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2 years ago

A piece of plastic with a mass of 15 g is

Physics

1 answer:

satela [25.4K]2 years ago

5 0

Answer:

The density of plastic is equal to 0.6 g/mL.

Explanation:

Given that,

The mass of piece of plastic, m = 15 g

It is placed in a graduated cylinder. The water level in the graduated cylinder rises from 30 mL to 55 mL when the plastic is added.

We need to find the density of plastic.

Rise in volume = 55 mL - 30 mL

= 25 mL

The density of an object is given by :

$d=\dfrac{m}{V}\\\\d=\dfrac{15\ g}{25\ mL}\\\\d=0.6\ g/mL$

So, the density of plastic is equal to 0.6 g/mL.

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Two cars A and B are 100m apart moving towards each other with

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Let car A's starting position be the origin, so that its position at time t is

A: x = (40 m/s) t

and car B has position at time t of

B: x = 100 m - (60 m/s) t

They meet when their positions are equal:

(40 m/s) t = 100 m - (60 m/s) t

(100 m/s) t = 100 m

t = (100 m) / (100 m/s) = 1 s

so the cars meet 1 second after they start moving.

They are 100 m apart when the difference in their positions is equal to 100 m:

(40 m/s) t - (100 m - (60 m/s) t) = 100 m

(subtract car B's position from car A's position because we take car A's direction to be positive)

(100 m/s) t = 200 m

t = (200 m) / (100 m/s) = 2 s

so the cars are 100 m apart after 2 seconds.

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2 years ago

The process of wind blowing sand from one location to another is called

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Answer:

weathering

Explanation:

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2 years ago

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ou purchase a rectangular piece of metal that has dimen- sions 5.0 * 15.0 * 30.0 mm and mass 0.0158 kg. The seller tells you tha

Natalija [7]

Answer: 7022.2kg/m³, yes, I was cheated

Explanation:

Density of an object is defined as the ratio of the mass of the object to its volume. Mathematically;

Density = Mass/Volume

Note that the unit of both mass and volume must be standard unit.

Given mass = 0.0158kg

Dimension of the metal = 5mm×15mm×30mm

Note that 1mm = 0.001m

The volume of the metal will be

0.005×0.015×0.03

= 0.00000225m³

Density = 0.0158/0.00000225

Average density of the metal = 7022.2kg/m³

Since the standard density of Gold is 19,320kg/m³ and is higher than the density prescribed for me, it shows the I was cheated.

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3 years ago

Which of the following electromagnetic waves is not used for communication?

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Microwaves and radio waves are employed in radio and satellite communications while infrared waves are used in remote controls and infrared features of new phones and other electrons. However, gamma rays have far too much energy and cause damage to the body. They are not used in communication.
The answer is A.

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2 years ago

Sphere A of mass 0.600 kg is initially moving to the right at 4.00 m/s. sphere B, of mass 1.80 kg is initially to the right of s

anzhelika [568]

A) The velocity of sphere A after the collision is 1.00 m/s to the right

B) The collision is elastic

C) The velocity of sphere C is 2.68 m/s at a direction of $-5.2^{\circ}$

D) The impulse exerted on C is 4.29 kg m/s at a direction of $-5.2^{\circ}$

E) The collision is inelastic

F) The velocity of the center of mass of the system is 4.00 m/s to the right

Explanation:

We can solve this part by using the principle of conservation of momentum. The total momentum of the system must be conserved before and after the collision:

$p_i = p_f\\m_A u_A + m_B u_B = m_A v_A + m_B v_B$

$m_A = 0.600 kg$ is the mass of sphere A

$u_A = 4.00 m/s$ is the initial velocity of the sphere A (taking the right as positive direction)

$v_A$ is the final velocity of sphere A

$m_B = 1.80 kg$ is the mass of sphere B

$u_B = 2.00 m/s$ is the initial velocity of the sphere B

$v_B = 3.00 m/s$ is the final velocity of the sphere B

Solving for vA:

$v_A = \frac{m_A u_A + m_B u_B - m_B v_B}{m_A}=\frac{(0.600)(4.00)+(1.80)(2.00)-(1.80)(3.00)}{0.600}=1.00 m/s$

The sign is positive, so the direction is to the right.

