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cricket20 [7]
3 years ago
13

Please view attached images. The first one is #2 and the second one is #1.

Physics
1 answer:
MakcuM [25]3 years ago
8 0
The proton would be 2 and d a part of 1 then calculate that hope this helped
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Which factor is not a type of spectra?
elena-s [515]
Discontinuous is not a spectra
4 0
4 years ago
A 0.40-kg cart with charge 4.0 x 10-5 C starts at rest on a horizontal frictionless surface 0.50 m from a fixed object with char
maw [93]

Answer:

26.82m/s

Explanation:

Given

Mass = m= 0.4kg

Initial Velocity = u = 0

Charge = 4.0E-5C

Distance= d = 0.5m

Object Charge = 2E-4C

First, we'll calculate the initial energy (E)

E = Potential Energy

PE = kQq / d

Where k = coulomb constant = 8.99E9Nm²/C²

Energy is then calculated by;

PE = 8.99E9 * 4E-5 * 2E-4 / 0.5

PE = 143.84J

Energy = Potential Energy = Kinetic Energy

K.E = ½mv² = 143.84J

½mv² = ½ * 0.40 * v² = 143.85

0.2v² = 143.85

v² = 143.85/0.2

v² = 719.25

v = √719.25

v = 26.81883666380777

v = 26.82m/s

Hence, the object is 26.82m/s fast when the cart moving is very far (infinity) from the fixed charge

4 0
4 years ago
Hi! Can you help me? :) A satellite's speed is 10,000 m/s. After 1 min, it is 5,000 m/s. What is the satellite's acceleration? T
Kruka [31]
<span>a = ΔV/Δt = (5000-10000)/60 = -500/6 = -83.(3) m/sec^2</span>
7 0
4 years ago
Read 2 more answers
Calculate the electric field intensity at a point 3 cm away from point charge of 3 x 10^-9 C.
lukranit [14]

Answer:

The electric field intensity is <u>30000 N/C.</u>

Explanation:

Given:

Magnitude of the point charge is, q=3\times 10^{-9}\ C

Distance of the given point from the point charge is, d=3\ cm=0.03\ m

Electric field intensity is directly proportional to the magnitude of point charge and inversely proportional to the square of the distance of the point and the given charge.

Therefore, electric field intensity 'E' at a distance of 'd' from a point charge 'q' is given as:

E=\frac{kq}{d^2}

Plug in k=9\times 10^9\ N\cdot m^2/C^2, q=3\times 10^{-9}\ C, d=0.03\ m. Solve for 'E'.

E=\frac{(9\times 10^9\ N\cdot m^2/C^2)(3\times 10^{-9}\ C)}{(0.03\ m)^2}\\\\E=\frac{27}{0.0009}\ N/C\\\\E=30000\ N/C

Therefore, the electric field intensity at a point 3 cm from the point charge is 30000 N/C.

3 0
3 years ago
How can the difference in the brightness of spectral lines be explained?
Ksenya-84 [330]
<span>more lines = a lot of electrons returning back to ground state from same level</span>
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4 years ago
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