a(child acceleration)=g(gravitational acceleration)*sin25=4.144 m/s2
.50g
A trajectory is the angle made with the horizontal when a projectile is fired. Suppose the projectile is a cannon ball. Assuming air is frictionless, that cannon ball will travel the greatest distance if the trajectory is 45 degrees from horizontal.
the force acting down the slope = sin 30 * 25 (kg) = 0.5 * 25 = 12.5 kg = (12.5 * 9.806 ) 122.575 newtons = resultant force of (122.575-120 ) 2.575 newtons downhill giving downhill acceleration of (using f=ma) 0.103 (m/s)/s
The magnitude of force f can be calculated using the equation f = mgsin(theta), where m is the mass of the object, g is the acceleration due to gravity, and theta is the angle of the incline. Given the angle of 30 degrees, the force can be calculated by plugging in the values of mass and acceleration due to gravity.
The acceleration of the boy can be found using the equation a = g * sin(θ), where g is the acceleration due to gravity (approximately 9.81 m/s^2) and θ is the angle of the slope (20 degrees). Therefore, the acceleration of the boy would be a = 9.81 m/s^2 * sin(20 degrees) ≈ 3.36 m/s^2.
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.
20 degrees is a measure of angular displacement. This cannot be converted to lateral (horizontal) displacement.
other horizontal lines? which would be 180 degrees i guess.
Horizontal beam width = 4.0 degrees Vertical beam width = 1.6 degrees
Extrapolate the experimental values of acceleration, vs. angle of the incline, to find the acceleration when the angle of inclination = 90 degrees. The acceleration at 90 degrees will equal 9.81 m/s/s, since this is the free-fall acceleration.
Fifteen degrees from the horizontal.
The magnitude of the vector at 45 degrees to the horizontal will be equal to the magnitude of its horizontal and vertical components. This is because the components are obtained by using trigonometric functions of the angle, and in this case, at 45 degrees, those functions yield the same value for both the horizontal and vertical components as the magnitude of the vector.