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2 years ago

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Two blocks are placed at the ends of a horizontal massless board, as in the drawing. The board is kept from rotating and rests o

n a support that serves as an axis of rotation. The block on the right has a mass of 2.8 kg. Determine the magnitude of the angular acceleration when the system is allowed to rotate.

1 answer:

Andrew [12]2 years ago

7 0

Answer:

The magnitude of the angular acceleration ∝ = $\frac{rxF}{2.8[tex]r^{2}$ }[/tex]

Explanation:

The angular acceleration ∝ is equal to the torque (radius multiplied by force) divided by the mass times the square of the radius. The magnitude of angular acceleration ∝ will have the equation above but we have to replace the mass in the equation by 2.8kg as stated.

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3 years ago

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2 years ago

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here is the question : A student throws a water balloon vertically downward from the top of a building. The balloon leaves the thrower's hand with a speed of 60.0m/s. Air resistance may be ignored,so the water balloon is in free fall after it leaves the throwers hand. a) What is its speed after falling for 2.00s? b) How far does it fall in 2.00s? c) What is the magnitude of its velocity after falling 10.0m?

Answer:

(A) 26 m/s

(B) 32.4 m

(C) v = 15.4 m/s

Explanation:

initial speed (u) = 6.4 m/s

acceleration due to gravity (a) = 9.9 m/s^[2}

time (t) = 2 s

(A) What is its speed after falling for 2.00s?

from the equation of motion v = u + at we can get the speed

v = 6.4 + (9.8 x 2) = 26 m/s

(B) How far does it fall in 2.00s?

from the equation of motion $s=ut+0.5at^{2}$ we can get the distance covered

s = (6.4 x 2) + (0.5 x 9.8 x 2 x 2)

s = 12.8 + 19.6 = 32.4 m

c) What is the magnitude of its velocity after falling 10.0m?

from the equation of motion below we can get the velocity

$v^{2} = u^{2} + 2as\\v^{2} = 6.4^{2} + (2x9.8x10)\\V^{2} = 236.96\\v = \sqrt{236.96}$

v = 15.4 m/s

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3 years ago

A 5 kg fish swimming at 1 m/s swallows an absentminded 500 g fish swimming toward it at a velocity that brings both fish to a ha

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To solve this problem we will apply the concepts related to the conservation of momentum. Momentum is defined as the product between mass and velocity of each body. And its conservation as the equality between the initial and final momentum. Mathematically described as

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Here

$m_1$ = Mass of big fish

$m_2$ = Mass of small fish

$v_1$ = Velocity of big fish

$v_2$ = Velocity of small fish

$v_F$ = Final Velocity

The big fish eats small fish and the final velocity is zero. Rearrange the equation for the initial velocity of small fish we have

$m_1u_1=-m_2u_2$

$u_2 = -\frac{m_1u_1}{m_2}$

Replacing we have,

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Therefore the speed of the small fish is 10m/s

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2 years ago

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Readme [11.4K]

Answer:

Explanation:

Given

radius of electron $(r)=5.3\times 10^{-11} m$

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we know

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$v=\sqrt{r\times a_c}$

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(b)For n=10

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$a_c=10^4\times 9\times 10^{22} m/s^2$

$a_c=9\times 10^{26} m/s^2$

$v=\sqrt{r\times a_c}$

$v=\sqrt{9\times 10^{26}\times 5.3\times 10^{-9}}$

$v=21.84\times 10^8 m/s$

8 0

3 years ago