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3 years ago

6

In what types of media would sound waves travel fastest

1 answer:

Sedaia [141]3 years ago

4 0

Answer:

solids

good day

or good night

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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

If a cat can exert 2000N Of Force to move a trailer 50m is 20 seconds how much power did the car use ?

inessss [21]

им putins брат, почему вы обманываете нашу систему образования, Это теперь запрещено в России.

8 0

3 years ago

A pendulum has 844 J of potential energy at the highest point of its swing. How much kinetic energy will it have at the bottom o

Stolb23 [73]

844J.
Assuming that there were no encumbrances during it's foreswing and it reached it's full potential at apogee.

8 0

3 years ago

If a roller coaster car had 40,000 J of gravitational potential energy when at rest on the top of a hill how much kinetic energy

Inessa [10]

Answer:

K.E = 30,000 J

Explanation:

Given,

The potential energy of the roller coaster car, P.E = 40000 J

The kinetic energy at height h/4, K.E = ?

According to the law of conservation of energy, the total energy of the system is conserved.

At height 'h', the total energy is,

P.E = mgh

K.E = 0

At height 'h/4', the total energy is

P.E + K.E = mgh

P.E = mgh/4

K.E = 1/2 mv²

Therefore,

mgh/4 + 1/2 mv² = mgh

gh/4 + v²/2 = gh

Hence,

v² = 3gh/2

Substituting in the K.E equation

K.E = 1/2 mv²

= 1/2 m (3gh/2)

= 3/4 mgh

= 3/4 x 40000

= 30000 J

Hence, the K.E of the roller coaster car is, K.E = 30000 J

6 0

3 years ago

What is the change in electric potential energy when the dipole moment of a molecule changes its orientation with respect to e⃗

AleksandrR [38]

Electric potential energy of a dipole is given as

$U = - P.E$

$U = - PE cos\theta$

so change in the potential energy is given by

$U_f - U_i = (-PEcos\theta_2)- (-PEcos\theta_1)$

$\Delta U = PE (cos\theta_1 - cos\theta_2)$

here initially it was parallel to electric field while finally it is perpendicular to the electric field

so we have

$\Delta U = PE(cos 0 - cos 90)$

$\Delta U = PE$

so above is the change in potential energy of dipole

8 0

3 years ago