On question 30, that is a displacement- time graph (DT). On this type of graph the gradient is equal to the velocity. B has the steepest gradient, then A and finally C
Now velocity is a vector quantity so it has a direction and speed ( speed doesn't have a fixed direction.)
on the DT graph im going to assume that movement B is a positive velocity with A and C being negative.
So by ranking these: A is the most negative, C is the least negative and B has to be the greatest as it is the only positive velocity.
Q31, The same type of graph is present, by looking at the gradients we can rank the largest and smallest velocities- speeds in the case of the question.
i'll skip my working out as its the same as before:
C, B, A and then D
the same idea as on Q30 applies to Q31 part b,
D,C,B then A
Well minerals are in the water we drink every Day, even the food we eat. Minerals are contained in rocks so there for both are important to life
Explanation:
Amplitude is your answer..
Mate..
hope you like it
Answer:
A) c₁ = m, c₂ = m/s
B) c₁ = m/s²
C) c₁ = m/s²
D) c₁ = m/s c₂ = °
E) c₁ = m/s , c₂ = /s
Explanation:
A) x = c₁ + c₂t
⇒m = m + (m/s)s (Only same units can be added)
⇒m = m
So, c₁ = m, c₂ = m/s
B) x = 0.5c₁t²
⇒m = 0.5 (m/s²)s²
⇒m = m
So, c₁ = m/s²
C) v² = 2c₁x
⇒m²/s² = 2 (m/s²)m
⇒m²/s² = m²/s²
So, c₁ = m/s²
D) x = c₁ cos(c₂)t
⇒m = (m/s) cos(°)s
⇒m = m
So, c₁ = m/s c₂ = °
E) v² = 2c₁v-(c₂x)²
⇒m²/s² = 2(m/s)(m/s)-(1/s²)(m²)
⇒m²/s² =m²/s²
So, c₁ = m/s , c₂ = /s
Answer:
6.0cm
Explanation:
Given
focal length = 15.0cm
object distance = 10.0cm
Required
Image distance v
Using the formula
1/f = 1/u + 1/v
1/15 = 1/10+1/v
1/v = 1/15 + 1/10
1/v = 2+3/30
1/v = 5/30
v = 30/5
v = 6.0cm
Hence the image distance is 6.0cm
