Answer:
Density of rock will be equal to 
Explanation:
It is given that mass of the rock 
Volume displaced bu rock 
We have to find the density of rock
Density is equal to ratio of mass and volume
Therefore density of rock 
So density of rock will be equal to
Answer:
Plot ln K vs 1/T
(a) -0.5004; (b) 0.002 539 K⁻¹; (c) -197.1 K⁻¹; (d) 1.64 kJ/mol
Explanation:
This is an example of the Arrhenius equation:
Thus, if we plot ln k vs 1/T, we should get a straight line with slope = -Eₐ/R and a y-intercept = lnA
Data:
Calculations:
(a) Rise
Δy = y₂ - y₁ = -0.9545 - (-0.4541) = -0.9545 + 0.4541 = -0.5004
(b) Run
Δx = x₂ - x₁ = 0.004 444 - 0.001 905 = 0.002 539 K⁻¹
(c) Slope
Δy/Δx = -0.5004/0.002 539 K⁻¹ = -197.1 K⁻¹
(d) Activation energy
Slope = -Eₐ/R
Eₐ = -R × slope = -8.314 J·K⁻¹mol⁻¹ × (-197.1 K⁻¹) = 1638 J/mol = 1.64 kJ/mol
Answer:
θ₂ ≈ 29.7°
Explanation:
Given:
n₁ = 1.33, n₂= 1.60, θ₁ = 36.6°, θ₂ = ?
Solution:
According to snell's Law
n₁ sin θ₁ = n₂ sin θ₂
1.33 × sin (36.6°) = 1.60 × sin θ₂
⇒ Sin θ₂ = 0.496
θ₂ ≈ 29.7°
Answer:
(D) Water is denser than oil but less than syrup.
Explanation:
Given:
A picture of a jar containing three liquids.
The position of the syrup is at the bottom of the jar. The position of water is in middle and the oil is at the top.
We know that, the oil is the lightest in density when compared to syrup and water. So, it is at the top.
Now, since water is in the middle of the two, it means that density of water is greater than that of oil but less dense than that of syrup.
So, the density of the syrup is the greatest. Hence, it sits at the bottom.
Therefore, the best explanation is that water being less dense than syrup but more dense than oil, will be in the middle of two.
This explains the position of all the three liquids. Water is in the middle, oil at top and syrup at bottom.
So, the last option is correct.
Answer:
76.3 J
Explanation:
I'm assuming the distance of 4.60 m is along the incline, not the vertical distance from the bottom. I'll call this distance d, so h = d sin θ.
Initial energy = final energy
Energy in spring = gravitational energy + kinetic energy + work by friction
E = mgh + 1/2 mv² + Fd
We need to find the force of friction. To do that, draw a free body diagram.
Normal to the incline, we have the normal force pointing up and the normal component of weight (mg cos θ).
Sum of the forces in the normal direction:
∑F = ma
N - mg cos θ = 0
N = mg cos θ
Friction is defined as:
F = Nμ
Plugging in the expression for N:
F = mgμ cos θ
Substituting:
E = mgh + 1/2 mv² + (mgμ cos θ) d
E = mg (d sin θ) + 1/2 mv² + (mgμ cos θ) d
E = mgd (sin θ + μ cos θ) + 1/2 mv²
Given:
m = 1.45 kg
g = 9.90 m/s²
d = 4.60 m
θ = 29.0°
μ = 0.45
v = 5.10 m/s
Solving:
E = mgd (sin θ + μ cos θ) + 1/2 mv²
E = (1.45) (9.80) (4.60) (sin 29.0 + 0.45 cos 29.0) + 1/2 (1.45) (5.10)²
E = 76.3 J
