5.4 x 1014Hz
wavelength x frequency = the speed of light
Answer:
The launching point is at a distance D = 962.2m and H = 39.2m
Explanation:
It would have been easier with the drawing. This problem is a projectile launching exercise, as they give us data after the window passes and the wall collides, let's calculate with this data the speeds at the point of contact with the window.
X axis
x = Vox t
t = x / vox
t = 7.1 / 340
t = 2.09 10-2 s
In this same time the height of the window fell
Y = Voy t - ½ g t²
Let's calculate the initial vertical speed, this speed is in the window
Voy = (Y + ½ g t²) / t
Voy = [0.6 + ½ 9.8 (2.09 10⁻²)²] /2.09 10⁻² = 0.579 / 0.0209
Voy = 27.7 m / s
We already have the speed at the point of contact with the window. Now let's calculate the distance (D) and height (H) to the launch point, for this we calculate the time it takes to get from the launch point to the window; at this point the vertical speed is Vy2 = 27.7 m / s
Vy = Voy - gt₂
Vy = 0 -g t₂
t₂ = Vy / g
t₂ = 27.7 / 9.8
t₂ = 2.83 s
This is the time it also takes to travel the horizontal and vertical distance
X = Vox t₂
D = 340 2.83
D = 962.2 m
Y = Voy₂– ½ g t₂²
Y = 0 - ½ g t2
H = Y = - ½ 9.8 2.83 2
H = 39.2 m
The launching point is at a distance D = 962.2m and H = 39.2m
Answer:
B = 9.16 10⁻² T
Explanation:
The speed selector is a configuration where the electric and magnetic force has the opposite direction, which for a specific speed cancel
q v B = q E
v = E / B
B = E / v
Let's calculate
B = 4.4 10⁵ / 4.8 10⁶
B = 9.16 10⁻² T
<span>This pivot is called refraction. It happens because different materials have different densities. The more dense the material the more atoms the light collides with and the slower it travels; the less dense, the fewer the collisions and hence a faster velocity. This pivoting, or refraction, is caused by the light either slowing down or speeding up.</span>
Answer:
The amount of work we could expect to get out of the system per second = 28,000J/s
Explanation:
Given the power supplied to the system as 28kW;
Energy = power / time
At very best, the amount of work we could expect to get out of the system per second = 28,000 W / 1 second = 28,000J/s
Therefore, for a a furnace which supplies 28kW of thermal power at 300C to an engine and exhausts waste energy at 20C.
At the very best, the amount of work we could expect to get out of the system per second = 28,000J/s
