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4 years ago

MULTIPLE CHOICE

Physics

2 answers:

TiliK225 [7]4 years ago

6 0

Answer:

The correct answer is a single north-seeking pole of unit strength

Explanation:

Setler79 [48]4 years ago

3 0

I'm pretty sure that the only right answer is <span>B) a north-seeking pole of unit strength with a south-seeking pole of the same strength. Because there are no single magnetic poles.</span>

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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

A weather balloon is inflated to a volume of 26.8 LL at a pressure of 744 mmHgmmHg and a temperature of 31.2 ∘C∘C. The balloon r

elena-s [515]

Answer:

43.96 L

Explanation:

We are given that

$V_1=26.8 L$

$P_1=744mm Hg$

$T_1=31.2^{\circ} C=31.2+273=304.2K$

$P_2=385mmHg$

$T_2=-14.8^{\circ}=273-14.8=258.2K$

We know that

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

$V_2=\frac{P_1V_1T_2}{P_2T_1}$

Substitute the values

$V_2=\frac{744\times 26.8\times 258.2}{385\times 304.2}$

$V_2=43.96 L$

Hence, the volume of balloon at -14.8 degree Celsius=43.96 L

5 0

3 years ago

A. In your own words, define what a transverse wave is.

olasank [31]

Answer:

a) From definition a transverse wave is which one where the elements moves perpendicular to the direction of the wave. For example is a wave is moving from the left to the right the elements would be wibrating or moving upward or downward.

We have a lot examples for a transverse wave. For example water waves, strings on the musical instruments , light and radio waves.

b) We can identify a transverse wave if the particles are displaced perpendicular to the direction of the wave. Usually these types of wave occur in elastic solids. And we can identify it when we see a pattern perpendicular between the wave direction and the particles motion. In simple words we need to see that the wave is moving down and up.

Explanation:

Part a

From definition a transverse wave is which one where the elements moves perpendicular to the direction of the wave. For example is a wave is moving from the left to the right the elements would be wibrating or moving upward or downward.

We have a lot examples for a transverse wave. For example water waves, strings on the musical instruments , light and radio waves.

Part b

We can identify a transverse wave if the particles are displaced perpendicular to the direction of the wave. Usually these types of wave occur in elastic solids. And we can identify it when we see a pattern perpendicular between the wave direction and the particles motion. In simple words we need to see that the wave is moving down and up.

3 0

3 years ago

A 46.0-kg box is being pushed a distance of 8.80 m across the floor by a force P whose magnitude is 171 N. The force P is parall

dolphi86 [110]

Answer:

a) 1504.8 J

b) 991.76 J

c) 0J

d) 0J

Explanation:

(a) The work done by the force P on the box is given by the following formula:

$W_P=Px$