Answer:
Wmoon = 131 [N]
Explanation:
We know that the weight of a body is equal to the product of mass by gravitational acceleration.
Since we are told that the gravitational acceleration of the moon is equal to one-sixth of the acceleration of Earth's gravitation. Then we must multiply the value of Earth's gravitation by one-sixth.
![w_{moon}=\frac{1}{6} *m*g\\w_{moon}=\frac{1}{6} *80*9.81\\w_{moon}=130.8 [N] = 131 [N]](https://tex.z-dn.net/?f=w_%7Bmoon%7D%3D%5Cfrac%7B1%7D%7B6%7D%20%2Am%2Ag%5C%5Cw_%7Bmoon%7D%3D%5Cfrac%7B1%7D%7B6%7D%20%2A80%2A9.81%5C%5Cw_%7Bmoon%7D%3D130.8%20%5BN%5D%20%3D%20131%20%5BN%5D)
Answer:
The correct answer is inertia.
Explanation:
The heavy bag of groceries is initially within the inertia frame of the car. This indicates that the heavy bag acquires the same speed as the car.
When the car stops, the heavy bag continues to move forward with the speed it had due to the principle of inertia, which states the property that the bodies cannot modify by themselves the state of rest or movement in which they are.
Have a nice day!
Answer:
Acceleration of Sea Lion is 4.41 g
This is 49% of maximum jet acceleration given as a = 9g
Explanation:
As we know that the radius of the circular loop is given as
R = 0.37 m
The speed of the fish is given as
Now the centripetal acceleration of the sea lion is given as
as we know that
so we have
Now Percentage of this acceleration wrt maximum jet acceleration is given as
%
Answer:
maximum speed of the bananas is 18.8183 m/s
Explanation:
Given data
amplitude A = 23.195 cm
spring constant K = 15.2676 N/m
mass of the bananas m = 56.9816 kg
to find out
maximum speed of the bananas
solution
we know that radial oscillation frequency formula that is = √(K/A)
radial oscillation frequency = √(15.2676/23.195)
radial oscillation frequency is 0.8113125 rad/s
so maximum speed of the bananas = radial oscillation frequency × amplitude
maximum speed of the bananas = 0.8113125 × 23.195
maximum speed of the bananas is 18.8183 m/s
