All categories

3 years ago

How does Density depend on the type of object being measured ?

Physics

1 answer:

max2010maxim [7]3 years ago

5 0

Answer:

ways to exaplined

Explanation:

Density is calculated by

dividing the mass by the volume, so that density is measured as units of mass/volume, often g/mL. If both water samples are at the same temperature, their densities should be identical, regardless of the samples' volume.

You might be interested in

A charge Q is distributed uniformly on a non-conducting ring of radius R and mass M. The ring is dropped from rest from a height

mihalych1998 [28]

Answer:

Check below for the explanation

Explanation:

Since it is stated that the ring is dropped from a height, h, through a non uniform magnetic field, two kinds of force will act on the ring, namely:

A magnetic force (that is non uniform since the field is non uniform)
Gravitational force

A certain amount of torque is provided by the non uniform magnetic force on the ring while the force gravity pulls it down. Due to the downward pull by the force of gravity on the ring and the torque acting on it as a result of the non uniform magnetic force, the ring begins to rotate.

5 0

3 years ago

(An easy problem which will be graded). Later in the quarter we will spend some time solving the diffusion equation Op(r, t) = D

alexandr402 [8]

Answer:

Explanation:

Answer is in the attachment below:

7 0

3 years ago

As a bullet shot vertically upward rises, the kinetic energy of the bullet

algol [13]

The kenitc energy of the bullet lowers as it keeps going up.

Because gravity is pushing the bullet down as the bullet goes up.

I'm pretty sure that the way to put this answer.

3 0

2 years ago

The acceleration due to gravity on the moon is about 1/6 of the acceleration due to gravity on the earth. A net force F acts hor

uysha [10]

Answer:

c. $a_m$

Explanation:

We are given that

Acceleration due to gravity on the moon= $a_m$

Acceleration due to gravity on the earth= $a_e$

$g_m=\frac{1}{6}g_e$

Net force due to am on an object on moon= $F_{net}=ma_m$

There is no friction and no drag force and there is no gravity involved

Then, the force acting on an object on earth= $F=ma_e$

$F=F_{net}$ (given)

$ma_m=ma_e$

$a_e=a_m$

Hence, option c is true.

3 0

2 years ago

I need answers and solvings to these questions

den301095 [7]

1) The period of a simple pendulum depends on B) III. only (the length of the pendulum)

2) The angular acceleration is C) $15.7 rad/s^2$

3) The frequency of the oscillation is C) 1.6 Hz

4) The period of vibration is B) 0.6 s

5) The diameter of the nozzle is A) 5.0 mm

6) The force that must be applied is B) 266.7 N

Explanation:

The period of a simple pendulum is given by

$T=2\pi \sqrt{\frac{L}{g}}$

where

T is the period

L is the length of the pendulum

g is the acceleration of gravity

From the equation, we see that the period of the pendulum depends only on its length and on the acceleration of gravity, while there is no dependence on the mass of the pendulum or on the amplitude of oscillation. Therefore, the correct option is

B) III. only (the length of the pendulum)

The angular acceleration of the rotating disc is given by the equation

$\alpha = \frac{\omega_f - \omega_i}{t}$

where

$\omega_f$ is the final angular velocity

$\omega_i$ is the initial angular velocity

t is the time elapsed

For the compact disc in this problem we have:

$\omega_i = 0$ (since it starts from rest)

$\omega_f = 300 rpm \cdot \frac{2\pi rad/rev}{60 s/min}=31.4 rad/s$ is the final angular velocity

t = 2 s

Substituting, we find

$\alpha = \frac{31.4-0}{2}=15.7 rad/s^2$

For a simple harmonic oscillator, the acceleration and the displacement of the system are related by the equation

$a=-\omega^2 x$

where

a is the acceleration

x is the displacement

$\omega$ is the angular frequency of the system

For the oscillator in this problem, we have the following relationship

$a=-100 x$

which implies that

$\omega^2 = 100$

And so

$\omega = \sqrt{100}=10 rad/s$

Also, the angular frequency is related to the frequency $f$ by

$f=\frac{\omega}{2\pi}$

Therefore, the frequency of this simple harmonic oscillator is

$f=\frac{10}{2\pi}=1.6 Hz$

When the mass is hanging on the sping, the weight of the mass is equal to the restoring force on the spring, so we can write

$mg=kx$

where

m is the mass

$g=9.8 m/s^2$ is the acceleration of gravity

k is the spring constant

x = 8.0 cm = 0.08 m is the stretching of the spring

We can re-arrange the equation as

$\frac{k}{m}=\frac{g}{x}=\frac{9.8}{0.08}=122.5$

The angular frequency of the spring is given by

$\omega=\sqrt{\frac{k}{m}}=\sqrt{122.5}=11.1 Hz$

And therefore, its period is

$T=\frac{2\pi}{\omega}=\frac{2\pi}{11.1}=0.6 s$

According to the equation of continuity, the volume flow rate must remain constant, so we can write

$A_1 v_1 = A_2 v_2$

where

$A_1 = \pi r_1^2$ is the cross-sectional area of the hose, with $r_1 = 5 mm$ being the radius of the hose

$v_1 = 4 m/s$ is the speed of the petrol in the hose

$A_2 = \pi r_2^2$ is the cross-sectional area of the nozzle, with $r_2$ being the radius of the nozzle

$v_2 = 16 m/s$ is the speed in the nozzle

Solving for $r_2$ , we find the radius of the nozzle:

$\pi r_1^2 v_1 = \pi r_2^2 v_2\\r_2 = r_1 \sqrt{\frac{v_1}{v_2}}=(5)\sqrt{\frac{4}{16}}=2.5 mm$

So, the diameter of the nozzle will be

$d_2 = 2r_2 = 2(2.5)=5.0 mm$

According to the Pascal principle, the pressure on the two pistons is the same, so we can write

$\frac{F_1}{A_1}=\frac{F_2}{A_2}$

where

$F_1$ is the force that must be applied to the small piston

$A_1 = \pi r_1^2$ is the area of the first piston, with $r_1= 2 cm$ being its radius

$F_2 = mg = (1500 kg)(9.8 m/s^2)=14700 N$ is the force applied on the bigger piston (the weight of the car)

$A_2 = \pi r_2^2$ is the area of the bigger piston, with $r_2= 15 cm$ being its radius

Solving for $F_1$ , we find

$F_1 = \frac{F_2A_1}{A_2}=\frac{F_2 \pi r_1^2}{\pi r_2^2}=\frac{(14700)(2)^2}{(15)^2}=261 N$

So, the closest answer is B) 266.7 N.

Learn more about pressure:

brainly.com/question/4868239

brainly.com/question/2438000

#LearnwithBrainly

5 0

3 years ago