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Hatshy [7]
3 years ago
8

Please help with this​

Physics
1 answer:
hichkok12 [17]3 years ago
5 0

Answer:

3/5i + 4/5J

Explanation:

use the commtative property to reorder the terms

3i + 4 j

to

4j + 3i

The conjugate of a complex number a=b is the number a-bi

4j -3i

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Which should not be a part of scientific inquiry?
lisabon 2012 [21]

I would say;

a - bias.

8 0
3 years ago
In the parts that follow select whether the number presented in statement A is greater than, less than, or equal to the number p
Brums [2.3K]

Answer:

a) Therefore 2.6km is greater than 2.57km.

Statement A is greater than statement B.

b) Therefore 5.7km is equal to 5.7km

Statement A is equal to statement B

Explanation:

a) Statement A : 2.567km to two significant figures.

2.567km 2. S.F = 2.6km

Statement B : 2.567km to three significant figures.

2.567km 3 S.F = 2.57km

Therefore 2.6km is greater than 2.57km.

Statement A is greater than statement B.

b) statement A: (2.567 km + 3.146km) to 2 S.F

(2.567km + 3.146km) = 5.713km to 2 S.F = 5.7km

Statement B : (2.567 km, to two significant figures) + (3.146 km, to two significant figures).

2.567km to 2 S.F = 2.6km

3.146km to 2 S.F = 3.1km

2.6km + 3.1km = 5.7km

Therefore 5.7km is equal to 5.7km

Statement A is equal to statement B

5 0
3 years ago
An electromagnetic wave has a frequency of 4.0 x 10^18 Hz. What is the wavelength of the wave?
LenaWriter [7]

Answer:

7.5 × 10^-11 m

Explanation:

Hope this helps !

3 0
3 years ago
Read 2 more answers
The predictions of Rutherford's scattering formula failed to correspond with experimental data when the energy of the incoming a
Vera_Pavlovna [14]

Answer:

The radius of the gold nucleus is 7.1x10⁻¹⁵m

Explanation:

The nearest distance is:

r=\frac{kze^{2} }{mpV^{2} } (eq. 1)

Where

z = atomic number of gold = 79

e = electron charge = 1.6x10⁻¹⁹C

k = electrostatic constant = 9x10⁹Nm²C²

energy of the particle = 32 MeV = 5.12x10⁻¹²J

At the potential energy is zero, all the energy will be kinetic energy:

E_{k} =\frac{1}{2} m V^{2}

Where

m = 4 mp = mass of proton

5.12x10^{-12} =\frac{1}{2} *4*m_{p}* V^{2} \\m_{p}* V^{2} = 2.56x10^{-12}

Replacing in equation 1

r=\frac{9x10^{9}*79*(1.6x10^{-19})^{2}   }{2.56x10^{-12} } =7.1x10^{-15} m

8 0
3 years ago
What is the period of a wave traveling 5 m/s if its wavelength is 20 m/s
Ilia_Sergeevich [38]

Speed of wave is given as

v = 5 m/s

Wavelength of the wave is given as

\lambda = 20 m

now from the formula of wave time period we can say

speed = \frac{wavelength}{time period}

5 = \frac{20}{T}

T = \frac{20}{5}

T = 4 s

so it will have time period of T = 4 s

7 0
3 years ago
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