<span>Diamond slowdown light more than Quartz , because diamonds have a greater index of refraction. Light will bend when its move from one medium to another. The Index of Refraction of Material is found by comparing the speed of light in their respective mediums.</span>
Complete question is:
A 1200 kg car reaches the top of a 100 m high hill at A with a speed vA. What is the value of vA that will allow the car to coast in neutral so as to just reach the top of the 150 m high hill at B with vB = 0 m/s. Neglect friction.
Answer:
(V_A) = 31.32 m/s
Explanation:
We are given;
car's mass, m = 1200 kg
h_A = 100 m
h_B = 150 m
v_B = 0 m/s
From law of conservation of energy,
the distance from point A to B is;
h = 150m - 100 m = 50 m
From Newton's equations of motion;
v² = u² + 2gh
Thus;
(V_B)² = (V_A)² + (-2gh)
(negative next to g because it's going against gravity)
Thus;
(V_B)² = (V_A)² - (2gh)
Plugging in the relevant values;
0² = (V_A)² - 2(9.81 × 50)
(V_A) = √981
(V_A) = 31.32 m/s
Answer:
62.06 g/mol
Explanation:
We are given that a solution containing 10 g of an unknown liquid and 90 g
Given mass of solute =
=10 g
Given mass of solvent=
=90 g
Freezing point of solution =-3.33
C
Freezing point of solvent =
C
Change in freezing point =Depression in freezing point
=Freezing point of solvent - freezing point of solution=0+3.33=
Hence, molar mass of unknown liquid is 62.06g/mol.
Answer:
112 m/s², 79.1°
Explanation:
In the x direction, given:
x₀ = 0 m
x = 19,500 cos 32.0° m
v₀ = 1810 cos 20.0° m/s
t = 9.20 s
Find: a
x = x₀ + v₀ t + ½ at²
19,500 cos 32.0° = 0 + (1810 cos 20.0°) (9.20) + ½ a (9.20)²
a = 21.01 m/s²
In the y direction, given:
y₀ = 0 m
y = 19,500 sin 32.0° m
v₀ = 1810 sin 20.0° m/s
t = 9.20 s
Find: a
y = y₀ + v₀ t + ½ at²
19,500 sin 32.0° = 0 + (1810 sin 20.0°) (9.20) + ½ a (9.20)²
a = 109.6 m/s²
The magnitude of the acceleration is:
a² = ax² + ay²
a² = (21.01)² + (109.6)²
a = 112 m/s²
And the direction is:
θ = atan(ay / ax)
θ = atan(109.6 / 21.01)
θ = 79.1°
