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grandymaker [24]

3 years ago

10

A proton of mass 1.67×10-27 kg is being accelerated along a straight line at 7.01×1015 m/s2 in an accelerator. If the proton has

an initial speed of 6.63×107 m/s and travels 3.63 cm, what then is its speed?

1 answer:

Masja [62]3 years ago

5 0

Answer:

$v_f=7*10^7\frac{m}{s}$

Explanation:

The proton is under a linear motion with constant acceleration. So, we use the kinemtic equations to calculate its final speed. We know its acceleration, its initial speed and its traveled distance. Thus, we use the following equation:

$v_f^2=v_0^2+2ax\\v_f=\sqrt{v_0^2+2ax}\\v_f=\sqrt{(6.63*10^7\frac{m}{s})^2+2(7.01*10^{15}\frac{m}{s^2})(3.63*10^{-2}m)}\\v_f=7*10^7\frac{m}{s}$

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Fudgin [204]

Answer:

$|\vec A + \vec B| = 50 units$

Explanation:

As we know that magnitude of two vectors is given as

$|\vec A + \vec B| = \sqrt{A^2 + B^2 + 2AB cos\theta}$

here we know that

A = magnitude of vector A

B = magnitude of vector B

$\theta$ = angle between two vectors

so here we know that

A = 30 units

B = 40 units

angle = 90 degree

so we have

$|\vec A + \vec B| = \sqrt{30^2 + 40^2 + 2(30)(40)cos90}$

$|\vec A + \vec B| = \sqrt{30^2 + 40^2}$

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3 years ago

Two small plastic spheres are given positive electrical charges. When they are a distance of 14.8cm apart, the repulsive force b

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Answer:

Explanation:

Case I: They have same charge.

Charge on each sphere = q

Distance between them, d = 14.8 cm = 0.148 m

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$F=\frac{Kq_{1}q_{2}}{d^{2}}$

By substituting the values

$0.235=\frac{9\times10^{9}q^{2}}{0.148^{2}}$

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Thus, the charge on each sphere is $q=7.56\times10^{-7}C$ .

Case II:

Charge on first sphere = 4q

Charge on second sphere = q

distance between them, d = 0.148 m

Force between them, F = 0.235 N

Use Coulomb's law in electrostatics

$F=\frac{Kq_{1}q_{2}}{d^{2}}$

By substituting the values

$0.235=\frac{9\times 10^{9}\times 4q^{2}}{0.148^{2}}$

$q=3.78\times10^{-7}C$

Thus, the charge on second sphere is $q=3.78\times10^{-7}C$ and the charge on first sphere is $4q = 4\times 3.78\times 10^{-7}=1.51 \times 10^{-6} C$ .

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3 years ago

Yung's experiment is performed with light of wavelength 502 nm from excited helium atoms. Fringes are measured carefully on a sc

Lady bird [3.3K]

Answer:

1.082 mm

Explanation:

From the question, we can see that we were given The following

Wavelength of the atoms, λ = 502 nm = 502*10^-9 m

Radius of the screen away from the double slit, r = 1.1 m

We know that Y(20) = 10.2 mm = 10.2*10^-3 m

d = (20 * R * λ) / Y(20)

d = (20 * 1.1 * 502*10^-9)/10.2*10^-3

d = 1.1*10^-5 / 10.2*10^-3

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What is the radius of a tightly wound solenoid of circular cross-section that has 180turns if a change in its internal magnetic

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Answer:

Radius of cross section, r = 0.24 m

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It is given that,

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Since, E = IR

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or

$A=0.19\ m^2$

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$r=\sqrt{\dfrac{A}{\pi}}$

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r = 0.24 m

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Answer:

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Hope it will help :)

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