Answer:
a) 11 m/s
b) 0.0564 s
Explanation:
Given:
m = 2100 kg
vi = 22 ..... m/s before collision
vf = 0 ......after collision to stop
Δs = 0.62 distance traveled after collision .. crumpling of truck
Part a
Part b
A car moves along a curved road of diameter 2 km. If the maximum velocity for safe driving on this path is 30 m/s, at what angle
1 answer:
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Answer:
The plane would need to travel at least (
.)
The runway should be sufficient.
Explanation:
Convert unit of the the take-off velocity of this plane to :
.
Initial velocity of the plane: .
Take-off velocity of the plane .
Let denote the distance that the plane travelled along the runway. Since acceleration is constant but unknown, make use of the SUVAT equation
.
Notice that this equation does not require the value of acceleration. Rather, this equation make use of the fact that the distance travelled (under constant acceleration) is equal to duration times average velocity
.
The distance that the plane need to cover would be:
.
Answer:
The shortest de Broglie wavelength for the electrons that are produced as photoelectrons is 0.81 nm
Explanation:
Given;
wavelength of ultraviolet light, λ = 270 nm
work function of the metal, φ = 2.3 eV = 2.3 x 1.602 x 10⁻¹⁹ J = 3.685 x 10⁻¹⁹ J
The energy of the ultraviolet light is given by;
The energy of the incident light is related to kinetic energy of the electron and work function of the metal by the following equation;
E = φ + K.E
K.E = E - φ
K.E = (7.362 x 10⁻¹⁹ J) - (3.685 x 10⁻¹⁹ J )
K.E = 3.677 x 10⁻¹⁹ J
K.E = ¹/₂mv²
mv² = 2K.E
velocity of the electron is given by;
the shortest de Broglie wavelength for the electrons is given by;
Therefore, the shortest de Broglie wavelength for the electrons that are produced as photoelectrons is 0.81 nm