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Norma-Jean [14]

3 years ago

11

75 POINTS AND BRAINLY

2 answers:

kramer3 years ago

8 0

Answer:

B=work=power/time

Explanation:

pentagon [3]3 years ago

4 0

Answer:

54

Explanation:

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As a laudably skeptical physics student, you want to test Coulomb's law. For this purpose, you set up a measurement in which a p

kherson [118]

Explanation:

The Coulomb's law states that the magnitude of each of the electric forces between two point-at-rest charges is directly proportional to the product of the magnitude of both charges and inversely proportional to the square of the distance that separates them:

$F=\frac{kq_1q_2}{d^2}$

In this case we have an electron (-e) and a proton (e), so:

$F=-\frac{ke^2}{d^2}\\F=-\frac{8.99*10^9\frac{N\cdot m^2}{s^2}(1.6*10^{-19}C)^2}{(933*10^{-9}m)^2}\\F=-2.64*10^{-16}N$

In this case, the electric force is negative, therefore, the force is repulsive and its magnitude is:

$F=2.64*10^{-16}N$

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3 years ago

According to the law of conservation of matter,the number of ______ is not changed by a chemical reaction

abruzzese [7]

The answer is mass. I have to comment more than 20 characters.

6 0

3 years ago

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You want to photograph a circular diffraction pattern whose central maximum has a diameter of 1.0 cm. You have a helium-neon las

topjm [15]

Answer:

0.776 m far Pinhole should be placed before the viewing screen

Explanation:

For circular aperture of diameter D will have a bright central maximum of diameter, width is given by

$w=\frac{2.44 \lambda L}{D}$

where $\lambda$ is wavelength of helium neon laser = 633 nm, D=10.cm, w=0.12 mm

Pinhole should be placed before the viewing screen is

$L=\frac{wD}{2.44\lambda}\\L=\frac{0.12\times 10^{-3}\times 0 .01}{2.44\times 633 \times 10^{-9}}\\L=0.776 m$

4 0

3 years ago

The area under the curve of the net external force vs time graph is equal to __________ or ________.

Sophie [7]

Impulse delivered

or

Change in momentum.

7 0

3 years ago

Cyclotrons are widely used in nuclear medicine for producing short-lived radioactive isotopes. These cyclotrons typically accele

lyudmila [28]

Answer:

Part a)

$v = 3.16 \times 10^7 m/s$

Part b)

r = 0.166 m

Explanation:

Part a)

As we know that the energy of the Hydride ion is given as

$E = 5 MeV$

here we have

$\frac{1}{2}mv^2 = 5\times 10^6(1.6 \times 10^{-19})$

also we know that

$m = 1.6 \times 10^{-27} kg$

now we have

$v = \sqrt{\frac{2 \times 5 \times 10^6(1.6 \times 10^{-19}}{1.6\times 10^{-27}}$

$v = 3.16 \times 10^7 m/s$

Part b)

As we know that magnetic force on the charge is centripetal force

so we have

$qvB = \frac{mv^2}{r}$

so we have

$r = \frac{mv}{qB}$

so we have

$r = \frac{1.6 \times 10^{-27}(3.16 \times 10^7)}{(1.6 \times 10^{-19}) 1.9}$

$r = 0.166 m$

4 0

3 years ago