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2 years ago

This force on compass dials is an example of a force that _______.

2 answers:

4 0

D. Acts at a distance, because magnetic forcefield is in two dots, Earth's north and south magnetic poles, and magnet is on some distance from them.

TEA [102]2 years ago

3 0

acts at a distance

is the answer

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You start with spring that's already been stretched an unknown amount from equilibrium. After stretching it an additional 2.0 cm

maxonik [38]

Answer: 35*10^3 N/m

Explanation: In order to explain this problem we know that the potential energy for spring is given by:

Up=1/2*k*x^2 where k is the spring constant and x is the streching or compresion position from the equilibrium point for the spring.

We also know that with additional streching of 2 cm of teh spring, the potential energy is 18J. Then it applied another additional streching of 2 cm and the energy is 25J.

Then the difference of energy for both cases is 7 J so:

ΔUp= 1/2*k* (0.02)^2 then

k=2*7/(0.02)^2=35000 N/m

7 0

2 years ago

Un pendule est constitue par une masse ponctuelle m= 0,1kg accrocher a un fil sans masse de longueur L = 0,4 m on ecarte ce pend

BabaBlast [244]

I could help
If I know what I was reading I’m sorry I need to translate

3 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

Consider the reaction data. A ⟶ products T ( K ) k ( s − 1 ) 225 0.385 525 0.635 What two points should be plotted to graphicall

lutik1710 [3]

Answer:

Plot ln K vs 1/T

(a) -0.5004; (b) 0.002 539 K⁻¹; (c) -197.1 K⁻¹; (d) 1.64 kJ/mol

Explanation:

This is an example of the Arrhenius equation:

$k = Ae^{-E_{a}/RT}\\\text{Take the ln of each side}\\\ln k = \ln A - \dfrac{E_{a}}{RT}\\\\\text{We can rearrange this to give}\\\ln k = - \dfrac{E_{a}}{R}\dfrac{1}{T} + \ln A\\\\y = mx + b$

Thus, if we plot ln k vs 1/T, we should get a straight line with slope = -Eₐ/R and a y-intercept = lnA

Data:

$\begin{array}{cccc}\textbf{k/s}\mathbf{^{-1}} &\mathbf{\ln k} & \textbf{T/K} & \mathbf{1/T(K^{-1})}\\0.285 & -0.9545 & 225 &0.004444\\0.635 & -0.4541 & 525 & 0.001905\\\end{array}$