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3 years ago

GUESS WHAT??? I'LL MAKE YOU BRAINLIEST!! PLEASE ANSWER THIS FAST!

Physics

2 answers:

MrRa [10]3 years ago

5 0

Hello!

I’m pretty sure your answer is D

lana [24]3 years ago

5 0

I’m pretty sure the answer is d

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A wad of clay of mass m1 = 0.49 kg with an initial horizontal velocity v1 = 1.89 m/s hits and adheres to the massless rigid bar

notka56 [123]

Answer:

<h2>The angular velocity just after collision is given as</h2><h2> $\omega = 0.23 rad/s$ </h2><h2>At the time of collision the hinge point will exert net external force on it so linear momentum is not conserved</h2>

Explanation:

As per given figure we know that there is no external torque about hinge point on the system of given mass

So here we will have

$L_i = L_f$

now we can say

$m_1v_1\frac{L}{2} = (m_2L^2 + m_1(\frac{L}{2})^2)\omega$

so we will have

$0.49(1.89)(0.45) = (2.13(0.90)^2 + 0.49(0.45)^2)\omega$

$\omega = 0.23 rad/s$

Linear momentum of the system is not conserved because at the time of collision the hinge point will exert net external force on the system of mass

So we can use angular momentum conservation about the hinge point

6 0

3 years ago

Which of the following phrases best describes a physical model?

Shtirlitz [24]

Answer:

option D

...............

3 0

3 years ago

In the periodic table, the most reactive metals are found a. in Group 1, the first column on the left. b. in Period 1, the first

Gnesinka [82]

Answer:

Bottom left corner for whatever group that is

Lithium, sodium, and potassium all react with water

3 0

3 years ago

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A metal block of mass 3 kg is falling downward and has velocity of 0.44 m/s when it is 0.8 m above the floor. It strikes the top

Anton [14]

Answer:

$y_{max} = 0.829\,m$

Explanation:

Let assume that one end of the spring is attached to the ground. The speed of the metal block when hits the relaxed vertical spring is:

$v = \sqrt{(0.8\,\frac{m}{s})^{2} + 2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (0.4\,m)}$

$v = 2.913\,\frac{m}{s}$

The maximum compression of the spring is calculated by using the Principle of Energy Conservation:

$(3\,kg)\cdot (9.807\,\frac{m}{s^{2}})\cdot (0.4\,m) + \frac{1}{2}\cdot (3\,kg)\cdot (2.913\,\frac{m}{s} )^{2} = (3\,kg) \cdot (9.807\,\frac{m}{s^{2}})\cdot (0.4\,m-\Delta s) + \frac{1}{2}\cdot (2000\,\frac{N}{m})\cdot (\Delta s) ^{2}$

After some algebraic handling, a second-order polynomial is formed:

$12.728\,J = \frac{1}{2}\cdot (2000\,\frac{N}{m} )\cdot (\Delta s)^{2} - (3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot \Delta s$

$1000\cdot (\Delta s)^{2}-29.421\cdot \Delta s - 12.728 = 0$

The roots of the polynomial are, respectively:

$\Delta s_{1} \approx 0.128\,m$

$\Delta s_{2} \approx -0.099\,m$

The first root is the only solution that is physically reasonable. Then, the elongation of the spring is:

$\Delta s \approx 0.128\,m$

The maximum height that the block reaches after rebound is:

$(3\,kg) \cdot (9.807\,\frac{m}{s^{2}} )\cdot (0.4\,m-\Delta s) + \frac{1}{2}\cdot (2000\,\frac{N}{m})\cdot (\Delta s)^{2} = (3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot y_{max}$

$y_{max} = 0.829\,m$

4 0

3 years ago

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Which of the following statements is true about hydrothermal vents?

irina [24]

A because its about an opening in the sea floor and tectonic plates are the only thing.

7 0

4 years ago

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