Question

27. The engine of a car exerts a force magnitude 50000 N on it. The frictional force between the car and the surface of the road is
3000 N. The net force on the car is​

Answer 1

Given :

Magnitude of the force exerted by engine of a car = 50,000 N

The frictional force b/w the car and the surface of the road = 3,000 N

To Find :

Net force on the car

Solution :

Net force acting on the body ,

$\sf \Sigma F_x=F-F_f$

$:\implies \sf \Sigma F_x=50,000-3,000$

$:\implies \sf \Sigma F_x=47,000\ N\ \; \bigstar$

More Info :

Newton 2nd Law :

It states that Force is defined as mass times the acceleration .

$\sf \bigstar\ \; \huge{\pink{F=ma}}$

Relation b/w Pressure and force is given by :

$\bigstar\ \; \sf \huge{\green{P=\dfrac{F}{A}}}$

Relation b/w Density and volume is given by :

$\bigstar\ \; \sf \huge{\blue{\rho =\dfrac{m}{V}}}$

Answer 2

ANSWER

Total drawings= 7500 * 4

                       =30000

Rate= 10%

Case 1: If he withdrew in 7500 at the beginning of each quarter-  

Note: When an equal amount of drawings is withdrawn at the beginning of each quarter, interest is calculated for a period of 7.5 months.

Hence,

Interest on drawings= 30000 * 10/100 * 7.5/12

                                 = 1875

Case 2: If he withdrew in 7500 at the end of each quarter-

Note: When an equal amount of drawings is withdrawn at the end of each quarter, interest is calculated for a period of 4.5 months.

Hence,

Interest on drawings= 30000 * 10/100 * 4.5/12

                                 = 1125

Case 3: If he withdrew in 7500 during the middle of each quarter-

Note: When an equal amount of drawings is withdrawn during the middle of each quarter, interest is calculated for a period of 6 months.  

Hence,

Interest on drawings= 30000 * 6/12 * 10/100

                                 = 1500

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