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Sladkaya [172]
3 years ago
11

Now it's time to begin the simulation. You will use this density simulation to answer all of the questions in this assignment. K

eep the simulation open and change it only as necessary for each task.
Select the Compare button on the bottom toolbar.

Select "Same Mass" from the Blocks menu.

Place the smallest red block labeled and the largest yellow block in the water tank. Which observations are correct? Check all that apply.

The larger yellow block floats.
The smaller red block floats.
The larger yellow block is denser than the red block.
The smaller red block is denser than water.
The larger yellow block and the smaller red block have the same density.
Physics
1 answer:
dmitriy555 [2]3 years ago
5 0

Answer:

it is a and c

Explanation:

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Two asteroids collide and stick together. The first asteroid has a mass of 15\times 10^3\,\mathrm{kg}15×10 3 kg and is initially
statuscvo [17]

Answer:

Final speed is 900.06 m/s at 0.2215^{\circ}  

Solution:

As per the question:

Mass of the first asteroid, m = 15\times 10^{3}\kg

Mass of the second asteroid, m' = 20\times 10^{3}\kg

Initial velocity of the first asteroid, v = 770 m/s

Initial velocity of the second asteroid, v' = 1020 m/s

Angle between the two initial velocities, \theta = 20^{\circ}

Now,

Since, the velocities and hence momentum are vector quantities, then by the triangle law of vector addition of 2 vectors A and B, the resultant is given by:

\vec{R} = \sqrt{A^{2} + 2ABcos\theta + B^{2}}

Thus applying vector addition and momentum conservation, the final velocity is given by:

(m + m')v_{final} = \sqrt{(mv)^{2} + 2(mv)(m'v')cos20^{\circ} + (m'v')^{2}}                               (1)

Now,

(m +m')v_{final} = (35\times 10^{3})v_{final}

(mv)^{2} = (15\times 10^{3}\times 770)^{2} = 1.334\times 10^{14}

(m'v')^{2} = (20\times 10^{3}\times 1020)^{2} = 4.16\times 10^{14}

2(mv)(m'v')cos20^{\circ} = 2(15\times 10^{3}\times 770)(20\times 10^{3}\times 1020)cos20^{\circ} = 4.43\times 10^{14}

Now, substituting the suitable values in eqn (1), we get:

v_{final} = 900.06\ m/s

Now, the direction for the two vectors is given by:

\theta = sin^{- 1} \frac{m'v'sin20^{\circ}}{(m + m')v_{final}}

\theta = sin^{- 1} \frac{20\times 10^{3}\times 1020sin20^{\circ}}{(35\times 10^{3})\times 900.06} = 0.2215^{\circ}

5 0
3 years ago
An alpha particle is identical to a _____.
djyliett [7]
An alpha particle is identical to a helium nucleus. The correct option among all the options given in the question is option "A". In case of both the alpha particle nucleus and helium nucleus, there are two protons and two neutrons. Alpha particles are generally formed during the process of alpha decay. They can also be produced in other ways.
8 0
4 years ago
Read 2 more answers
What is the strength of the electric field of a point charge of magnitude +4.8 × 10^-19 C at a distance of 4.0 × 10^-3 m? A. 3.6
soldi70 [24.7K]
Answer is D -  from the equation of electric field for point charge
8 0
3 years ago
A spring hangs from the ceiling with an unstretched length of x 0 = 0.35 m . A m 1 = 6.3 kg block is hung from the spring, causi
Jlenok [28]

Answer:

x₂=0.44m

Explanation:

First, we calculate the length the spring is stretch when the first block is hung from it:

\Delta x_1=0.50m-0.35m=0.15m

Now, since the stretched spring is in equilibrium, we have that the spring restoring force must be equal to the weight of the block:

k\Delta x_1=m_1g

Solving for the spring constant k, we get:

k=\frac{m_1g}{\Delta x_1}\\\\k=\frac{(6.3kg)(9.8m/s^{2})}{0.15m}=410\frac{N}{m}

Next, we use the same relationship, but for the second block, to find the value of the stretched length:

k\Delta x_2=m_2g\\\\\Delta x_2=\frac{m_2g}{k}\\\\\implies \Delta x_2=\frac{(3.7kg)(9.8m/s^{2})}{410N/m}=0.088m

Finally, we sum this to the unstretched length to obtain the length of the spring:

x_2=0.35m+0.088m=0.44m

In words, the length of the spring when the second block is hung from it, is 0.44m.

3 0
4 years ago
A wire is joined to points X and Y in the circuit diagram shown. A diagram of a circuit with a power source on the left. Directl
Sidana [21]

The circuit change when the wire is added will see a short circuit occur and makes bulbs 1 and 2 turn off but keeps bulbs 3 and 4 lit. Option D. This is further explained below.

<h3>How does the circuit change when the wire is added?</h3>

Generally, Electronic circuits consist of a series of interconnected parts that form a closed loop through which electricity may flow.

In conclusion, If two wires are linked together, a short circuit will develop, cutting power to bulbs 1 and 2. But there is no impact on bulbs 3 and 4. There is no problem with bulbs 3 and 4.

Read more about circuit

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7 0
2 years ago
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