Please help me on this!

Why does it dangerous to jump from a moving vehicle?

Kaylis [27]

Why is it dangerous*btw.....And because you can easily crash into another vehicle also because it's just plain stupid to jump anyway ...

4 0

3 years ago

A 3.30 x 10-2-kg block is resting on a horizontal frictionless surface and is attached to a horizontal spring whose spring const

avanturin [10]

Answer:

0.17547 m

Explanation:

m = Mass of block = $3.3\times 10^{-2}\ kg$

v = Velocity of block = 10.8 m/s

k = Spring constant = 125 N/m

A = Amplitude

The kinetic energy of the system is conserved

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kA^2\\\Rightarrow \\\Rightarrow A=\sqrt{\dfrac{mv^2}{k}}\\\Rightarrow A=\sqrt{\dfrac{3.3\times 10^{-2}\times 10.8^2}{125}}\\\Rightarrow A=0.17547\ m$

The amplitude of the resulting simple harmonic motion is 0.17547 m

6 0

3 years ago

Two very small charged particles exert an electrostatic force F on each other. If the distance between them is doubled, the forc

Solnce55 [7]

Answer:

The answer is D

Explanation:

Your welcome

4 0

3 years ago

Which of the following describes the relationship of conductance to a conductor?

Brilliant_brown [7]

<span>The answer is The conductance of a conductor is inversely proportional to the cross-sectional area of the conductor.</span> <span>Conductance is directly related to the ease offered by any material to the passage of electric current. Conductance is the opposite of resistance. The higher the conductance, the lower the resistance and vice versa, the greater the resistance, the less conductance, so both are inversely proportional</span>

4 0

3 years ago

Two large parallel conducting plates carrying opposite charges of equal magnitude are separated by 2.20 cm. Part A If the surfac

alukav5142 [94]

Answer:

5308.34 N/C

Explanation:

Given:

Surface density of each plate (σ) = 47.0 nC/m² = $47\times 10^{-9}\ C/m^2$

Separation between the plates (d) = 2.20 cm

We know, from Gauss law for a thin sheet of plate that, the electric field at a point near the sheet of surface density 'σ' is given as:

$E=\dfrac{\sigma}{2\epsilon_0}$

Now, as the plates are oppositely charged, so the electric field in the region between the plates will be in same direction and thus their magnitudes gets added up. Therefore,

$E_{between}=E+E=2E=\frac{2\sigma}{2\epsilon_0}=\frac{\sigma}{\epsilon_0}$

Now, plug in $47\times 10^{-9}\ C/m^2$ for 'σ' and $8.85\times 10^{-12}\ F/m$ for $\epsilon_0$ and solve for the electric field. This gives,

$E_{between}=\frac{47\times 10^{-9}\ C/m^2}{8.854\times 10^{-12}\ F/m}\\\\E_{between}= 5308.34\ N/C$

Therefore, the electric field between the plates has a magnitude of 5308.34 N/C

5 0

3 years ago