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DerKrebs [107]
3 years ago
12

Laurie is moving a dresser with a mass of 250 kg. She does 126 J of work with a force if 14 N. How far does she move the dresser

?
Physics
2 answers:
DanielleElmas [232]3 years ago
8 0

Answer: 9 m

Explanation:

Work is said to be done when an unbalanced force causes displacement of the body.

Force is the product of mass (m) and acceleration (a).

Work = Force × Displacement

⇒W = F.s = ma.s

It is given that mass of the dresser is, m = 250 kg

work done, W = 126 J

Force acting on the dresser, F = 14 N

we need to find displacement, s

⇒126 J = 14 N × s

⇒ s = 126 J/ 14 N = 9 m

Hence, Laurie is able to move the dresser to about 9 m.

galben [10]3 years ago
6 0

<u>Answer:</u>

9 m

<u>Explanation:</u>

When an object moves or displaces along the direction of the applied force application of force, it is said that work is done.

We know the formula of work:

<em>Work = force x displacement </em>

where force = mass x acceleration

So this formula can be further broken down to :

Work = mass x acceleration x d

Putting in the given values to get:

126 = 14 x d

d = 126/14

d = 9 m

Therefore, the dresser was moved by 9m.


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The ratio of time of flight for inelastic collision to elastic collision is 1:2

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Considering inelastic collision, the final velocity of the system is calculated as;

m_1u_1 + m_2u_2 = v(m_1 + m_2)\\\\m_1u_1 + 0 = v(m_1 + m_2)\\\\v= \frac{m_1u_1}{m_1 + m_2} \ -- (1)\\\\

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v = u + gt\\\\v = 0 + gt\\\\v = gt\\\\t= \frac{v}{g} \\\\t_A = \frac{m_1u_1}{g(m_1 + m_2)} \ \ ---(2)

Considering elastic collision, the final velocity of the system is calculated as;

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Apply one-directional velocity

u_1 + (-v_1) = u_2 + v_2\\\\u_1 -v_1 = 0 + v_2\\\\v_1 = v_2 -u_1

Substitute the value of v_1 into the above equation;

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v_2 is the final velocity of the block after collision

<em>Since the</em><em> bullet bounces off</em><em>, we assume that </em><em>only the block fell </em><em>to the ground from the table.</em>

The time of motion of the block is calculated as follows;

v_2 = v_0 + gt\\\\v_2 = 0 + gt\\\\t = \frac{v_2}{g} \\\\t_B = \frac{v_2}{g} \\\\ t_B = \frac{2m_1u_1}{g(m_1 + m_2)} \ \ ---(4)

The ratio of time of flight for inelastic collision to elastic collision is calculated as follows;

\frac{t_A}{t_B} = \frac{m_1u_1}{g(m_1 + m_2)} \times \frac{g(m_1 + m_2)}{2m_1u_1} \\\\\frac{t_A}{t_B} = \frac{1}{2} \\\\t_A:t_B = 1: 2

Learn more about elastic and inelastic collision here: brainly.com/question/7694106

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An extended body (not shown in the figure) has its center of mass at the origin of the reference frame. In the case below give t
Vera_Pavlovna [14]

The direction of torque τ this method is mathematically given as

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Option A is correct.

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Hence to decipher the torque direction with respect to the center of mass of the body due to force F acting on the body at a location indicated by the vector r

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It's important to note that momentum and velocity are vectors and direction matters, so if +x direction is the direction towards the wall and the -x direction away the wall \overrightarrow{v_{i}}=+2.2\, \frac{m}{s} and \overrightarrow{v_{f}}=-2.2\, \frac{m}{s} so (2) becomes:

\varDelta\overrightarrow{p}=m(-2.2- (+2.2))=-(4.5)(4.4)

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