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34kurt
3 years ago
14

1. The smallest unit of matter that cannot be broken down by ordinary means is

Physics
2 answers:
anastassius [24]3 years ago
6 0

Answer:

the cell

Explanation:

the cell theory states cells are the smallest unit of life

8090 [49]3 years ago
6 0

Answer:

The answer is B

Explanation:

The basic unit of matter of which can't be broken down into simpler or smaller fragments by any chemical reaction is an atom

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How does an octopus or a squid uses Newton's third law of motion to escape from enemies
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I think everytime they swim away, the water pushes the enemy back, which makes the octopus go faster. Every action has an equal and opposite reaction. Hope this helps.
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4 years ago
A woman (mass= 50.5 kg) jumps off of the ground, and comes back down to the ground at a velocity of -8.4 m/s.
Blizzard [7]

Answer:

Approximately 1.6\times 10^{3}\; \rm N.

Explanation:

By the Impulse-Momentum Theorem, the change in this woman's momentum  will be equal to the impulse that is applied to her.

The momentum p of an object is equal to the product of its mass m and velocity v. That is: p = m \cdot v.

Let v(\text{before}) and v(\text{after}) represent the velocity of the woman before and after the landing. Let m represent the woman's mass.

  • The woman's momentum before the landing would be m \cdot v(\text{before}).
  • The woman's momentum after the landing would be m \cdot v(\text{after}).

Therefore, the change in this woman's momentum would be:

\begin{aligned}& \Delta p \\ & = p(\text{after}) - p(\text{before}) \\ &= m \cdot (v(\text{after})- v(\text{before}))\end{aligned}.

On the other hand, impulse is equal to force multiplied by the duration of the force. Let F represent the average force on the woman. The impulse on her during the landing would be F \cdot t.

Apply the Impulse-Momentum Theorem.

  • Impulse: F\cdot t.
  • Change in momentum: m \cdot (v(\text{after})- v(\text{before})).

Impulse is equal to the change in momentum:

F \cdot t = m \cdot (v(\text{after})- v(\text{before})).

After landing, the woman comes to a stop. Her velocity would become zero. Therefore, v(\text{after}) = 0\; \rm m \cdot s^{-1}.

\begin{aligned}F &= \displaystyle \frac{m \cdot (v(\text{after})- v(\text{before}))}{t} \\ &= \frac{50.5\; \text{kg} \times \left(0 \; \mathrm{m \cdot s^{-1}}- 8.4\; \mathrm{m \cdot s^{-1}}\right)}{0.27\; \rm s} \\ &\approx 1.6 \times 10^{3}\; \rm N\end{aligned}.

3 0
4 years ago
You and your friends are doing physics experiments on a frozen pond that serves as a frictionless, horizontal surface. Sam, with
Misha Larkins [42]

Answer:

Sam's and Abigail speeds before colliding were a. 12.34 m/s and b. 2.86 m/s, respectively. Their total kinetic energy was diminished by c. 1484.42 J, approximately

Explanation:

By conservation of momentum, we have

\Delta \vec{p} = 0\\\\m\vec{v}_i = m\vec{v}_f\\m_S v_S\hat{i} + m_A v_A\hat{j} = m_S \times 7 \cos{(40)} \hat{i} + m_S \times 7 \sin{(40)} \hat{j} + m_A \times 9.4 \cos{(18)} \hat{i} - m_A \times 9.4 \sin{(18)} \hat{j}.

Writing for each direction at a time,

m_S v_S = 7m_S \cos{(40)} +9.4 m_A \cos{(18)}\\v_S = 7 \cos{(40)} + 9.4 \frac{m_A}{m_S} \cos{(18)} = 7 \cos{(40)} + 9.4\times\frac{57}{73} \cos{(18)} \approx \mathbf{12.34}\\m_A v_A = 7m_S\sin{(40)} - 9.4m_A\sin{(18)}\\v_A = 7\frac{m_S}{m_A}\sin{(40)} - 9.4\sin{(18)} = 7\times\frac{73}{57}\sin{(40)} - 9.4\sin{(18)} \approx \mathbf{2.86}.

Their kinetic energy changed by

K_f - K_i = \left( \frac{1}{2}m_s v_{fs}^2 + \frac{1}{2}m_a v_{fa}^2 \right) - \left( \frac{1}{2}m_s v_{is}^2 + \frac{1}{2}m_a v_{ia}^2 \right) =  \left( \frac{1}{2}\times 73 \times 7^2 + \frac{1}{2}\times 57 \times 9.4^2 \right) - \left( \frac{1}{2}\times 73 \times 12.34^2  + \frac{1}{2}\times 57 \times 2.86^2 \right) \approx \mathbf{-1484.418 J}.

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3 years ago
A golf club hits a 0.0459 kg golf ball at rest; the club is in contact with the ball for 0.00138 s. Afterwards the ball leaves a
Advocard [28]

Answer: 2120 N is correct

Explanation: correct answer for acellus

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3 years ago
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