Which of the following is an example of energy being transferred by sound waves? A loud clap of thunder causes a window to vibra
2 answers:
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Answer:
a) The takeoff speed is 10 m/s.
b) The maximum height above the ground is 1.2 m.
Explanation:
The position of the kangaroo and its velocity at any given time "t" can be calculated by the following equations:
r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)
v =(v0 · cos α, v0 · sin α + g · t)
Where:
r = position vector at time "t".
x0 = initial horizontal position.
v0 = initial velocity.
α = jumping angle.
y0 = initial vertical position.
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).
v = velocity vector at time "t"
a) Please see the attached figure for a better understanding of the problem. In red is depicted the position vector at the final time (r final). The components of r final are known:
r final = (8.7 m, 0 m)
Then at final time:
8.7 m = x0 + v0 · t · cos α
0 m = y0 + v0 · t · sin α + 1/2 · g · t²
(notice in the figure that the origin of the frame of reference is located at the jumping point so that x0 and y0 = 0). Then:
8.7 m = v0 · t · cos α
Solving for "v0":
8.7 m /(t · cos α) = v0
Replacing v0 in the equation of the y-component, we can obtain the final time:
0 m = 8.7 m · tan 29° - 1/2 · 9.8 m/s² · t² (remember: sin α / cos α = tan α)
- 8.7 m · tan 29° / -4.9 m/s² = t²
t = 0.99 s
Now, we can calculate the initial speed:
8.7 m /t · cos α = v0
v0 = 8.7 m / (0.99 s · cos 29°)
<u>v0 = 10 m/s</u>
The takeoff speed is 10 m/s
b) When the kangaroo is at its maximum height, the velocity vector is horizontal (see figure). That means that the y-component of the velocity at that time is 0:
0 = v0 · sin α + g · t
Solving for "t":
-v0 · sin α / g = t
t = - 10 m/s · sin 29° / 9.8 m/s²
t = 0.49 s
Notice that we could have halved the final time (0.99 s, calculated above) to obtain the time at which the kangaroo is at its maximum height. That´s because the trajectory is parabolic.
Now, let´s find the height of the kangaroo at that time:
y = y0 + v0 · t · sin α + 1/2 · g · t²
y = 10 m/s · 0.49 s · sin 29° - 1/2 · 9.8 m/s² · (0.49 s)²
<u>y = 1.2 m</u>
The maximum height above the ground is 1.2 m.
Answer:
Explanation:
Hello!
In this case, for this melting process, we can identify two sub-processes in order to take the stainless steel from solid to liquid:
1. Heat up from 298.15 K to 1673 K.
2. Undergo the phase transition.
Both process have an associated enthalpy as shown below:
Therefore, the required heat is:
Notice the problem is not providing neither the mass or volume, that is why we assumed the mass is 1 g; however, it can be changed to the mass you are given.
Best regards!
Answer:
The maximum radius the asteroid can have for her to be able to leave it entirely simply by jumping straight up is approximately 1782.45 meters
Explanation:
Whereby the height the astronaut can jump on Earth = 0.500 m, we have the following kinematic equation;
v² = u² - 2·g·h
Where;
v = The final velocity
u = The initial velocity
g = The acceleration due to gravity ≈ 9.8 m/s²
h = The height she jumps
At the maximum height, = 0.500 m, she jumps, v = 0, therefore, we have;
0² = u² - 2·g·
u² = 2 × 9.8 × 0.5 = 9.8
u = √9.8 ≈ 3.13
u = 3.13 m/s
Her initial jumping velocity ≈ 3.13 m/s
Escape velocity,
Where;
M = The mass of the asteroid
G = The Universal gravitational constant = 6.67408 × 10⁻¹¹ m³/(kg·s²)
r = The radius of the asteroid
The average density of the Earth = 5515 kg/m³
The mass of the asteroid, M = Density × Volume = 5515 kg/m³× 4/3 × π × r³
The escape velocity, she has, ≈ 3.13 m/s is therefore;
Therefore, the maximum radius of the asteroid can have for her jumping velocity to be equal to the escape velocity for her to be able to leave it entirely simply by jumping straight up = r ≈ 1782.45 meters.