To verify if the collision is elastic, we have to check if the total kinetic energy is conserved or not.

Before the collision:

$K_i = \frac{1}{2}m_A u_A^2 + \frac{1}{2}m_B u_B^2 =\frac{1}{2}(0.600)(4.00)^2 + \frac{1}{2}(1.80)(2.00)^2=8.4 J$

After the collision:

$K_f = \frac{1}{2}m_A v_A^2 + \frac{1}{2}m_B v_B^2 = \frac{1}{2}(0.600)(1.00)^2 + \frac{1}{2}(1.80)(3.00)^2=8.4 J$

The total kinetic energy is conserved: therefore, the collision is elastic.

Now we analyze the collision between sphere B and C. Again, we apply the law of conservation of momentum, but in two dimensions: so, the total momentum must be conserved both on the x- and on the y- direction.

Taking the initial direction of sphere B as positive x-direction, the total momentum before the collision along the x-axis is:

$p_x = m_B v_B = (1.80)(3.00)=5.40 kg m/s$

While the total momentum along the y-axis is zero:

$p_y = 0$

We can now write the equations of conservation of momentum along the two directions as follows:

$p_x = p'_{Bx} + p'_{Cx}\\0 = p'_{By} + p'_{Cy}$ (1)

We also know the components of the momentum of B after the collision:

$p'_{Bx}=(1.20)(cos 19)=1.13 kg m/s\\p'_{By}=(1.20)(sin 19)=0.39 kg m/s$

So substituting into (1), we find the components of the momentum of C after the collision:

$p'_{Cx}=p_B - p'_{Bx}=5.40 - 1.13=4.27 kg m/s\\p'_{Cy}=p_C - p'_{Cy}=0-0.39 = -0.39 kg m/s$

So the magnitude of the momentum of C is

$p'_C = \sqrt{p_{Cx}^2+p_{Cy}^2}=\sqrt{4.27^2+(-0.39)^2}=4.29 kg m/s$

Dividing by the mass of C (1.60 kg), we find the magnitude of the velocity:

$v_c = \frac{p_C}{m_C}=\frac{4.29}{1.60}=2.68 m/s$

And the direction is

$\theta=tan^{-1}(\frac{p_y}{p_x})=tan^{-1}(\frac{-0.39}{4.27})=-5.2^{\circ}$

The impulse imparted by B to C is equal to the change in momentum of C.

The initial momentum of C is zero, since it was at rest:

$p_C = 0$

While the final momentum is:

$p'_C = 4.29 kg m/s$

So the magnitude of the impulse exerted on C is

$I=p'_C - p_C = 4.29 - 0 = 4.29 kg m/s$

And the direction is the angle between the direction of the final momentum and the direction of the initial momentum: since the initial momentum is zero, the angle is simply equal to the angle of the final momentum, therefore $-5.2^{\circ}$ .

To check if the collision is elastic, we have to check if the total kinetic energy is conserved or not.

The total kinetic energy before the collision is just the kinetic energy of B, since C was at rest:

$K_i = \frac{1}{2}m_B u_B^2 = \frac{1}{2}(1.80)(3.00)^2=8.1 J$

The total kinetic energy after the collision is the sum of the kinetic energies of B and C:

$K_f = \frac{1}{2}m_B v_B^2 + \frac{1}{2}m_C v_C^2 = \frac{1}{2}(1.80)(1.20)^2 + \frac{1}{2}(1.60)(2.68)^2=7.0 J$

Since the total kinetic energy is not conserved, the collision is inelastic.

Here we notice that the system is isolated: so there are no external forces acting on the system, and this means the system has no acceleration, according to Newton's second law:

$F=Ma$

Since F = 0, then a = 0, and so the center of mass of the system moves at constant velocity.

Therefore, the centre of mass after the 2nd collision must be equal to the velocity of the centre of mass before the 1st collision: which is the velocity of the sphere A before the 1st collision (because the other 2 spheres were at rest), so it is simply 4.00 m/s to the right.

Learn more about momentum and collisions:

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brainly.com/question/2990238

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