Ask question

Ask question

All categories

3 years ago

15

In pottery class, you throw a pot from a lump of wet clay. your pot's mass is 5.5 kg. after the pot is fired, it's mass is 4.9 k

g. the density of wet clay is 1.60 g/cm3 and the density of fired clay is 1.36 g/cm3. what was the volume of your pot before it was fired? what was the volume of the pot after it was fired? express your answer in significant digits.

1 answer:

kondor19780726 [428]3 years ago

8 0

We can calculate for the volume by taking the ratio of mass and density, that is:

volume = mass / density

A. when mass = 5.5 kg = 5500 g; density = 1.60 g/cm^3

volume = 5500 g / (1.60 g/cm^3)

volume = 3,437.5 cm^3

Taking the significant digits:

volume = 3,400 cm^3 = 3.4 L

B. when mass = 4.9 kg = 4900 g; density = 1.36 g/cm^3

volume = 4900 g / (1.36 g/cm^3)

volume = 3,602.94 cm^3

Taking the significant digits:

<span>volume = 3,600 cm^3 = 3.6 L</span>

You might be interested in

classic physics problem states that if a projectile is shot vertically up into the air with an initial velocity of 128 feet per

ddd [48]

Answer:

$t_1 = 0.28 s$

$t_2 = 7.72 s$

Explanation:

Given that height of the projectile as a function of time is

$h = -16 t^2 + 128 t + 112$

here we know that

h = 147 ft

so from above equation

$147 = -16 t^2 + 128 t + 112$

$16 t^2 - 128 t + 35 = 0$

now by solving above quadratic equation we know that

$t_1 = 0.28 s$

$t_2 = 7.72 s$

6 0

3 years ago

1.)Two objects, one of m=20,000 kg, and another of 12,500 kg, are placed at a distance of 5 meters apart. What is the force of g

Delvig [45]

1) $6.67\cdot 10^{-4} N$

The force of gravitation between the two objects is given by:

$F=G\frac{m_1 m_2}{r^2}$

where

$G=6.67\cdot 10^{-11} kg^{-1} m^{3} s^{-2}$ is the gravitational constant

m1 = 20,000 kg is the mass of the first object

m2 = 12,500 kg is the mass of the second object

r = 5 m is the distance between the two objects

Substituting the numbers inside the equation, we find

$F=(6.67\cdot 10^{-11})\frac{(20,000 kg)(12,500 kg)}{(5 m)^2}=6.67\cdot 10^{-4} N$

2) $2.7\cdot 10^{-3} N$

From the formula in exercise 1), we see that the force is inversely proportional to the square of the distance:

$F \sim \frac{1}{r^2}$

this means that if we cut in a half the distance without changing the masses, the magnitude of the forces changes by a factor

$F'\sim \frac{1}{(r/2)^2}=4 \frac{1}{r^2}=4F$

So, the gravitational force increases by a factor 4. Therefore, the new force will be

$F' = 4 F=4(6.67\cdot 10^{-4} N)=2.7\cdot 10^{-3} N$

3) 12.5 Nm

The torque is equal to the product between the magnitude of the perpendicular force and the distance between the point of application of the force and the centre of rotation:

$\tau=Fd$

Where, in this case:

F = 25 N is the perpendicular force

d = 0.5 m is the distance between the force and the center

By using the equation, we find

$\tau=(25 N)(0.5 m)=12.5 Nm$

4) 0.049 kg m^2/s

The relationship between angular momentum (L), moment of inertia (I) and angular velocity ( $\omega$ ) is:

$L=I\omega$

In this problem, we have

$I=0.007875 kgm^2$

$\omega=6.28 rad/s$

So, the angular momentum is

$L=I\omega=(0.007875 kgm^2)(6.28 rad/s)=0.049 kg m^2/s$

6 0

3 years ago

According to Newton’s first law of motion, what will an object in motion do when no external force acts on it?

KonstantinChe [14]

By definition, we have to:

Newton's first law states that any object will remain in a state of rest or with a uniform rectilinear motion unless an external force acts on it.

Therefore, according to the first law of Newton, if the object is already in motion and has no force acting on it then, it will remain with a uniform rectilinear motion.

Answer:

The object will remain with a uniform rectilinear movement when the external force does not act on it.

4 0

4 years ago

Read 2 more answers

Nitroglycerin flows through a pipe of diameter 3.0 cm at 2.0 m/s. If the diameter narrows to 0.5 cm, what will the velocity be?

Korolek [52]

Answer:

72 m/s

Explanation:

D1 = 3 cm, v1 = 2 m/s

D2 = 0.5 cm,

Let the velocity at narrow end be v2.

By use of equation of continuity

A1 v1 = A2 v2

3.14 × 3 × 3 × 2 = 3.14 × 0.5 ×0.5 × v2

v2 = 72 m/s

8 0

3 years ago

3. Si usted duplicara la amplitud de un M.A.S. ¿cómo cambiaría la frecuencia, velocidad máxima, la aceleración máxima y la energ

WITCHER [35]

Answer:

Explanation:

la frecuencia = ω/2π, nada cambio

v(max) = ωA → ω2Α = 2ωA duplicara velocidad máxima

a(max) = ω²Α → ω²2Α = 2ω²Α duplicara la aceleración máxima

la energía total ½kA² → ½k(2Α)² = 4(½kA²) cuatro veces la energía

8 0

3 years ago