Answer:
doubled as well.
Explanation:
Using what I call "Method of Simplest Integers," insert all charges and distance as just 1 (ignore k, because it is a constant), so
Fe=k(1*1)/(1^2)=k(1)/1=k(1)
Since one of the charges is doubled,
Fe'=k(2*1)/(1^2)'k(2)/1=k(2)
Set Fe and Fe' side by side and compare their outcomes. cancel out k, because it's constant, and find out what multiple makes Fe becomes Fe'. In this case, it is 2. So the force would be doubled if one of the
Explanation:
Acceleration. Angular acceleration: Is the rate of change of the angular velocity of a body with respect to time.
Force. Torque: Is also called rotational force, since an applied torque will change the rotational motion of a body.
Mass. Moment of inertia: It is the resistance that opposes a body to rotates.
Work. Work: In a rotational motion, the work is done by the torque.
Translational kinetic energy. Rotational kinetic energy: is the kinetic energy due to the rotational motion of a body.
Linear momentum. Angular momentum: Represents the quantity of rotational motion of a body.
Impulse. Angular impulse: Is the change in angular momentum of a body.
Answer:
The resultant velocity of the plane relative to the ground is;
150 kh/h north
Explanation:
The flight speed of the plane = 210 km/h
The direction of flight of the plane = North
The speed at which the wind is blowing = 60 km/h
The direction of the wind = South
Therefore, representing the speed of the plane and the wind in vector format, we have;
The velocity vector of the plane = 210.
The velocity vector of the wind = -60.
Where, North is taken as the positive y or direction
The resultant velocity vector is found by summation of the two vectors as follows;
Resultant velocity vector = The velocity vector of the plane + The velocity vector of the wind
Resultant velocity vector = 210. + (-60.) = 210. - 60. = 150.
The resultant velocity vector = 150.
Therefore, the resultant velocity of the plane relative to the ground = 150 kh/h north.
Answer:
a) The heat input per cycle is 2857.143 joules.
b) The temperature of the low-temperature reservoir is 49.655 °C.
Explanation:
a) The efficiency of the Carnot engine is defined by the following formula:
(1)
Where:
- Low temperature reservoir, in Kelvin.
- High temperature reservoir, in Kelvin.
- Heat output, in joules.
- Heat input, in joules.
- Engine efficiency, no unit.
If we know that and , the heat input of the Carnot engine is:
The heat input per cycle is 2857.143 joules.
b) If we know that and , then the temperature of the low-temperature reservoir:
The temperature of the low-temperature reservoir is 49.655 °C.