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3 years ago

Doc Brown has calculated his Delorean can accelerate at a rate of 2.52 m/s/s. How

Physics

1 answer:

GREYUIT [131]3 years ago

6 0

Answer:

304.89m

Explanation:

Given

acceleration a = 2.52m/s²

final speed v = 39.2m/s

initial speed = 0m/s (car accelerates from rest)

Using the equation of motion below to get the distance of Doc brown from Marty;

v² = u²+2as

substitute the given parameters

39.2² = 0²+2(2.52)s

1536.64 = 0+5.04s

divide both sides by 5.04

1536.64/5.04 = 5.04s/5.04

rearrange the equation

5.04s/5.04 = 1536.64/5.04

s = 304.89m

Hence He and Marty must stand at 304.89m to allow the car to accelerate from rest to a speed of 39.2 m/s?

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The electric potential at the dot in the figure is 3160 V. What is charge q?

PSYCHO15rus [73]

Hi there!

Recall the equation for electric potential of a point charge:

$V = \frac{kQ}{r}$

V = Electric potential (V)
k = Coulomb's Constant(Nm²/C²)

Q = Charge (C)
r = distance (m)

We can begin by solving for the given electric potentials. Remember, charge must be accounted for. Electric potential is also a SCALAR quantity.

Upper right charge's potential:

$V = \frac{(8.99*10^9)(-5 * 10^{-9})}{0.04} = -1123.75 V$

Lower left charge's potential:

$V = \frac{(8.99*10^9)(5*10^{-9})}{0.02} = 2247.5 V$

Add the two, and subtract from the total EP at the point:

$3160 + 1123.75 - 2247.5 = 2036.25$

The remaining charge must have a potential of 2036.25 V, so:

$2036.25 = \frac{(8.99*10^9)(Q)}{\sqrt{0.02^2 + 0.04^2}}\\\\2036.25 = \frac{(8.99*10^9)Q}{0.0447} \\\\Q = 0.000000010127 = \boxed{10.13nC}$

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1 year ago

By what potential difference must a proton [m_0 = 1.67E-27 kg) be accelerated to have a wavelength lambda = 4.23E-12 m? By what

Vinil7 [7]

Explanation:

1. Mass of the proton, $m_p=1.67\times 10^{-27}\ kg$

Wavelength, $\lambda_p=4.23\times 10^{-12}\ m$

We need to find the potential difference. The relationship between potential difference and wavelength is given by :

$\lambda=\dfrac{h}{\sqrt{2m_pq_pV}}$

$V=\dfrac{h^2}{2q_pm_p\lambda^2}$

$V=\dfrac{(6.62\times 10^{-34})^2}{2\times 1.6\times 10^{-19}\times 1.67\times 10^{-27}\times (4.23\times 10^{-12})^2}$

V = 45.83 volts

2. Mass of the electron, $m_p=9.1\times 10^{-31}\ kg$

Wavelength, $\lambda_p=4.23\times 10^{-12}\ m$

We need to find the potential difference. The relationship between potential difference and wavelength is given by :

$\lambda=\dfrac{h}{\sqrt{2m_eq_eV}}$

$V=\dfrac{h^2}{2q_em_e\lambda^2}$

$V=\dfrac{(6.62\times 10^{-34})^2}{2\times 1.6\times 10^{-19}\times 9.1\times 10^{-31}\times (4.23\times 10^{-12})^2}$

$V=6.92\times 10^{34}\ V$

V = 84109.27 volt

Hence, this is the required solution.

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