3 years ago

Plans used for work that has to do with construction in or around earth are called

Physics

1 answer:

attashe74 [19]3 years ago

7 0

Plans used for work that has to do with construction in or around Earth are called, “Civil Plans.”

Hope this helped!

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Recall that the blocks can only move along the x axis. the x components of their velocities at a certain moment are v1x and v2x.

Contact [7]

The center of mass is given with this formula:
$x_c=\frac{\sum_{n=1}^{n=i}m_ix_i}{M}$
Velocity is:
$v=\frac{dv}{dt}$
So, for the velocity of the center of mass we have:
$\frac{dx_c}{dt}=\frac{\sum_{n=1}^{n=i}d(m_ix_i)}{Mdt}\\
v_c=\frac{\sum_{n=1}^{n=i}p_i}{M}\\$
In our case it is:
$v_{xc}=\frac{m_1v_{x1}+m_2v_{x2}}{m_1+m_2}$

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3 years ago

A weightlifter liftsa 1,250-N barbell 2 m in 3 s. how much power was used to lift the barbell?

STatiana [176]

Power = Force * Distance/ time
P = 1,250 * 2/3
P = 2,500/3
P = 833.33 Watts

So, your final answer is 833.33 Watts

5 0

3 years ago

Two nuclei join to form a larger nucleus during which process

pentagon [3]

nuclear fusion- produces huge amounts of energy.

4 0

3 years ago

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An elevator is being lowered at steadily decreasing speed by a steel cable attached to an electric motor. There is no air resist

shtirl [24]

Answer:

F = W + ma a> 0

Explanation:

For this exercise let's use Newton's second law

we assume the upward direction as positive

F - W = m a

F = W + ma

F = m (g + a)

In this case they indicate that the speed is less and less as it goes down, therefore the acceleration must be opposite to the speed, that is, the acceleration is upwards, consequently it is positive

We can see that since a> 0 the force F must have greater than the weight of the elevator

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2 years ago

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A particle of charge Q is fixed at the origin of an xy coordinate system. At t = 0 a particle (m = 0.727 g, q = 2.73 µC is locat

Montano1993 [528]

Answer:

Q = 12.466μC

Explanation:

For the particle to execute a circular motion, the electrostatic force must be equal to the centripetal force:

$Fe = \frac{K*q*Q}{r^{2}} = \frac{m*V^{2}}{r}$