The gravitational potential energy will increase
Explanation:
The gravitational potential energy (GPE) of an object is the energy possessed by the object due to its position in a gravitational field.
Near the Earth's surface, the GPE of an object is given by
where
m is the mass of the object
g is the acceleration of gravity
h is the heigth of the object above the ground
From the equation, we see that the GPE is directly proportional to the mass: therefore, if the mass increases, the GPE will increase as well.
So, for the beanbag in this problem, when its mass increases, the GPE will increase as well.
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Answer:
The magnitude of the force required to move the electron through the given field is 2.203 N
Explanation:
Given;
The field strength of the electron, E = 1.375 x 10¹⁹ N/C
charge of electron, q = 1.602 x 10⁻¹⁹ C
The magnitude of the force required to move the electron through the given field is calculated as follows;
F = Eq
F = (1.375 x 10¹⁹ N/C) (1.602 x 10⁻¹⁹ C)
F = 2.203 N
Therefore, the magnitude of the force required to move the electron through the given field is 2.203 N
Answer:
Work is the energy transferred to or from an object via the application of force along a displacement.
Answer:
Newton’s Three Laws of Motion has a great impact.
Explanation:
Newton’s Three Laws of Motion has a great impact on the bowling game for the 2 students. When the student one throw ball to the student 2, the ball decrease its speed due to the gravity and opposing air. If these forces are removed from the system the ball will continue its motion till another force is applied on it. When the force applied to the ball it produces acceleration in the direction to the applied force. If the ball touches the ground it bounce back with equal force which is a reaction of the ground.
For an ideal transformer power loss is assumed to be zero
i.e. the power in primary coil due to input voltage must be equal to power in secondary coil due to output voltage
this can be written in form of equation
here we know that
![i_1 = 10 A{/tex]now we will use above equation[tex]140*3.5 = 10 * V_1](https://tex.z-dn.net/?f=i_1%20%3D%2010%20A%7B%2Ftex%5D%3C%2Fp%3E%3Cp%3Enow%20we%20will%20use%20above%20equation%3C%2Fp%3E%3Cp%3E%5Btex%5D140%2A3.5%20%3D%2010%20%2A%20V_1)
So primary coil voltage is 49 Volts
