The branch of science concerning heat flow and energy conversions is called
2 answers:
You might be interested in
Answer:
θ = 12.60°
Explanation:
In order to calculate the angle below the horizontal for the velocity of the hockey puck, you need to calculate both x and y component of the velocity of the puck, and also you need to use the following formula:
(1)
θ: angle below he horizontal
vy: y component of the velocity just after the puck hits the ground
vx: x component of the velocity
The x component of the velocity is constant in the complete trajectory and is calculated by using the following formula:
vo: initial velocity = 28.0 m/s
The y component is calculated with the following equation:
(2)
voy: vertical component of the initial velocity = 0m/s
g: gravitational acceleration = 9.8 m/s^2
y: height
You solve the equation (2) for vy and replace the values of the parameters:
Finally, you use the equation (1) to find the angle:
The angle below the horizontal is 12.60°
Answer:
a.) magnitude __49.7__ unit(s)
b.) direction __123.6°_ counterclockwise from the +x axis
Explanation:
Let Vector is v
x-component of Vector v = x = -27.5 units (minus sign indicate that x-component is along the minus x-axis )
y-component of Vector v = y = 41.4 units
Magnitude of v = ?
Direction of v = ?
To find the magnitude of the vector
v =
v =
v = 49.7 units
To find direction
θ = tan⁻¹(y/x)
θ = tan⁻¹(41.4/-27.5)
θ = -56.4°
This Angle is in the clockwise direction with respect to -x axis.
We need to find Angle counterclockwise from the +x axis.
So,
θ = 180° - 56.4°
θ = 123.6°
The given vector is in 2nd quadrant