When a 83.0 kg. person climbs into an 1600 kg. car, the car's springs compress vertically 1.6 cm. what will be the frequency of vibration when the car hits a bump? (ignore damping.)

the correct answer is f= 0.8747 oscillations/sec

Explanation:

The first bit of info allows you to find the spring constant;  

K = force of compression/compression  

where the force of compression is the "added" weight on the spring;  

K = (83)(9.8)/(.016)              K= 50837.5

Once you have ,K , you can use the frequency relation between ,K, and total mass on spring ,      M= 1600Kg + 83Kg   M=1683 kg, as;  

f =(1/2Pi)SqRt[K/M] oscillations/sec

f= (1/2 (3.1416)SqRt (50837.5/1683)

f=1/6.2832)SqRt 30.2065

f=  0.1592 *5.4960            f= 0.8747 oscillations/sec

weather forecasts are done by collecting data about the current state of the atmosphere mainly the temperature, humidity and wind and using meteorology, to determine the atmosphere that evolves over time. the h and l mean the high and low pressure. the red strip is a warm front. the blue strip is a cold front. the blue and red strip is a stationary front. the purple strip is a occluded front.

explanation: