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Natali [406]
4 years ago
13

The magnetic field at the equator points north. if you throw a positively charged object (for example, a baseball with some elec

trons removed) to the east, what is the direction of the magnetic force on the object
Physics
1 answer:
asambeis [7]4 years ago
7 0
Recall the equation for magnetic force:

F = qv x B          *x is cross product, not separate variable!

If the magnetic field points towards N and you throw E, then the magnetic force would point up, or out of the page. Use the right-hand rule. You point your finger towards the direction of the object, and curl your finger to the magnetic field. Your thumb is the direction of the magnetic force.

Hope this helps!
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Find the vector sum of the 3 vectors. A= -3i - 2j + 7k, B= i + 3j + 3k, B= -5k​
DENIUS [597]

Answer:

34k+B plus 9 :)

Explanation:

7 0
3 years ago
A car is moved 26km/hr due east for 4minute.what is its average velocity in m/s?​
fenix001 [56]

Answer:

25m/sec

Explanation:

Speed of car in first 15 minutes = 40 km/h

Distance covered = speed × Time taken

60 minutes = 1 hour

15 minutes =

60

15

hour

Distance covered = 240×

60

15

=10km

Speed of car in next 15 minutes = 60 km/h

Distance covered = 60×

60

15

=15km

∴ Total distance covered = (10+15)km = 25 km

6 0
3 years ago
A series of lines involving a common level in the spectrum of atomic hydrogen lies at 656.46 nm, 486.27 nm, 434.17 nm, and 410.2
julsineya [31]

Answer:

The wavelength of next line in the series will be 397.05 nm

Explanation:

From Rydberg equation;

\frac{1}{\lambda} = R_H(\frac{1}{n_1^2} -\frac{1}{n_2^2})

where;

λ is the wavelength

n lines in the series

RH is Rydberg constant = 1.097 x 10⁷ m⁻¹

Also at a given maximum wavelength, we can determine the first line n₁ in the series

\frac{1}{\lambda_{max}R_H} = \frac{1}{n_1^2} -\frac{1}{(n_1 +1)^2} \\\\

\frac{1}{\lambda_{max}R_H} = \frac{2n_1+1}{n_1^2(n_1 +1)^2} \\\\\lambda_{max}R_H} = \frac{n_1^2(n_1 +1)^2}{2n_1+1}

Given;

maximum wavelength = 656.46 nm

\lambda_{max}R_H} = \frac{n_1^2(n_1 +1)^2}{2n_1+1}\\\\656.46 *10^{-9}*1.097*10^7 = \frac{n_1^2(n_1 +1)^2}{2n_1+1}\\\\7.2 = \frac{n_1^2(n_1 +1)^2}{2n_1+1}

Now, test for different values of n that will be equal to 7.2

let n₁ = 1

\frac{n_1^2(n_1 +1)^2}{2n_1+1} =  \frac{(1)^2(1 +1)^2}{2(1)+1} = 1.3\\

n₁(1) ≠ 7.2

Again, let n₁ = 2

\frac{n_1^2(n_1 +1)^2}{2n_1+1} =  \frac{(2)^2(2 +1)^2}{2(2)+1} = 7.2\\

∴ n₁(2) = 7.2

For the least wavelength given as 410.29 nm, n = ?

\frac{1}{\lambda} = R_H(\frac{1}{n_1^2} -\frac{1}{n_2^2})\\\\\frac{1}{410.29*10^{-9}} = 1.097*10^7(\frac{1}{2^2} -\frac{1}{n_2^2})\\\\\frac{1}{4.5} =\frac{1}{4} -\frac{1}{n_2^2}\\\\\frac{1}{n_2^2} =\frac{1}{4}  -\frac{1}{4.5} \\\\\frac{1}{n_2^2}  = 0.0277778\\\\n_2^2 = \frac{1}{0.0277778} \\\\n_2^2 = 36\\\\n_2= \sqrt{36}\  = 6

next line in the series will be 7

The wavelength of next line in the series will be;

\frac{1}{\lambda} = R_H(\frac{1}{n_1^2} -\frac{1}{n_2^2})\\\\\frac{1}{\lambda} = 1.097*10^7(\frac{1}{2^2} -\frac{1}{7^2})\\\\\frac{1}{\lambda}  =  1.097*10^7(0.22959)\\\\\frac{1}{\lambda}   = 2518602.3\\\\\lambda = 397.05 \ nm

Therefore, the wavelength of next line in the series will be 397.05 nm

5 0
3 years ago
Question 4
zmey [24]
I think the answer is B but not too sure.
8 0
3 years ago
Calculate the wavelength of the electromagentic waves with the given frequencies, and determine the type of electromagnetic radi
ololo11 [35]

To develop this problem we require the concepts related to wavelength and its expression to calculate it.

The wavelength is given by

\lambda =\frac{c}{f}

Where,

c=3*10^8 m/s light velocity

f = frequency.

Our values are given by,

f_1=4.38*10^{14}Hz

f_2= 4.14*10^{20}Hz

f_3 = 3.24*10^{12}Hz

Then,

\lambda_1=\frac{3*10^8}{4.38*10^{14}}= 6.8493*10^{-7}m Visible

\lambda_2=\frac{3*10^8}{4.14*10^{20}}= 7.24*10^{-13}m Gamma Ray

\lambda_3=\frac{3*10^8}{3.24*10^{12}}= 9.259*10^{-5}m Infrared

<em>*Note the designation on the type of rays that are, can be found in consulted via On-line or in the optical books referring to the electromagnetic spectrum table with their respective ranges.</em>

5 0
4 years ago
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