What type of rays would you expect to be used frequently at a hospital to make medical diagnoses?

An electron has a velocity of 1.50 km/s (in the positive x direction) and an acceleration of 2.00 ✕ 1012 m/s2 (in the positive z

olga nikolaevna [1]

Answer:

see explanation

Explanation:

Given that,

velocity of 1.50 km/s = 1.50 × 10³m/s

acceleration of 2.00 ✕ 1012 m/s2

electric field has a magnitude of strength of 18.0 N/C

$\bar F= q[\bar E + \bar V \times \bar B]\\\\\bar F = [\bar E + \bar V \times ( B_x \hat i +B_y \hat j +B_z \hat z )]\\\\\\m \bar a = [\bar E + \bar V \times ( B_x \hat i +B_y \hat j +B_z \hat z )]$

$9.1 \times 10^-^3^1 \times 2\times 10^1^2 \hat k=-1.6\times10^-^1^9 \hat k [18\hat k+ 1.5\times 10^3 \hat i \times (B_x \hat i +B_y \hat j +B_z \hat k)]$ $42.2 \times 10^-^1^9 \hat k = -2.4 \times 10^1^6B_y \hat k + 2.4 \times 10 ^1^6 \hat j B_z\\$

$B_x = undetermined$

$B_y = \frac{42.2 \times 10^-^1^9}{-2.4 \times 10^-^1^6} \\\\= - 0.0176 T$

$B_z = 0T$

8 0

4 years ago

Two objects attract each other gravitationally. If the mass of each object doubles, how does the gravitational force between the

madam [21]

Answer:

The gravitational force between them quadruples

Explanation:

According to law of gravitation, the force of attraction (F) between two masses m1 and m2 is directly proportional to the product of the masses and inversely proportional to the square of the distance(r) between them. Mathematically,

F1 = Gm1m2/r²... 1

If their masses doubles, the formula becomes;

F2 = G(2m1)(2m2)/r²

F2 = 4Gm1m2/r² ... 2

Dividing equation 2 by 1, we have;

F2/F1 = {4Gm1m2/r²}÷{Gm1m2/r²}

F2/F1 = 4Gm1m2/r²×r²/Gm1m2

F2/F1 = 4

F2 = 4F1

The gravitational force between the masses when they doubles quadruples.

6 0

4 years ago

If two students push a box with a force of 3N in the same direction, and there is no other unbalanced force on the box, what is

FrozenT [24]

It’s 9N because unbalanced force on the box

6 0

3 years ago

Read 2 more answers

Light of wavelength 650 nm is normally incident on the rear of a grating. The first bright fringe (other than the central one) i

koban [17]

Answer:

A

$N = 1340.86 \ slits / cm$

B

$\theta = 15.7^o$

Explanation:

From the question we are told that

The wavelength is $\lambda = 650 \ nm = 650 *10^{-9} \ m$

The angle of first bright fringe is $\theta = 5^o$

The order of the fringe considered is n =1

Generally the condition for constructive interference is

$dsin (\theta ) = n * \lambda$

=> $d = \frac{1 * 650 *10^{-9 }}{ sin(5)}$

=> $d = 7.458 *10^{-6} \ m$

Converting to cm

$d = 7.458 *10^{-6} \ m = 7.458 *10^{-6} * 100 = 0.0007458 \ cm$

Generally the number of grating pre centimeter is mathematically represented as

$N = \frac{1}{d}$

=> $N = \frac{1}{0.0007458}$

=> $N = 1340.86 \ slits / cm$

Considering question B

From the question we are told that

The first wavelength is $\lambda_1 = 650 \ nm = 650 *10^{-9} \ m$

The second wavelength is $\lambda_2 = 429 \ m = 420 *10^{-9 } \ m$

The order of the fringe is $n = 2$

The grating is $N = 5000 \ slits / cm$

Generally the slit width is mathematically represented as

$d = \frac{1}{N }$

=> $d = \frac{1}{ 5000 }$

=> $d = 0.0002 \ c m = 2.0 *10^{-6} \ m$

Generally the condition for constructive interference for the first ray is mathematically represented as

$d sin(\theta_1) = n * \lambda_1$

=> $\theta_1 = sin^{-1} [\frac{ 2 * \lambda }{d}]$

=> $\theta_1 = sin^{-1} [\frac{ 2 * 650 *10^{-9} }{ 2*10^{-6}}]$

=> $\theta_1 = 40.5 ^o$

Generally the condition for constructive interference for the second ray is mathematically represented as

$d sin(\theta_2) = n * \lambda_2$

=> $\theta_2 = sin^{-1} [\frac{ 2 * \lambda_1 }{d}]$

=> $\theta_2 = sin^{-1} [\frac{ 2 * 420 *10^{-9} }{ 2*10^{-6}}]$

=> $\theta_2 = 24.8 ^o$

Generally the angular separation is mathematically represented as

$\theta = \theta_1 - \theta_1$

=> $\theta = 42.5^o - 24.8^o$

=> $\theta = 15.7^o$

4 0

3 years ago

(Please help asap)

UNO [17]

Answer:

mass yes

Explanation:

8 0

3 years ago

Read 2 more answers