Your spaceship lands on an unknown planet. To determine the characteristics of this planet, you drop a wrench from 3.00 m above the ground and measure that it hits the ground 0.814 s later What is the acceleration of gravity near the surface of this planet?

The gravitational acceleration near the surface of this planet is 9.056 m/s^2

Explanation:

When you drop an object, the only acceleration acting on the object is the gravitational acceleration, so we can writhe the acceleration of the wrench as

a(t) = -g

where the minus sign is because the wrench will move downwards.

for the velocity of the object, we integrate over time and get:

v(t) = -g*t + v0

where v0 is the initial velocity, in this case is 0, because the wrench is droped.

integrating again, we will get the position of the wrench:

p(t) = -(g/2)*t^2 + p0

where p0 is the initial position, in this case is 3m above the ground.

p(t) = -(g/2)*t^2 + 3m

now, afther 0.814seconds the wrench hits the ground, so we have:

p(0.814 s) = 0 =  -(g/2)*(0.814s)^2 + 3m

now we need to solve this for g.

-(g/2)*(0.814s)^2 + 3m = 0

(g/2)*(0.814s)^2 = 3m  

(g/2) = 3m/*(0.814s)^2 = 4.528m/s^2

g = 2*4.528m/s^2 = 9.056 m/s^2

when a device is used in a circuit in which the voltage is 81 v, the current flowing through the device is 3 a. what is the resistance of the device?

a. 27 ω

b. 40.5 a

c. 243 ω

d. 243 a

the correct answer is a. 27