Glucose turns into glycogen
Answer:
m = 0.1 [kg]
Explanation:
To solve this problem we must use Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration. In this way, we have the following equation.

where:
F = force = 16[N]
a = acceleration = 160 [m/s²]
m = mass [kg]
Now replacing:
![m=F/a\\m = 16/160\\m = 0.1 [kg]](https://tex.z-dn.net/?f=m%3DF%2Fa%5C%5Cm%20%3D%2016%2F160%5C%5Cm%20%3D%200.1%20%5Bkg%5D)
Answer:
T = (99.78 ±0.05) ºC
Explanation:
This problem refers to the procedure to perform the measurements and their correct reporting.
Suppose that the thermometer used has an appreciation of 0.1ºC and the values of 5 measurements resulted in
# T (ºC)
1 99.8
2 99.7
3 99.7
4 99.8
5 99.9
the processing calculate mean value
T_average = ∑
/ n
T_average = (99.8 + 99.7 + 99.7 + 99.8 + 99.9) / 5
T_average = 99.78ºC
the absolute error or uncertainty of the measurement is half the appreciation of the instrument
ΔT = ± 0.05ºC
so the result must be written
T = (99.78 ±0.05) ºC
Answer:
What is the normal force on the object (draw a diagram if needed)
784 N
How much force is required to get the object to start to move from rest (max static friction)?
470.4 N
When the object starts moving, what is the force of kinetic friction?
274.4 N
If the moving object has a tension force of 300N to the right pulling it, what is the net force on the object in the horizontal direction? 784 N
What is the acceleration (with direction) of the object based on your answer for part d? Remember that a = net force/m.
Explanation: