The scale is a rest scale reads the support for is not enough net force.
Answer:
20.4m/s²
Explanation:
Given parameters:
Initial velocity = 0m/s
Distance = 53m
Time = 5.2s
Unknown:
Acceleration = ?
Solution:
This is a linear motion and we use the right motion equation;
S = ut +
at²
S is the distance
u is the initial velocity
a is the acceleration
t is the time
Insert the parameters and solve;
53 = (0x 5.2) +
x a x 5.2
53 = 2.6a
a =
= 20.4m/s²
1) the weight of an object at Earth's surface is given by

, where m is the mass of the object and

is the gravitational acceleration at Earth's surface. The book in this problem has a mass of m=2.2 kg, therefore its weight is

2) On Mars, the value of the gravitational acceleration is different:

. The formula to calculate the weight of the object on Mars is still the same, but we have to use this value of g instead of the one on Earth:

3) The weight of the textbook on Venus is F=19.6 N. We already know its mass (m=2.2 kg), therefore by re-arranging the usual equation F=mg, we can find the value of the gravitational acceleration g on Venus:

4) The mass of the pair of running shoes is m=0.5 kg. Their weight is F=11.55 N, therefore we can find the value of the gravitational acceleration g on Jupiter by re-arranging the usual equation F=mg:

5) The weight of the pair of shoes of m=0.5 kg on Pluto is F=0.3 N. As in the previous step, we can calculate the strength of the gravity g on Pluto as

<span>6) On Earth, the gravity acceleration is </span>

<span>. The mass of the pair of shoes is m=0.5 kg, therefore their weight on Earth is
</span>

<span>
</span>
What you know:
Vi=0m/s
Vf=143.8m/s
A=-9.8m/s
d=???
Use the equation Vf^2=Vi^2+2A(d)
Rearrange to isolate d: d=Vf^2/2A
d=(143.8)^2/2(-9.8)
d=20678.4/-19.6
d=-1055m
The tank was released from a height of 1055m