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otez555 [7]
3 years ago
13

How is the density of ice compare with the density of water? explain to me pplz

Physics
1 answer:
Viefleur [7K]3 years ago
3 0
Lower. Water expands on lower temperatures, meaning less molecules in 1 m3, thus making it less dense
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What is the acceleration of a 4,000 kg car pushed with a<br> force of 12,000 N?
jek_recluse [69]

Answer:

3 m/s

Explanation:

A= F/m

12,000/ 4000 = 3

8 0
2 years ago
Read 2 more answers
While finding the spring constant, if X1 = 12 cm, X2 = 15 cm, and hanging mass = 22 grams, the value of spring constant K would
Pavel [41]

Answer:

If x₁=12 cm then k=1.7985 N/m

If x₂=15 cm then k=1.4388 N/m

Explanation:

Hanging mass= 22 g=0.022 kg

Acceleration due to gravity g=9.81 m/s²

If x₁=displacement= 12 cm=0.12 m

k= spring constant

F=ma\\\Rightarrow F=0.022\times 9.81\\\Rightarrow F=0.21582\ N

\text {For spring}\\F=kx\\\Rightarrow 0.21582=k\times 0.012\\\Rightarrow k=1.7985\ N/m\\

∴k = 1.7985 N/m

If x₂=15 cm=0.15 m

Force of the hanging mass is same however the spring constant will change

\text {For spring}\\F=kx\\\Rightarrow 0.21582=k\times 0.015\\\Rightarrow k=1.4388\ N/m\\

∴k = 1.4388 N/m

As the mass is not changing the spring constant has to change. That means that here there are two spring one with k=1.7985 N/m and the other with k= 1.4388 N/m

4 0
3 years ago
My Notes Ask Your Teacher 6. Six blocks with different masses, m, each start from rest at the top of smooth, frictionless inclin
Tresset [83]

Answer:

Kf > Ka = Kb > Kc > Kd > Ke

Explanation:

We can apply

E₀ = E₁

where

E₀: Mechanical energy at the beginning of the motion (top of the incline)

E₁: Mechanical energy at the end (bottom of the incline)

then

K₀ + U₀ = K₁ + U₁

If v₀ = 0  ⇒  K₀

and  h₁ = 0   ⇒    U₁ = 0

we get

U₀ = K₁    

U₀ = m*g*h₀ = K₁

we apply the same equation in each case

a) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J

b) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J

c) U₀ = K₁ = m*g*h₀ = 35 Kg*9.81 m/s²*4m = 1373.40 J

d) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*16m = 1098.72 J

e) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*4m = 274.68 J

f) U₀ = K₁ = m*g*h₀ = 105 Kg*9.81 m/s²*6m = 6180.30 J

finally, we can say that

Kf > Ka = Kb > Kc > Kd > Ke

8 0
3 years ago
Below is the data from a gas law experiment comparing the pressure and the volume of a gas at a given temperature.
Wewaii [24]

Answer:

The combined gas equation relates three variables pressure, temperature and volume when the number of moles is constant.

The equation is PV / T = constant. Which is valid for a fixed number of moles of the gas.

You can derive the combined gas equation from the combination of Bolye's law, Charles' law and Gay-Lussac's law, which needs some algebra.

Explanation:

9 0
3 years ago
Read 2 more answers
A stone is thrown vertically up. What kind of energy did the stone have initially? What happens to this energy as the stone asce
Alex

Answer:

kinetic energy at first

Explanation:

kinetic turns to potential as it gains height

7 0
3 years ago
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