Which scientific design has the fewest limitations?

1 answer:

4 0

USING DISSECTION TO STUDY THE GENERAL STRUCTURE OF LILY FLOWERS.

Other choices are very hard to do. Maintaining a salt-water aquarium is very hard while going to an arctic tundra or space is impractical.

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when the c4 key on a piano keyboard is pressed, a string inside the piano is struck by a hammer and begins vibrating back and fo

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The correct answer is f= 260 Hz

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Two 2.0kg bodies A and B collide The velocities before the collision are U1=15i+30j and U2=-10j+5.0j After the collision V1=-5.0

Svetllana [295]

Use the law of conservation of momentum. Since the momentum is a linear measure, we can treat each of the dimension separately:

i-direction:

$m_1u_{1i}+m_2u_{2i}=m_1v_{1i}+m_2v_{2i}\\v_{2i} = \frac{m_1u_{1i}+m_2u_{2i}-m_1v_{1i}}{m_2}=\frac{(2\cdot 15-2\cdot10+2\cdot5)kg\frac{m}{s}}{2kg}=10\frac{m}{s}$

j-direction:

$m_1u_{1j}+m_2u_{2j}=m_1v_{1j}+m_2v_{2j}\\v_{2j} = \frac{m_1u_{1j}+m_2u_{2j}-m_1v_{1j}}{m_2}=\frac{(2\cdot 30+2\cdot5-2\cdot20)kg\frac{m}{s}}{2kg}=15\frac{m}{s}$

Answer: Final velocity is: (10i + 15j) m/s

Change in the kinetic energy:

$\Delta E_k = E_{ku}-E_{kv} = \frac{1}{2}m(u_1^2+u_2^2-v_1^2-v_2^2)=\\=\frac{1}{2}2kg(1125+125-425-325)\frac{m^2}{s^2}=500J$

Answer: The system lost 500J worth of kinetic energy in the collision

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The smallest possible part of an element that can still be identified as the element is called

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Answer:

BLM

Explanation:

ACAB

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The structure of which substances tends to exhibit the greatest hardness?

bagirrra123 [75]

The answer is C. ionic compounds, compared to another hard compound which is the covalent bond, the structure of the ionic compound is lattice or crystalline in form thus less room for imperfections on its structure compared to the latter choices.

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3 years ago

A very strong, but inept, shot putter puts the shot straight up vertically with an initial velocity of 11.0 m/s. how long does h

noname [10]

The shot putter should get out of the way before the ball returns to the launch position.

Assume that the launch height is the reference height of zero.
u = 11.0 m/s, upward launch velocity.
g = 9.8 m/s², acceleration due to gravity.

The time when the ball is at the reference position (of zero) is given by
ut - (1/2)gt² = 0
11t - 0.5*9.8t² = 0
t(11 - 4.9t) = 0
t = 0 or t = 4.9/11 = 0.45 s

t = 0 corresponds to when the ball is launched.
t = 0.45 corresponds to when the ball returns to the launch position.

Answer: 0.45 s

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