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V = 1/2at^2
a = 9.8 m/2^2 (constant)
t = 4.0 s
1/2 • 9.8 • 4^2
1/2 • 9.8 • 16
= 78.4 m/s
<h3>Answer:- </h3><h3>16000 </h3><h3> Explanation:-</h3><h3 /><h3>force = 800N</h3><h3>area of 1 feet = 0.025</h3><h3>Area of 2 feet= 0.025+0.025 = 0.050m²</h3><h3>presure = force / area </h3><h3> = 800 N / 0.050</h3><h3> = 16000</h3>
Answer:
check the image above please
The period of a spring-cart system motion when the system is in orbit inside the international space station will not change because in conditions of weightlessness it is not possible to measure the mass of a body from its weight and the period depends on the mass of the body.