The magnitude of the force acting on the object lying on a flat surface without moving is 10 N.
The given parameters;
- magnitude of force on the object, F = 10 N
- angle between the object and the horizontal flat surface = 0⁰
Apply Newton's second law of motion to determine the magnitude of the force on the object.
Due to the position of the object, the magnitude of the force acting on it is calculated as;
Therefore, the magnitude of the force acting on the object is 10 N.
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The formula for potential energy is mass x height x gravitational force. Your mass is 5.3 kg and your height is 6.6 meters. The gravitational force on earth is 9.8 m/s. That means your answer is 5.3 x 6.6 x 9.8 and that equals 342.804
Answer:
Vf = 41.6 [m/s]
Explanation:
To solve this problem we must use the equations of kinematics.
Vf² = Vo² + (2*g*y)
where:
Vf = final velocity [m/s]
Vo = initial velocity = 0
g = gravity acceleration = 9.81 [m/s²]
y = height = 88.2 [m]
Note: The positive sign of the equation tells us that the acceleration of gravity goes in the direction of motion.
Vf² = Vo² + (2*g*y)
Vf² = 0 + (2*9.81*88.2)
Vf = (1730.48)^0.5
Vf = 41.6 [m/s]
The wavelength gets longer.
Answer:
Speed of the ball in position D 
<u>Explanation:</u>
Given:
Position of B=11.3
Position of D=1.9
Velocity of B=6.2
To Find:
Velocity of D
Solution:
According to the formula, Velocity is given as
![V d=\sqrt{\left[V b^{2}+(2 \times g \times d y)\right]}](https://tex.z-dn.net/?f=V%20d%3D%5Csqrt%7B%5Cleft%5BV%20b%5E%7B2%7D%2B%282%20%5Ctimes%20g%20%5Ctimes%20d%20y%29%5Cright%5D%7D)
=Velocity of B
=Velocity of D
g=acceleration due to gravity=9.8 m/s^2
=Change in position of B and D
Substitute the all values in the above equation we get
![V d=\sqrt{\left[6.2^{2}+(2 \times 9.8 \times(11.3-1.9))\right]}](https://tex.z-dn.net/?f=V%20d%3D%5Csqrt%7B%5Cleft%5B6.2%5E%7B2%7D%2B%282%20%5Ctimes%209.8%20%5Ctimes%2811.3-1.9%29%29%5Cright%5D%7D)
![=\sqrt{[38.44+(2 \times 9.8 \times(9.4))]}](https://tex.z-dn.net/?f=%3D%5Csqrt%7B%5B38.44%2B%282%20%5Ctimes%209.8%20%5Ctimes%289.4%29%29%5D%7D)
![=\sqrt{[38.44+(19.6 \times 9.4)]}](https://tex.z-dn.net/?f=%3D%5Csqrt%7B%5B38.44%2B%2819.6%20%5Ctimes%209.4%29%5D%7D)
Result :
The velocity of D is
