All categories

3 years ago

Physical Science- Please help, mate.

Physics

1 answer:

Elis [28]3 years ago

5 0

a) Mercury (Hg) is the most dense

b) Titanium (Ti) is the least dense

c) We can tell it by comparing the slope of the lines

Explanation:

The density of an object is defined as:

$d=\frac{m}{V}$

where

m is the mass of the sample

V is the volume of the sample

In the graph in this problem, the mass of different samples of different materials is plotted against the volume of the samples. Therefore, we have:

- the mass on the y-axis

- the volume on the x-axis

This means that the slope of each curve is equal to the ratio between the value of the y-variable and the value of the x-variable, and therefore, it is equal to the ratio between mass and volume, which is the density:

$m=\frac{y}{x}=\frac{m}{V}=d$

So, we can compaare the density of the different materials by comparing the slopes of their lines. In particular:

- The line with greatest slope corresponds to the material with greatest density (Mercury)

- The line with the least slope corresponds to the material with least density (titanium)

Learn more about density:

brainly.com/question/5055270

brainly.com/question/8441651

#LearnwithBrainly

You might be interested in

Answer the following questions for a mass that is hanging on a spring and oscillating up and down with simple harmonic motion. N

LiRa [457]

Answer:

1. equilibrium

2. bottom

3. bottom

4. nowhere

5. bottom

6. top & bottom

7. equilibrium

8. equilibrium

1. No

2. Yes

Explanation:

According to the following equation of motion for SHM:

$x(t) = A\cos(\omega t + \phi)$

where A is the amplitude, ω is the angular frequency, and ∅ is the phase angle.

Furthermore, the velocity and acceleration functions are as follows:

$y(t) = -\omega A\sin(\omega t + \phi)\\a(t) = -\omega^2 A\cos(\omega t + \phi)$

1. The acceleration is zero at the equilibrium. At the equilibrium, the net force on the object is zero. And according to Newton's Second Law, if the net force is zero, then the acceleration is zero as well.

2. The forces on the object in a vertical spring are the weight of the object and the spring force.

F = mg - kx

Since mg is constant along the motion, then the net force is maximum at the amplitude. For the special case in this question, the mass is always below the rest length of the spring. So the net force is maximum at the lower amplitude, because x is greater in magnitude at the lower amplitude. According to Newton's Second Law, acceleration is proportional to the net force, hence the acceleration is at a maximum at the bottom.

3. As explained above, the magnitude of the net force is at a maximum at the lower amplitude, that is bottom.

4. The spring force is defined by Hooke's Law: F = -kx. Since the oscillation is small enough so that the mass is always below the rest length of the spring, then x is always greater than zero, hence nowhere in the motion will the spring force becomes zero.

5. As explained above, the force of gravity is constant and the spring force is proportional to the displacement, x. Therefore, the spring force is at a maximum at the lower amplitude, that is bottom.

6. The speed is zero when the mass is instantaneously at rest, that is the amplitude.

7. The net force on the mass is zero at the equilibrium.

8. The speed is at a maximum at the equilibrium.

1. We will use the equation of motions given above. For simplicity, let's take ∅ = 0. At half its amplitude:

$\frac{A}{2} = A\cos(\omega t)\\\frac{1}{2} = \cos(\omega t)\\\omega t = \pi / 3$

Then the velocity at that point is

$v(t) = -\omega A\sin(\pi /3) = -\omega A (0.866)$

The maximum speed is where the acceleration is equal to zero:

$0 = -\omega^2 A\cos(\omega t)\\\omega t = \pi / 2\\v_{max} = -\omega A\sin(\pi /2) = -\omega A$

Comparing the maximum velocity to the velocity at A/2 yields that it is not half the maximum velocity:

$-\omega A(0.866) \neq -\omega A$

2. The maximum acceleration is at the amplitude.

$A = A\cos(\omega t)\\\omega t = 2\pi\\a_{max} = -\omega^2 A\cos(2\pi) = -\omega^2 A$

And the acceleration at A/2 is

$\frac{A}{2} = A\cos(\omega t)\\\omega t = \pi / 3\\a(t) = -\omega^2 A\cos(\pi / 3) = -\omega^2 A (0.5)$

Comparing these two results yields that the acceleration at half the amplitude is half the maximum acceleration.

5 0

3 years ago

What do generators do

777dan777 [17]

Answer:

generate electricity or power

3 0

4 years ago

Read 2 more answers

A powered winch is used to pull a sailboat to shore. The winch uses a 900 W motor. If the motor is used for 30 s, how much work

kolbaska11 [484]

Answer:

they have given us

power = 900W

time = 30s

$p = \frac{wd}{t} \\ p \times t = wd \\ 900 \times 30 = w.d \\ 27000j$

4 0

3 years ago

Read 2 more answers

How density of substance change with change in temperature?

Anna71 [15]

The answer is above but I don't know if it's correct.

7 0

2 years ago

The figure shows an object moving from point B to point A and the gravitational potential energy of the object-Earth system at d

Natasha2012 [34]

The amount of kinetic energy that will be gained from the particle as it moves from point B to point A is 12U₀.

Since the particle moves through in a projectile path (trajectory). The gravitational potential energy acts on the particle in the downward direction.

A resistive force refers to a force that resists and acts in the direction opposite to the velocity of a body.

$\mathbf{F_r' = 12 F}$