At one instant in time, a rock falling off a cliff experiences an upward force from air resistance at a magnitude of 10 n. the rock itself has a mass of 7 kg. what is the magnitude and direction of the acceleration of the rock at this instant? (take the magnitude of acceleration due to gravity to be 9.8 m/s2 and indicate direction by using a value for downward and a (+) value for upward.)
(hint: what is the net force acting on the rock? )
note: be careful entering your units! follow the instructions in the guide posted in the homework section!

a=-8.4 m/s^2

Explanation:

We can use (-) value for downward and a (+) value for upward

The weight of the rock is calculated as follows:

W=mass*gravity

W=7kg*(-9.8m/s^2)

W=-68.8N

The air resistance force makes sense opposite to the weight of the rock:

Fr=+10N

The total force on the rock is:

Ft= Fr+W

Ft=10N-68.8N

Ft=-58.8N

We also know that force is the product of mass by acceleration:

Ft= m*a

Thus:

a= Ft/m

a=-58.8N/7kg

a=-8.4 m/s^2 (downward)

spare some change and give it to the poor

explanation:

plz : ((

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg