The ________ and ________ are used to resolve vectors into their components.

Help! (Look in the pic)

hjlf

Answer:

turtle

Explanation:

5 0

3 years ago

A rod of length L and electrical resistance R moves through a constant uniform magnetic field ; both the magnetic field and the

CaHeK987 [17]

Answer:

don't know what class are you you are using which mobile or laptop

6 0

3 years ago

What is true of transverse waves?

topjm [15]

<h2>Answer:D</h2>

Explanation:

Option A:

Surface waves are neither transverse nor longitudinal.They traverse perpendicularly or parallel to the wave's motion along the interface between different media.

Option B:

Transverse waves vibrate perpendicularly to the direction of the propagation of the wave.

Option C:

Sound is a longitudinal wave.Not a transverse wave.

Option D:

Transverse waves don't require a medium for propagation.But they propagate in medium too.

7 0

3 years ago

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3. A model rocket takes 0.05 seconds to speed up from rest to its maximum velocity of 80 m/s.

nikklg [1K]

Answer:

$1600 \frac{m}{s^2}$

Explanation:

Acceleration is defined as the change in velocity divided by the time it took to produce such change. The formula then reads:

$a = \frac{change-in-velocity}{time} = \frac{Vf-Vi}{t}$

Where Vf is the final velocity of the object, (in our case 80 m/s)

Vi is the initial velocity of the object (in our case 0 m/s because the object was at rest)

and t is the time it took to change from the Vi to the Vf (in our case 0.05 seconds.

Therefore we have:

$a = \frac{80 m/s - 0 m/s}{0.05 sec} = 1600 \frac{m}{s^2}$

Notice that the units of acceleration in the SI system are $\frac{m}{s^2}$ (meters divided square seconds)

7 0

3 years ago

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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