

<span>Discuss the relationship between electric and magnetic fields. How are they related?</span>

The magnetic field is related to the electric field in such a way that magnetic fields moves perpendicular the electric field. This relationship exists between a magnetic field and an electric field is a consequence of associating elementary particles.

Hope This Helped! :3

4 0

2 years ago