Find the constant (N/m) of a spring that is stretched 10.53 cm by a 641 g hanging object.

59.65 N/m

(or if we round and use 3 sig figs it is 60.0 N/m)

Explanation:

F=kx × F=mg

Step one:

we have to convert the centimeters to meters and the grams to kilograms. When converting cm to m you divide by 100. When converting g to kg, you divide by 1000. so:

  10.53÷100=0.1053 m    and  641÷1000= 0.641 kg

also since the problem mentions "hanging" we can assume this object is in earth and therefore this object has the acceleration of gravity. Which is 9.8m/s²

step 2:

time to plug in our variables, so,

( N/m x)(0.1053m)×(0.641 kg)(9.8m/s²)

step 3:

since we are missing a variable on the left side of this equation we have to multiply our variables on the right side and then work backwards.

so:(0.641 kg)(9.8m/s²)= 6.2818

and then we divide by our only given variable from the left (0.1053m) to both sides. and it'll end up looking something like this on your paper when showing work:

( N/m x)(0.1053m)×(0.641 kg)(9.8m/s²)

0.1053m                       0.1053m

so: 6.2818÷0.1053= 59.65 N/m or 60.0 N/m

hope this helped :)

answer: the exterior temperature of commercial jet planes as they leave the stratosphere and return to the ground increases steadily.

explanation:

the lower most layer of atmosphere is troposphere which extends from 0 to 12 km above the surface of the earth. the temperature initially decreases as distance is increased from the earth's surface. commercial jets fly in lower part of the stratosphere. the stratosphere extends upto 50 km. the temperature in stratosphere starts rising again due to presence of ozone which traps the uv rays.

thus, when the planes leave the stratosphere and return to the ground, the exterior temperature increases steadily.

like poles repel so keep pushing north to north and repels funny to see this