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3 years ago

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Two people carry a heavy electric motor by placing it on a light board 1.80 m long. One person lifts at one end with a force of

350 N, and the other lifts the opposite end with a force of 620 N.
What is the weight of the motor?

1 answer:

bixtya [17]3 years ago

6 0

Answer:

$W=970\ N$

Explanation:

When two people carry a weight placing it on a very light board then we neglect the weight of the board and the board acts like a beam with two mass placed over it and two end point supports are here provided by men.

Then balancing the forces in the vertical direction, weight of the motor is:

$W=350+620$

$W=970\ N$

These men are able to carry the motor in equilibrium condition because they have placed the motor (considering it a point mass) at a position such that the moment due to the lifting forces are equal and opposite.

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A diffraction grating is a type of optical instrument obtained with a continuous pattern. The pattern of the diffracted light by a grating depends on the structure and number of elements present.

The given data in the problem is

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$n \lambda = d sin\theta \\\\ d = \frac{n \lambda}{sin \theta } \\\\ d = \frac{1 \times 4.90 \times 10^{-7}}{sin 9.2^0 } \\\\ d=3.06 \times 10^{-6} \\\\ d=3.06 \times 10^{-3} \ mm$

The number of lines per mm is found as;

$\rm N= \frac{1}{d} \\\\ N= \frac{1}{3.06 \TIMES 10^{-3}} \\\\ N=326.8 /mm$

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$\sf{\pink{\underline{\underline{\blue{GIVEN:-}}}}}$

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$\sf{\pink{\underline{\underline{\blue{SOLUTION:-}}}}}$

⚡ Let $\rm{\vec{a}}$ and $\rm{\vec{b}}$ are the two vectors .

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$\orange\bigstar\:\rm{\pink{\boxed{\green{\vec{a}\:.\:\vec{b}\:=\:ab\cos{\theta}\:}}}}$

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3 years ago

Read 2 more answers

If measurements of gas are 75 L and 300 kIlopascals and then the gas is measured is second time and found to be 50 L, describe w

Ann [662]

Answer:

The pressure must have increased in the process

Explanation:

The State Equation for gasses reads: $P*V=n*R*T$

where P is the gas' pressure, V its volume, n the number of moles of gas, R the gas constant and T the temperature in degrees Kelvin.

If the temperature of the gas doesn't change in the described process, the right hand side of the equation stays the same. If that is the case, given that when the Volume of the gas diminishes from 75 liters to 50 liters, then the pressure must have increased to keep that product "P * V" constant:

$P_i*V_i=P_f*V_f\\75 *300=50*X\\X=\frac{75*300}{50} =450$

So the pressure must have gone up to 450 kilopascals.

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