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scoundrel [369]
3 years ago
12

A car initally at rest accelerates in a straight line at 3 m/s2 what will be its speed after 2 seconds

Physics
1 answer:
Ainat [17]3 years ago
5 0

Answer

In this Question we have given

acceleration, a=3m/s^2

and time, t=2s

we have to find speed, v=?

Apply newton's Ist equation of motion which goes as

[tex]v =u + at\\ Here   v  is final velocity\\u - initial velocity\\a-acceleration\\t- time[/tex]

In this problem the car starts from rest it means initial velocity, u =0.

Putting values of u, a and t in first equation of motion

v=0+3*2\\v=6m/s

Therefore speed of car after 2 second will be 6m/s

Note: In this type of problems. First check  the units of all quatities. As all the quantities must be first changed into matching units i.e cgs or s.i. units before applying the equation.

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8 0
2 years ago
Two identical conducting spheres are placed 80.0 cm apart. One is given a charge of 5.8 C and the other is given a charge of 6.4
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Answer: 5.214(10)^{11} N

Explanation:

According to <u>Coulomb's Law:</u>  

<em>"The electrostatic force F_{E} between two point charges q_{1} and q_{2} is proportional to the product of the charges and inversely proportional to the square of the distance d that separates them, and has the direction of the line that joins them".</em>

<em />

Mathematically this law is written as:

F_{E}=K\frac{q_{1}.q_{2}}{d^{2}}  

Where:

F_{E}  is the electrostatic force

K=8.99(10)^{9} Nm^{2}/C^{2} is the Coulomb's constant  

q_{1}=5.8 C and q_{2}=6.4 C are the electric charges

d=80 cm \frac{1 m}{100 cm}=0.8 m is the separation distance between the charges

Solving:

F_{E}=8.99(10)^{9} Nm^{2}/C^{2}\frac{(5.8 C)(6.4 C)}{(0.8 m)^{2}}  

F_{E}=5.214(10)^{11} N    

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