Answer:
6) 2.6 m/s, 31°
7) 9.2 m/s
8) 1.2 s
Explanation:
I'll do #6, #7, and #8 as examples. You can solve #9 using the equation from #7, and #10 using the equation from #8.
6) Take north to be +y and east to be +x.
Given:
vₓ = 2.2 m/s
vᵧ = 1.3 m/s
Find: v
v² = vₓ² + vᵧ²
v² = (2.2 m/s)² + (1.3 m/s)²
v ≈ 2.6 m/s
θ = atan(vᵧ / vₓ)
θ = atan(1.3 / 2.2)
θ ≈ 31°
7) Given:
Δy = -4.3 m
v₀ = 0 m/s
a = -9.8 m/s²
Find: v
v² = v₀² + 2aΔy
v² = (0 m/s)² + 2 (-9.8 m/s²) (-4.3 m)
v ≈ 9.2 m/s
8) Given:
Δy = -6.7 m
v₀ = 0 m/s
a = -9.8 m/s²
Find: t
Δy = v₀ t + ½ at²
-6.7 m = (0 m/s) t + ½ (-9.8 m/s²) t²
t = 1.2 s
Answer:
The refractive index of fluid 2 is 1.78
Explanation:
Refractive index , n = real depth/apparent depth
For the first fluid, n = 1.37 and apparent depth = 9.00 cm.
The real depth of the container is thus
real depth = n × apparent depth = 1.37 × 9.00 cm = 12.33 cm
To find the refractive index of fluid index of fluid 2, we use the relation
Refractive index , n = real depth/apparent depth.
Now,the real depth = 12.33 cm and the apparent depth = 6.86 cm.
So, n = 12.33 cm/6.86 cm = 1.78
So the refractive index of fluid 2 is 1.78
Answer: 60 seconds
Explanation:
The following information can be derived from the question:
rt=d,
150r = 100
Divide through by 150
r = 100/150
r = 0.67
r = 0.67 feet/second
Fir the second part of the question,
100r=100
r = 100/100
r = 1
r= 1 feet/second
combined rate will now be:
= 0.67+1
= 1.67 feet/second
We then slot it back into the formula
rt = d
1.67t = 100
t = 100/1.67
t= 59.88 seconds
t = 60 seconds approximately
Answer:
The SI unit for length is meters(m), for mass is kilograms(kg)
Explanation:
hope it helps