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Angelina_Jolie [31]
3 years ago
15

During Mr. Nye's science class, students were expected to identify various substances using physical properties they could easil

y measure. Mr. Nye gave each of the five groups a green, metal cube. The students had to identify what metal the cube was made of. The length of each side was 3.0 cm and the mass was 270 grams.
What property should Mr. Nye's science students use to identify the metal in the cubes?

A) color

B) density

C) mass

D) volume
Physics
2 answers:
Zepler [3.9K]3 years ago
6 0
Hey there

The students should use density. 


nydimaria [60]3 years ago
4 0
Density is the best property to use, as while multiple different metals could create cubes with the same color, mass, or volume, no different metal could create a cube with the same mass and volume.  Density is based on mass and volume, and as a result no two different metals will have the same density.
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A model rocket is launched straight upward with an initial speed of 52.0 m/s. It accelerates with a constant upward acceleration
JulsSmile [24]

Explanation:

(a) After the engines stop, the rocket reaches a maximum height at which it will stop and begin to descend in free fall due to gravity.

(b) We must separate the motion into two parts, when the rocket's engines is on  and when the rocket's engines is off.

First we must find the rocket speed when the engines stop:

v_f^2=v_0^2+2ay_1\\v_f^2=(52\frac{m}{s})^2+2(1\frac{m}{s^2})(160m)\\v_f^2=3024\frac{m^2}{s^2}\\v_f=\sqrt{3024\frac{m^2}{s^2}}=54.99\frac{m}{s}

This final speed is the initial speed in the second part of the motion, when engines stop until reach its maximun height. Therefore, in this part the final speed its zero and the value of g its negative, since decelerates the rocket:

v_f^2=v_0^2+2gy_{2}\\y_{2}=\frac{v_f^2-v_0^2}{2g}\\y_{2}=\frac{0^2-(54.99\frac{m}{s})^2}{2(-9.8\frac{m}{s^2})}=154.28m

So, the maximum height reached by the rocket is:

h=y_1+y_2\\h=160m+154.28m=314.28m

(c) In the first part we have:

v_f=v_0+at_1\\t_1=\frac{v_f-v_0}{a}\\t_1=\frac{54.99\frac{m}{s}-52\frac{m}{s}}{1\frac{m}{s^2}}\\t_1=2.99s

And in the second part:

t_2=\frac{v_f-v_0}{g}\\t_2=\frac{0-54.99\frac{m}{s}}{-9.8\frac{m}{s^2}}\\t_2=5.61s

So,  the time it takes to reach the maximum height is:

t_3=t_1+t_2\\t_3=2.99s+5.61s=8.60s

(d) We already know the time between the liftoff and the maximum height, we must find the rocket's time between the maximum height and the ground, therefore, is a free fall motion:

v_f^2=v_0^2+2ay\\v_f^2=0^2+2(9.8\frac{m}{s^2})(314.28m)\\v_f=\sqrt{6159.888\frac{m^2}{s^2}}=78.48\frac{m}{s}

t_4=\frac{v_f-v_0}{g}\\t_4=\frac{78.48\frac{m}{s}-0}{9.8\frac{m}{s^2}}\\t_4=8.01s

So, the total time is:

t=t_3+t_4\\t=8.60s+8.01s\\t=16.61s

7 0
3 years ago
Matthew is waterskiing. As the boat starts moving, he is at an angle of 8.0° to the right of the boat. The boat applies 250 newt
Citrus2011 [14]

The work done is B. 1.2\cdot 10^4 J

Explanation:

The work done by a force on an object is given by

W=Fd cos \theta

where

F is the magnitude of the force

d is the displacement of the object

\theta is the angle between the direction of the force and of the displacement

For the boat in this problem, we have:

F = 250 N (force applied)

d = 50 m (displacement)

\theta=8.0^{\circ}

Substituting, we find the work done:

W=(250)(50)(cos 8^{\circ})=1.2\cdot 10^4 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

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3 years ago
Why is friction a problem in space travellers​
marin [14]
The answer is gravitational:)
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3 years ago
Explain the statement: “Cells are the basic units of structure and function in living things.”
babymother [125]

All living things are made up of cells.  Each cell's individual function helps the organism collectively function.  They are the most basic as they can be self-contained.  And there is no other particle smaller that can function on it's own.

6 0
3 years ago
An inductor in an LC circuit has a maximum current of 2.4 A and a maximum energy of 56 mJ.
Harrizon [31]

Answer:

The energy stored in the capacitor, when the current in the inductor is 1.2 A, is 41.6 mJ.

Explanation:

In a LC oscillating circuit, the energy is stored in the electric field (between the plates of the capacitor) and in the magnetic field (surrounding the wires of the inductor).

At any time, the sum of both energies can be expressed as follows:

E = 1/2 Q² / C   +  1/2 L I²

In this type of circuit, energy oscillates, which means that it is exchanging between both fields all time.

When the capacitor is completely discharged, all the energy is stored in the magnetic field, and at that time, the current is maximum.

The total energy, when I is maximum, can be written as follows:

E = 1/2 L I² (1)

In our case, when I= 2.4A, E= 56 mJ.

So, we can find out the value of L, which will allow us to know the value of the magnetic energy at any time, having the value of the instantaneous current.

Solving for L in (1):

L = 2 *.56 mJ / (2.4)² A² = 20 mH

The next step is getting the value of the energy stored in the inductor, when I = 1.2 A, as follows:

Em = 1/2 *20 mH.* (1.2)² A² = 14.4 mJ

As the total energy must be always the same, i.e., 56 mJ, the energy stored in the capacitor, assuming no losses, must be the difference between the total energy and the one stored in the magnetic field:

Ec = 56 mJ - 14.4 mJ = 41.6 mJ

3 0
3 years ago
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