We Know, F = m*a
Here, F = 34 N
m = 213 Kg
Substitute their values in the equation,
34 = 213 * a
a = 34/213
a = 0.159 m/s²
So, your final answer & the acceleration of the object would be 0.159 m/s²
Hope this helps!
Answer:
17.64 km/h
Explanation:
mass of car, m = 1000 kg
Kinetic energy of car, K = 1.2 x 10^4 J
Let the speed of car is v.
Use the formula for kinetic energy.

By substituting the values

v = 4.9 m/s
Now convert metre per second into km / h
We know that
1 km = 1000 m
1 h = 3600 second
So, 
v = 17.64 km/h
Thus, the reading of speedometer is 17.64 km/h.
Answer:
The necessary separation between the two parallel plates is 0.104 mm
Explanation:
Given;
length of each side of the square plate, L = 6.5 cm = 0.065 m
charge on each plate, Q = 12.5 nC
potential difference across the plates, V = 34.8 V
Potential difference across parallel plates is given as;

Where;
d is the separation or distance between the two parallel plates;

Therefore, the necessary separation between the two parallel plates is 0.104 mm
Answer:
The table can be used to predict the properties of elements, even those that have not yet been discovered. Columns (groups) and rows (periods) indicate elements that share similar characteristics.
The table makes trends in element properties apparent and easy to understand.
The table provides important information used to balance chemical equations. Atoms are important because they form the basic building blocks of all visible matter in the universe. There are 92 types of atoms that exist in nature, and other types of atoms can be made in the lab. The different types of atoms are called elements. Hydrogen, gold and iron are examples of elements comprised of unique types of a single kind of atom.
Explanation: