Answer:
a) i₈ = 0.5 i₄, b) i₁₀ = 0.3 i₃, i₁₀ = 0.8 i₈
Explanation:
For this exercise we use ohm's law
V = i R
i = V / R
we assume that the applied voltage is the same in all cases
let's find the current for each resistance
R = 4 Ω
i₄ = V / 4
R = 8 Ω
i₈ = V / 8
we look for the relationship between these two currents
i₈ /i₄ = 4/8 = ½
i₈ = 0.5 i₄
R = 3 Ω
i₃ = V3
R = 10 Ω
i₁₀ = V / 10
we look for relationships
i₁₀ / 1₃ = 3/10
i₁₀ = 0.3 i₃
i₁₀ / 1₈ = 8/10
i₁₀ = 0.8 i₈
Answer:
Their number should increase
Explanation:
The photoelectric effect is a phenomenon that causes the ejection of electrons from that metal as light shined onto a metal surface. Only certain frequencies of light can cause the ejection of electrons. However, if the frequency of the incident light is too low then no electrons were ejected even if the intensity of the light was very high. If the frequency of the light was higher then electrons were able to be ejected from the metal surface even if the intensity of the light was very low.
According to the accepted wave theory, light of any frequency will cause electrons to be emitted. Kinetic energy emitted by the electrons depends upon the intensity of light.
According to the accepted wave theory, number of electrons being ejected by the metal should increase
Answer:
Explanation:
The tidal current flows to the east at 2.0 m/s and the speed of the kayaker is 3.0 m/s.
Let Vector
is the tidal current velocity as shown in the diagram.
In order to travel straight across the harbor, the vector addition of both the velocities (i.e the resultant velocity,
must be in the north direction.
Let
is the speed of the kayaker having angle \theta measured north of east as shown in the figure.
For the resultant velocity in the north direction, the tail of the vector
and head of the vector
must lie on the north-south line.
Now, for this condition, from the triangle OAB




Hence, the kayaker must paddle in the direction of
in the north of east direction.
Frequensey or hertz, I looked this up on the internet!
Answer:
Its A. Walking on the beach