Using V= vo +at with Vo = 0 and a= 4m/sec2.
V= 0+ 4x8= 32m/s
Answer:
shown in the attachment
Explanation:
The detailed step by step and necessary mathematical application is as shown in the attachment.
Answer:
![g_{moon}=1.67 [m/s^{2} ]](https://tex.z-dn.net/?f=g_%7Bmoon%7D%3D1.67%20%5Bm%2Fs%5E%7B2%7D%20%5D)
Explanation:
The weight of some mass is defined as the product of mass by gravitational acceleration. In this way using the following formula we can find the weight.

where:
w = weight [N]
m = mass = 0.06 [kg]
g = gravity acceleration = 10 [N/kg]
Therefore:
![w=0.06*10\\w=0.6[N]](https://tex.z-dn.net/?f=w%3D0.06%2A10%5C%5Cw%3D0.6%5BN%5D)
By Hooke's law we know that the force in a spring can be calculated by means of the following expression.

where:
k = spring constant [N/m]
x = deformed distance = 6 [cm] = 0.06 [m]
We can find the spring constant.
![k= F/x\\k=0.6/0.06\\k=10 [N/m]](https://tex.z-dn.net/?f=k%3D%20F%2Fx%5C%5Ck%3D0.6%2F0.06%5C%5Ck%3D10%20%5BN%2Fm%5D)
Since we use the same spring on the moon and the same mass, the constant of the spring does not change, the same goes for the mass.
![F_{moon}=k*0.01\\F = 10*0.01\\F=0.1[N]](https://tex.z-dn.net/?f=F_%7Bmoon%7D%3Dk%2A0.01%5C%5CF%20%3D%2010%2A0.01%5C%5CF%3D0.1%5BN%5D)
Since this force is equal to the weight, we can now determine the gravitational acceleration.
![F=m*g_{moon}\\g=F/m\\g = 0.1/0.06\\g_{moon} = 1.67[m/s^{2} ]](https://tex.z-dn.net/?f=F%3Dm%2Ag_%7Bmoon%7D%5C%5Cg%3DF%2Fm%5C%5Cg%20%3D%200.1%2F0.06%5C%5Cg_%7Bmoon%7D%20%3D%201.67%5Bm%2Fs%5E%7B2%7D%20%5D)
Answer:

Explanation:
To determine the final temperature of the sample, we use the specific heat formular as follows:

Finally, the temperature of the aluminium sample has raised 18 K.
The force exerted by student A with his scooter is 306 N and that of student B is 204 N.
<h3>
Force applied by each student</h3>
The force exerted by each student is calculated from Newton's second law of motion.
F = ma
where;
- m is mass
- a is acceleration
F(A) = 127.5 x 2.4
F(A) = 306 N
F(B) = 120 x 1.7
F(B) = 204 N
Thus, the force exerted by student A with his scooter is 306 N and that of student B is 204 N.
Learn more about force here: brainly.com/question/12970081
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