The work done to pull the sled up to the hill is given by

where
F is the intensity of the force
d is the distance where the force is applied.
In our problem, the work done is

and the distance through which the force is applied is

, so we can calculate the average force by re-arranging the previous equation and by using these data:
Answer:
2.5 m/s²
Explanation:
Given,
Initial speed ( u ) = 10 m/s
Final speed ( v ) = 20 m/s
Time ( t ) = 4 seconds
To find : Acceleration ( a ) = ?
Formula : -
a = ( v - u ) / t
a = ( 20 - 10 ) / 4
= 10 / 4
= 5 / 2
a = 2.5 m/s²
Therefore,
The acceleration of the scooter is 2.5 m/s²
Answer: a) 139.4 μV; b) 129.6 μV
Explanation: In order to solve this problem we have to use the Ohm law given by:
V=R*I whre R= ρ *L/A where ρ;L and A are the resistivity, length and cross section of teh wire.
Then we have:
for cooper R=1.71 *10^-8* 1.8/(0.001628)^2= 11.61 * 10^-3Ω
and for silver R= 1.58 *10^-8* 1.8/(0.001628)^2=10.80 * 10^-3Ω
Finalle we calculate the potential difference (V) for both wires:
Vcooper=11.62* 10^-3* 12 * 10^-3=139.410^-6 V
V silver= 10.80 10^-3* 12 * 10^-3=129.6 10^-6 V
Answer:
The first statement is false, the Sun has a stronger gravitational pull.
:
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