What is the metric unit of measuring the volume of liquids?

You drop a rock off the top of a 100 m tall building. How long does it take to hit the ground?

Vitek1552 [10]

It would mostly depend on its weight

4 0

3 years ago

Which two substances have no fixed shape and no fixed volume?

fenix001 [56]

Answer:

Gas like oxygen, nitrogen etc

4 0

2 years ago

A gas contained within a piston-cylinder assembly, initially at a volume of 0.1 m3 , undergoes a constant-pressure expansion at

Gnom [1K]

Answer:

Work: 4.0 kJ, heat: 4.25 kJ

Explanation:

For a gas transformation at constant pressure, the work done by the gas is given by

$W=p(V_f -V_i)$

where in this case we have:

$p = 2 bar = 2\cdot 10^5 Pa$ is the pressure

$V_i = 0.1 m^3$ is the initial volume

$V_f = 0.12 m^3$ is the final volume

Substituting,

$W=(2\cdot 10^5)(0.12-0.10)=4000 J = 4.0 kJ$

The 1st law of thermodynamics also states that

$\Delta U = Q-W$

where

$\Delta U$ is the change in internal energy of the gas

Q is the heat absorbed by the gas

Here we know that

$\Delta U = +0.25 kJ$

Therefore we can re-arrange the equation to find the heat absorbed by the gas:

$Q=\Delta U + W = 0.25 kJ + 4.0 kJ = 4.25 kJ$

7 0

3 years ago

Two go-carts, A and B, race each other around a 1.0 track. Go-cart A travels at a constant speed of 20.0 /. Go-cart B accelerate

maria [59]

Complete Question

Q. Two go-carts, A and B, race each other around a 1.0km track. Go-cart A travels at a constant speed of 20m/s. Go-cart B accelerates uniformly from rest at a rate of 0.333m/s^2. Which go-cart wins the race and by how much time?

Answer:

Go-cart A is faster

Explanation:

From the question we are told that

The length of the track is $l = 1.0 \ km = 1000 \ m$

The speed of A is $v__{A}} = 20 \ m/s$

The uniform acceleration of B is $a__{B}} = 0.333 \ m/s^2$

Generally the time taken by go-cart A is mathematically represented as

$t__{A}} = \frac{l}{v__{A}}}$

=> $t__{A}} = \frac{1000}{20}$

=> $t__{A}} = 50 \ s$

Generally from kinematic equation we can evaluate the time taken by go-cart B as

$l = ut__{B}} + \frac{1}{2} a__{B}} * t__{B}}^2$

given that go-cart B starts from rest u = 0 m/s

So

$1000 = 0 *t__{B}} + \frac{1}{2} * 0.333 * t__{B}}^2$

=> $1000 = 0 *t__{B}} + \frac{1}{2} 0.333 * t__{B}}^2$

=> $t__{B}} = 77.5 \ seconds$

Comparing $t__{A}} \ and \ t__{B}}$ we see that $t__{A}}$ is smaller so go-cart A is faster

3 0

2 years ago

Calculate the percentage of an iceberg submerged beneath the surface of the ocean given that the density of ice is 916.3kg/m3 an

pogonyaev

Answer:

The percentage of an iceberg submerged beneath the surface of the ocean = 89.67%

Explanation:

Let V be the total volume of the iceberg

Let x be the volume of iceberg submerged

According to Archimedes principle,

weight of the iceberg = weight of the water displaced (that is, weight of x volume of water)

Weight of the iceberg = mg= ρ(iceberg) × V × g

Weight of water displaced = ρ(fluid) × x × g

We then have

ρ(iceberg) × V × g = ρ(fluid) × x × g

(x/V) = ρ(iceberg) ÷ ρ(fluid) = 916.3 ÷ 1021.9 = 0.8967 = 89.67%

Hope this Helps!!!!

6 0

2 years ago