All machines are not 100% efficient because of <span>C. Friction</span>
Answer:
1.) 113500J
2.) 237m
Explanation:
Hello!
To solve this exercise follow the following steps, the description and complete process is in the attached image
1. Draw the full sketch of the problem.
2. The work is defined as the product of the trajectory by the force that is parallel to this direction, for this reason to find the work done we multiply the horizontal distance (250m) by the applied force (454N)
3. The potential energy is equal to the product of mass, gravity and height and is equal to the work done by the force applied by the cyclist, of this relationship and using algebra we can find the height that the cyclist climbed
4. We use the sine function to find the diagonal distance using the height and angle of the slope
Answer:
1.12 m
0.08291 m
Explanation:
u = Upstream velocity = 0.4 m/s
Re = Reynold's number = 
(turbulent)
= Viscosity of water = 
Here the flow is turbulent so we have the relation
The approximate location downstream from the leading edge where the boundary layer becomes turbulent is 1.4 m
Boundary layer thickness relation is given by
The boundary layer thickness is 0.08291 m
<span>A transverse wave is a moving wave that consists of oscillations occurring perpendicular (or right angled) to the direction of energy transfer. If a transverse wave is moving in the positive x-direction, its oscillations are in up and down directions that lie in the y–z plane. Light is an example of a transverse wave.</span>
Answer:
a) K = 3 MeV b) K= 1.5 MeV
Explanation:
We can solve this experiment using the equation of the magnetic force with Newton's second law, where the acceleration is centripetal.
F = q v x B
We can also write this equation based on the modules of the vectors
F = qv B sin θ
With Newton's second law
F = ma
F = m v² / r
q v B = m v² / r
v = q B r / m
The kinetic energy is
K = ½ m v²
Substituting
K = ½ m (q B r/ m)²
K = ½ B² r² q² / m
K = (½ B² R²) q²/m
The amount in brackets does not change during the experiment
K = A q² / m
For the proton
K = 3.0 10⁶eV (1.6 10⁻¹⁹ J / 1eV) = 4.8 10⁻¹³ J
With this data we can find the amount we call A
A = K m/q²
A = 4.8 10⁻¹³ 1.67 10⁻²⁷ /(1.6 10⁻¹⁹)²
A = 3.13 10⁻²
With this value we can write the equation
K = 3.13 10⁻² q² / m
Alpha particle
m = 4 uma = 4 1.66 10⁻²⁷ kg
K = 3.13 10⁻² (2 1.6 10⁻¹⁹)² / 4.0 1.66 10⁻²⁷
K = 4.82 10⁻¹³ J ((1 eV / 1.6 10⁻¹⁹ J) = 3 10⁶ eV
K = 3 MeV
Deuteron
K = 3.13 10⁻² (1.6 10⁻¹⁹)²/2 1.66 10⁻²⁷
K = 2.4 10⁻¹³ J (1eV / 1.6 10⁻¹⁹J)
K = 1.5 10⁶ eV
K= 1.5 MeV
