Answer:
Explanation:
The first number is .
The second number is .
We must multiply these two numbers together.
In scientific notation:
Therefore, this is the solution you are looking for.
A clever inventor has created a device that can launch water balloons with an initial speed of 85.0 m/s. Her goal is to pass a b
alloon through a small hoop mounted on the observation platform at the top of the Eiffel Tower, 276 m above the ground. If the balloon is to pass through the hoop at the peak of its flight, at what angle above horizontal should she launch the balloon? Please explain step by step
1 answer:
Answer:
She must launch the balloon at a 59.9° angle above the horizontal plane.
Explanation:
Refer to the accompanying diagram for a visual explanation of the scenario.
The position and velocity vectors of the balloon at time "t" can be computed using these formulas:
r = (x0 + v0 · t · cos θ, y0 + v0 · t · sin θ + 1/2 · g · t²)
v = (v0 · cos θ, v0 · sin θ + g · t)
Where:
r = position vector at time "t".
x0 = initial horizontal position.
v0 = initial velocity.
t = time.
θ = launch angle.
y0 = initial vertical position.
g = acceleration due to gravity (-9.81 m/s² if upward is treated as positive).
v = velocity vector at time "t".
We can establish the origin of the reference frame at the launch point, making x0 and y0 = 0.
At the maximum altitude (276 m), the balloon's velocity vector is horizontal (refer to v1 in the diagram). Thus, the y-component of the velocity vector is zero. We can express this using the velocity vector's y-component equation:
At maximum altitude:
vy = v0 · sin θ + g · t
0 = v0 · sin θ + g · t
Also, at maximum height, the y-component of the position vector is 276 m (see r1y in the diagram). Therefore:
At peak height:
y = y0 + v0 · t · sin θ + 1/2 · g · t²
276 m = y0 + v0 · t · sin θ + 1/2 · g · t²
We have two equations involving two unknowns (θ and t):
276 m = y0 + v0 · t · sin θ + 1/2 · g · t²
0 = v0 · sin θ + g · t
To solve for the variables, we can rearrange the velocity equation for sin θ and substitute it back into the position equation to find t and eventually θ:
0 = v0 · sin θ + g · t
0 = 85.0 m/s · sin θ - 9.81 m/s² · t
9.81 m/s² · t / 85.0 m/s = sin θ
Now, substituting sin θ in the position equation:
276 m = y0 + v0 · t · sin θ + 1/2 · g · t² (y0 = 0)
276 m = 85.0 m/s · t · (9.81 m/s² · t /85.0 m/s) - 1/2 · 9.81 m/s² · t²
276 m = 9.81 m/s² · t² - 1/2 · 9.81 m/s² · t²
276 m = 1/2 · 9.81 m/s² · t²
276 m / (1/2 · 9.81 m/s²) = t²
t = 7.50 s
With t known, we can find the angle θ using the previously derived formula:
9.81 m/s² · t / 85.0 m/s = sin θ
9.81 m/s² · 7.50 s / 85.0 m/s = sin θ
θ = 59.9°
She should launch the balloon at a 59.9° angle above the horizontal.
You might be interested in
assuming north-south is along the Y-axis and east-west along the X-axis
X = total X-displacement
from the graph, total displacement in the X-direction is computed as
X = 0 - 20 + 60 Cos45 + 0
X = 42.42 - 20
X = 22.42 m
Y = total Y-displacement
from the graph, total displacement in the Y-direction is computed as
Y = 40 + 0 + 60 Sin45 + 50
Y = 90 + 42.42
Y = 132.42 m
To calculate the magnitude of the net displacement vector, we apply the Pythagorean theorem, yielding
magnitude: Sqrt(X² + Y²) = Sqrt(22.42² + 132.42²) = 134.31 m
Direction: tan⁻¹(Y/X) = tan⁻¹(132.42/22.42) = 80.4 deg north of east
Answer:
The force required has a magnitude of 2601.9 N
Explanation:
m = 450 kg
Static friction coefficient μs = 0.73
Kinetic friction coefficient μk = 0.59
The force necessary to initiate movement of the crate is .
Once the crate begins to move, the frictional force decreases to .
To maintain the motion of the crate at a steady velocity, we must lower the pushing force to .
Subsequently, the pushing force aligns with the frictional force stemming from kinetic friction, enabling balanced forces and consistent velocity.
<pMagnitude of the forceTo address this issue, we will utilize the principles related to Gauss' law, which states that the electric flux across a surface corresponds to the object's charge divided by the permittivity of vacuum. In mathematical terms, this can be expressed as
It's crucial to remember that the net charge equals the difference between the two specified charges, so upon substitution,
The negative sign indicates that the flux is directed into the surface