Yes.
5
A 5-degree fall over 2 meters corresponds to a vertical drop of approximately 0.174 meters, or about 17.4 centimeters. This can be calculated using basic trigonometry, where the vertical drop (rise) is the sine of the angle multiplied by the horizontal distance. In this case, ( \text{Drop} = 2 \times \sin(5^\circ) ).
A fall of 4 degrees over 1 meter refers to a slope or incline where the vertical drop is 4 degrees relative to the horizontal. To calculate the vertical drop, you can use the tangent function: the vertical drop is approximately 0.07 meters (or 7 centimeters) over 1 meter of horizontal distance. This represents a gentle slope, as 4 degrees is a small angle.
The area of a square, with 7 meter sides is 49 square meters. Multiply one horizontal side by one vertical side to find the square meters. This formula will be true for a square or rectangle.
Neglecting air resistance, the components of acceleration of an object that's dropped, tossed, pitched, flung, lobbed, heaved, launched, or shot are constant. The horizontal component is zero. The vertical component is 9.8 meters per second2, directed downward. These are both constant throughout the object's trajectory.
Divide the motion into a horizontal and a vertical component. The horizontal component won't be affected by gravity. The vertical component will get a downward acceleration of 9.8 meters per second per second.
Area = 0.5*14.5*7 = 50.75 sq metres.
Horizontal, else you would have to have a kettle 6 meters deep
5
To calculate the vertical fall over a horizontal distance at a given angle, you can use trigonometry. In this case, the fall at 2 degrees over 6 meters can be calculated using the formula: vertical fall = horizontal distance * tan(angle). Plugging in the values, the vertical fall would be approximately 0.21 meters, or 21 centimeters.
1 kilometre = 1000 metres. that is true whether the distance is horizontal, vertical or in any other direction.
The horizontal clearance for 33kV and 11kV phase to phase is typically around 4.5-5 meters, while the vertical clearance is usually about 5-6 meters. For phase to ground, the horizontal clearance is around 3-3.5 meters and the vertical clearance is around 4 meters for both 33kV and 11kV lines. These values may vary based on local regulations and standards.
In the absence of air resistance, the force of gravity has no effect on the horizontal component of a projectile's velocity, and causes the vertical component of its velocity to increase by 9.8 meters (32.2 feet) per second downward for every second of its flight.
Ignoring air resistance, its horizontal speed is still 9 meters per second, its vertical speed is approx. 9.81 m/s, as the acceleration of gravity is 9.81 meters per second per second.
A 5-degree fall over 2 meters corresponds to a vertical drop of approximately 0.174 meters, or about 17.4 centimeters. This can be calculated using basic trigonometry, where the vertical drop (rise) is the sine of the angle multiplied by the horizontal distance. In this case, ( \text{Drop} = 2 \times \sin(5^\circ) ).
13
The vertical motion is accelerated downwards, at a rate of about 9.8 meters per square second. The horizontal motion is not affected. This all assumes that air resistance is insignificant, and can therefore be ignored.