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3 years ago

37 Questions - All Questions Are Required

Physics

2 answers:

Ahat [919]3 years ago

6 0

A= dv / dt
a= (22 - 10)/3
a= (12)/3
a= 4 m/s²

ANTONII [103]3 years ago

5 0

Acceleration = vf-vi /t
10-22/3=2.6m/s^2

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A wave transfers _____ from the source outward in the direction of wave motion.

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A wave transfers energy from the source outward in the direction of wave motion.

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A student is late and running to class. A hall monitor tells them to slow down, so the student slows to a walking speed of 1.4 m

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8.9m/s

Explanation:

Final velocity = initial velocity + acceleration *time

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Consider a system consisting of the Earth, a vertical spring attached to the Earth, and a mass attached to the other end of the

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Answer:

The external force leads to an increase on gravitational and spring potential energies.

Explanation:

The system consists of a mass, resort and Earth. According to the Principle of Energy Conservation there is a potential energy as a consequence of the interaction between Earth and the mass and spring potential energy because of the spring deformation and, besides, the existence of work due to an external force:

$U_{g,A} + U_{k,A} + W_{ext} = U_{g,B} + U_{k,B}$

$W_{ext} = (U_{g,B}-U_{g,A}) + (U_{k,B}-U_{k,A})$

$F\cdot \Delta r = \frac{G\cdot m\cdot M}{r_{o}-\Delta r} - \frac{G\cdot m \cdot M}{r_{o}} +\frac{1}{2}\cdot k \cdot (x_{o}+\Delta r)^{2} - \frac{1}{2} \cdot k \cdot (x_{o})^{2}$

$F\cdot \Delta r = G\cdot m \cdot M \cdot \left(\frac{1}{r_{o}-\Delta r}-\frac{1}{r_{o}} \right)+\frac{1}{2}\cdot k \cdot [(x_{o}+\Delta r)^{2} -x_{o}^{2}]$

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3 years ago

Read 2 more answers

A block is pulled across a flat surface at a constant speed using a force of 50 newtons at an angle of 60 degrees above the hori

vladimir2022 [97]

The magnitude of the friction force is 25 N

Explanation:

To solve this problem, we just have to analyze the forces acting on the block along the horizontal direction. We have:

The horizontal component of the pulling force, $F cos \theta$ , where F = 50 N is the magnitude and $\theta=60^{\circ}$ is the angle between the direction of the force and the horizontal; this force acts in the forward direction
The force of friction, $F_f$ , acting in the backward direction

According to Newton's second law, the net force acting on the block in the horizontal direction must be equal to the product between the mass of the block and its acceleration:

$\sum F_x = ma_x$