A rock rolls down a hill. Which form of energy is this an example of? (2 points)

V125BC [204]

<span>Basically, the variable n is equal to the amount of P out of the system divided by the about of P into the system

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7 0

3 years ago

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A car speedometer has a 4% uncertainty. What is the range of possible speeds (in km/h) when it reads 110 km/h?

Kruka [31]

4% of 110 is 4.4. So the possible range of speeds is the interval from 110-4.4 till 110+4.4.
105.6 till 114.4

4 0

3 years ago

A 180 g model airplane charged to 18 mC and traveling at 2.2 m/s passes within 8.6 cm of a wire, nearly parallel to its path, ca

viva [34]

Answer:

$a=0.2*10^{-5}g$

Explanation:

From the question we are told that:

Mass $M=180=>0.18kg$

Charge $Q=18mC=18*10^-^3C$

Velocity $v=2.2m/s$

Length of Wire $L=8.6cm=>0.086$

Current $I=30A$

Generally the equation for Magnetic Field of Wire B is mathematically given by

$B=\frac{\mu_0*I}{2\pi*l}$

$B=\frac{4*3.14*10^-^7*I}{2*3.14*8.6}$

$B=6.978*10^{-5}T$

Generally the equation for Force on the plane F is mathematically given by

$F=qvB$

Therefore

$ma=qvB$

$a=\frac{qvB}{m}$

$a=\frac{18*10^{-5}83.4*6.978*10^{-5}}{0.18kg}$

$a=2.37*10^{-5}$

Therefore in Terms of g's

$a=\frac{2.37*10^{-5}}{9.8}$

$a=0.2*10^{-5}g$

8 0

3 years ago

How can we show the magnetic force of a magnet illustrate with an example

morpeh [17]

Answer:

take some sharp mixture of iron spread it from up the magnet you will see the magnetic field of magnet

3 0

3 years ago

A block–spring system vibrating on a frictionless, horizontal surface with an amplitude of 7.0 cm has an energy of 14 J. If the

Bingel [31]

Answer:

$E_T= 28J$

Explanation:

The energy of Mass-Spring System the sum of the potential energy of the block plus the kinetic energy of the block:

$E_T=U+K=\frac{1}{2} k \Delta x^2+\frac{1}{2} mv^2$

Where:

$\Delta x=Amplitude\hspace{3}or\hspace{3}d eformation\hspace{3} of\hspace{3} the\hspace{3} spring\\m=Mass\hspace{3}of\hspace{3}the\hspace{3}block\\k=Constant\hspace{3}of\hspace{3}the\hspace{3}spring\\v=Velocity\hspace{3}of\hspace{3}the\hspace{3}block$

There are two cases, the first case is when the spring is compressed to its maximum value, in this case the value of the kinetic energy is zero, since there is no speed, so:

$E_T=\frac{1}{2} k \Delta x^2\\\\14=\frac{1}{2} k7^2\\\\Solving\hspace{3} for\hspace{3} k\\\\k=\frac{28}{49} =\frac{4}{7}$

The second case is when the block passes through its equilibrium position, in this case the elastic potential energy is zero since $\Delta x=0$ , so:

$E_T=\frac{1}{2} mv^2\\\\14=\frac{1}{2} mv^2\\\\Solving\hspace{3} for\hspace{3} v\\\\v^2=\frac{28}{m}$

Now, let's find the energy of the system when the block is replaced by one whose mass is twice the mass of the original block using the previous data:

$E_T=U+K=\frac{1}{2} k \Delta x^2+\frac{1}{2} m_2v^2$

Where in this case:

$m_2=New\hspace{3}mass=Twice\hspace{3} the\hspace{3} mass \hspace{3}of\hspace{3} the\hspace{3} original=2m$

Therefore:

$E_T=\frac{1}{2} (\frac{4}{7} ) (7^2)+\frac{1}{2} (2m)(\frac{28}{m_2})=\frac{1}{2} (\frac{4}{7} ) (7^2)+\frac{1}{2} (2m)(\frac{28}{2m})=14+14=28J$

8 0

4 years ago