The Volume of the ice block is 5376.344 cm^3.
The density of a material is define as the mass per unit volume.
Here, the density of ice given is 0.93 g/cm^3
Mass of the ice block given is 5 kg or 5000 g
Now calculate the volume of the ice block
density=mass/volume
0.93=5000/Volume
Volume =5376.344 cm^3
Therefore the volume of ice block is 5376.344 cm^3
Answer:
small car since they weigh less than a bus
Explanation:
Answer:
a) True. There is dependence on the radius and moment of inertia, no data is given to calculate the moment of inertia
c) True. Information is missing to perform the calculation
Explanation:
Let's consider solving this exercise before seeing the final statements.
We use Newton's second law Rotational
τ = I α
T r = I α
T gR = I α
Alf = T R / I (1)
T = α I / R
Now let's use Newton's second law in the mass that descends
W- T = m a
a = (m g -T) / m
The two accelerations need related
a = R α
α = a / R
a = (m g - α I / R) / m
R α = g - α I /m R
α (R + I / mR) = g
α = g / R (1 + I / mR²)
We can see that the angular acceleration depends on the radius and the moments of inertia of the steering wheels, the mass is constant
Let's review the claims
a) True. There is dependence on the radius and moment of inertia, no data is given to calculate the moment of inertia
b) False. Missing data for calculation
c) True. Information is missing to perform the calculation
d) False. There is a dependency if the radius and moment of inertia increases angular acceleration decreases
Answer:
Explanation:
a ) Between r = 0 and r = r₁
Electric field will be zero . It is so because no charge lies in between r = 0 and r = r₁ .
b ) From r = r₁ to r = r₂
At distance r , charge contained in the sphere of radius r
volume charge density x 4/3 π r³
q = Q x r³ / R³
Applying Gauss's law
4πr² E = q / ε₀
4πr² E = Q x r³ / ε₀R³
E= Q x r / (4πε₀R³)
E ∝ r .
c )
Outside of r = r₂
charge contained in the sphere of radius r = Q
Applying Gauss's law
4πr² E = q / ε₀
4πr² E = Q / ε₀
E = Q / 4πε₀r²
E ∝ 1 / r² .
