Apply the Pythagorean theorem to get the resultant velocity:
V = 
Given values:
Vx = 57.8m/s
Vy = 5.6m/s
Plug in and solve for V:
V = 58.1m/s
EDIT: Let's get the direction of the resultant velocity as well.
This equation will give the angle of the velocity as measured off of the ground:
θ = tan⁻¹(Vy/Vx)
Again, the given values are:
Vx = 57.8m/s
Vy = 5.6m/s
Plug in the values and solve for the angle θ:
θ = tan⁻¹(5.6/57.8)
θ = 5.5°
The resultant velocity is oriented 5.5° off the ground.
Answer:
The Equation of Newton's Law of Gravity. G is called the Gravitational Constant, and has the value 6.67×10-11 N ∙m2kg-2 (N is for Newton, the physicists' unit of force)or 1.5×10-11 lb∙m2kg-2
Explanation:
This best answer is d because if u idle you either waste gas or burn the engine out and what is painting the car going to do
So the answer is d
Answer
F = 124 N
Explanation:
given,
mass, m = 5 Kg
time, t = 4.1 s
displacement = y(t)=(2.80 m/s)t +(0.61 m/s³)t³
velocity
again differentiating to get the equation of acceleration
force at time t = 4.10 s
F = m a
F = 5 x 3.66 x 4.1
F = 75 N
the net force when crate is moving upward
F = Mg + Ma
F = 5 x 9.8 + 75
F = 124 N
the magnitude of force is equal to 124 N
Answer:
maximum horizontal distance = 10m
initial vertical velocity of the ball = 4.9m/s
Explanation:
Complete question
<em>A ball is launched with an initial horizontal velocity of 10.0 meters per second. It takes 500 milliseconds for the ball to reach its maximum height.
</em>
<em>Determine the maximum horizontal distance that the ball will travel.
</em>
<em>Calculate the initial vertical velocity of the ball.</em>
<em />
Maximum horizontal distance x is expressed as;
x = vT
T is total time of flight
T = 2t
Hence x = 2vt
v is the velocity
t is the time
Given
v = 10.0m/s
time t = 500ms = 0.5s
Horizontal distance = 2 * 10 * 0.5
Horizontal distance = 20 * 0.5
Horizontal distance = 10m
Hence the maximum horizontal distance that the ball will travel is 10m
To get the initial horizontal distance, we will use the equation of motion
v = u - gt
T maximum height, v = 0
Substitute
0 = u - 9.8(0.5)
-u = - 4.9
u = 4.9m/s
Hence the initial vertical velocity of the ball is 4.9m/s
