For the second question you’re solving for resistance. resistance= voltage/ current. 120/0.5= 240. the answer is 240 ohms
for the third question you would do 2*4 since it’s asking for voltage, the answer is 8 volts :)
Answer:
30 miles
Explanation:
<u>Step 1:</u>
Divide -> 45/60= .75 miles/minute
<u>Step 2:</u>
Multiply -> .75 x 40= 30
If it increased its speed steadily at a constant rate, then the average speed for the minute was
(1/2)(10m/s + 20m/s) = 15 m/s .
Rolling at an average speed of 15 m/s for 1 minute (60 seconds), it travels
(15 m/s) (60 sec) = 900 meters
Answer:
Explanation:
Given the following :
Speed (V) = speed of 2.30×10^7 m/s
Acceleration (a) = 1.70×10^13 m/s^2
Using the right hand rule provided by Lorentz law:
B = F / qvSinΘ
Where B = magnitude of the magnetic field
v = speed of the particle
Θ = 90° (perpendicular to the field)
q = charge of the particle
SinΘ = sin90° = 1
Note F = ma
Therefore,
B = ma / qvSinΘ
Mass of proton = 1.67 × 10^-27
Charge = 1.6 × 10^-19 C
B = [(1.67 × 10^-27) × (1.70 × 10^13)] / (1.6 × 10^-19) × (2.30 × 10^7) × 1
B = 2.839 × 10^-14 / 3.68 × 10^-12
B = 0.7715 × 10^-2
B = 7.72 × 10^-3 T
2) Magnetic field will be in the negative y direction according to the right hand thumb rule.
Since Velocity is in the positive z- direction, acceleration in the positive x - direction, then magnetic field must be in the negative y-direction.
Answer:
D. Top is emission; bottom absorption.
Explanation:
Emission and spectrum of elements are due to the element absorbing or emitting wavelength of e-m energy. Elementary particles of elements can absorb energy from a ground state to enter an excited state, creating an absorption spectrum, or they can lose energy and fall back to a lower energy state, creating an emission spectrum. A simple rule to differentiate between an emission and an absorption spectrum is that: "all absorbed wavelength is emitted, but not all emitted wavelength is absorbed."
From the image, the lines indicates wavelengths. We can see that all of the wavelengths of the bottom absorption spectrum coincides with some of the wavelength of the upper emission wavelengths.