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Zolol [24]
3 years ago
14

Mass number of an atom with 11 protons

Physics
1 answer:
Ratling [72]3 years ago
8 0
The atomic mass of sodium (Na) is 22.98977
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An example of an action and reaction is sitting on a chair. Your body pushes down on the chair because of the force of what?
vaieri [72.5K]

Your body is pushing down on a chair because it is being attracted by gravity, the reason that your body is not moving down because there is a normal force acting on your body; together the net force of weight (m*g) and normal force is equal to zero

8 0
3 years ago
A car is moving from rest and the velocity increases to 30 m/s in 4 seconds. Calculate its acceleration.
vaieri [72.5K]

Answer:

7.5 m/s²

Explanation:

Given:

v₀ = 0 m/s

v = 30 m/s

t = 4 s

Find: a

v = at + v₀

(30 m/s) = a (4 s) + (0 m/s)

a = 7.5 m/s²

8 0
3 years ago
The question states: two large, parallel conducting plates are 12cm
AnnZ [28]

Answer:

1. 24375 N/C

2. 2925 V

Explanation:

d = 12 cm = 0.12 m

F = 3.9 x 10^-15 N

q = 1.6 x 10^-19 C

1. The relation between the electric field and the charge is given by

F = q E

So, E=\frac{F}{q}

E=\frac{3.9 \times 10^{-15}}{1.6 \times 10^{-19}}

E = 24375 N/C

2. The potential difference and the electric field is related by the given relation.

V = E x d

where, V be the potential difference, E be the electric field strength and d be the distance between the electrodes.

By substituting the values, we get

V = 24375 x 0.12 = 2925 Volt

6 0
2 years ago
A simple harmonic oscillator has amplitude 0.43 m and period 3.9 sec. What is the maximum acceleration?
rjkz [21]

Answer:

Maximum acceleration in the simple harmonic motion will be 0.854rad/sec^2              

Explanation:

We have given amplitude of simple harmonic motion is A = 0.43 m

Time period of the oscillation is T = 3.9 sec

We have to find the maximum acceleration

For this we have to find the angular frequency

Angular frequency will be equal to \omega =\frac{2\pi }{T}=\frac{2\times 3.14}{3.9}=1.61rad/sec

Maximum acceleration is given by a_{max}=\omega ^2A=1.61^2\times 0.43=0.854rad/sec^2

So maximum acceleration in the simple harmonic motion will be 0.854rad/sec^2

8 0
3 years ago
Read 2 more answers
If an object accelerates from rest, with a constant of 8 m/s2, what will its velocity be after 35s?
AlladinOne [14]

Answer:

<em>Its speed will be 280 m/s</em>

Explanation:

<u>Constant Acceleration Motion</u>

It's a type of motion in which the speed of an object changes by an equal amount in every equal period of time.

If a is the constant acceleration, vo the initial speed, vf the final speed, and t the time, vf can be calculated as:

v_f=v_o+at

The object accelerates from rest (vo=0) at a constant acceleration of a=8\ m/s^2. The final speed at t=35 seconds is:

v_f=0+8*35

v_f=280\ m/s

Its speed will be 280 m/s

5 0
2 years ago
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