PHYSICS HIGH SCHOOL

An object completes 10 cycles in 50 s what is the period of its rotation?​

Answers

Answer 1
Answer: 5 second period rotation

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The law of suggests that the orbit of planets is not circular but .

Answers

Answer:

Kepler's first law suggests that the orbit of planets is not circular but elliptical

Explanation:

The three Kepler's laws explain the motion of the planets orbiting the Sun:

- The first law tells that the orbits of the planets around the Sun are ellipses, with the Sun located at one of the two focii

- The second law tells that a line connecting the Sun with the planet sweeps out equal areas in equal amounts of time

- The third law tells that the square of the orbital periods of the planets is proportional to the cube of their average distance from the Sun

As we can read, the first law tells us that the orbit of the planets is not circular, but elliptical.

6+6=8 move one match puzzle game

Answers

Answer:

=

Explanation:

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Calculate the force needed to bring a 1,019-kg car to rest from a speed of 29 m/s in a distance of 118 m

Answers

Acceleration = ▵v/▵t
Time = d/v
Fisrt calculate time : ( 118/29 ) = 4 seconds
Then calculate acceleration
A = 29/4 = 7.25 m/s²
Now the force.
Force = mass * acceleration.
F= 1,019 * 7.25
F= 7,387 N

How does a supernova occur?

Answers

Supernova occurswhen black holes stretch so much running out of nuclear fuel. When a starcollapses to form a black hole, its mass increases. Supernovas have greatermasses than black holes. One example application of this is the theory of BigBang. Fromtheories such as Ekpyrotic theory, White Holes, Matrix theory, and Quantumtheory, the Big Bang theory is the theory mostly accepted by scientists.According to the theory, the universe started as a singularity, that is, from ablack hole under extreme gravitational pressure and expanded (instead ofexploded) and cooled. Its cooling, according to the theory, is still happeningnow as of the moment.

Sound is a example of what kind of wave

Answers

Answer: The correct answer is longitudinal wave.

Explanation:

Longitudinal wave:  The vibration of the particle is parallel to the direction of the propagation of the wave.

Longitudinal wave consists of the compression and the rarefaction.

Compression is the region where the density of the particles are more. Rarefaction is the region where the density of the particles are less.

The example of the longitudinal wave is Sound wave.
Sound is an example of a longitudinal wave.

A box of books weighing 328 N moves witha constant velocity across the floor when it is
pushed with a force of 378 N exerted downward at an angle of 25.6◦ below the horizontal.
Find µk between the box and the floor.

Answers

Answer:

Approximately 0.694, assuming that the floor is level.

Explanation:

Between two surfaces that are moving relative to one another, the coefficient of kinetic friction \mu_(k) is equal to the ratio between friction and normal force.

Since the box in this question is moving at a constant speed, the box would be in a translational equilibrium. Forces on this box should be balanced in both the horizontal component and the vertical component.

The value of \mu_(k) in this question can be found in the following steps:

  • Decompose the external force into horizontal and vertical components.
  • Balance forces in the horizontal direction to obtain an expression for the friction on this object.
  • Balance forces in the vertical direction to find an expression for the normal force on this object.
  • Divide the magnitude of friction by the magnitude of the normal force to find the coefficient of kinetic friction, \mu_(k).

At an angle of \theta = 25.6^(\circ) from the horizontal, magnitude of the vertical and horizontal components of the external force would be:

  • Horizontal component: F\, \cos(\theta) = (378\; {\rm N})\, \cos(25.6^(\circ)).
  • Vertical component: F\, \sin(\theta) = (378\; {\rm N})\, \sin(25.6^(\circ)).

Assume that the floor is level. Forces on this object in the horizontal direction would include:

  • Horizontal component of the external force, pointing in the direction of motion.
  • Friction from the ground, pointing backward opposite to the direction of motion.

Forces on this object are balanced in the horizontal direction. Hence, the magnitude of friction would be equal to that of the horizontal component of the external force:

(\text{magnitude of friction}) = F\, \cos(\theta).

Forces on this object in the vertical direction would include:

  • Weight of this object, pointing downward.
  • Vertical component of the external force, pointing downward.
  • Normal force from the ground, pointing upward.

Forces on this object in the vertical direction are also balanced. The magnitude of the normal force (pointing upward) should be equal to the sum of the magnitude of the two forces pointing downward:

\begin{aligned} & (\text{magnitude of normal force}) \n =\; & (\text{magnitude of weight}) + F\, \sin(\theta)\end{aligned}.

It is given that the magnitude of the weight of this object is 328\; {\rm N}. To find the coefficient of kinetic friction, divide the magnitude of friction by the magnitude of the normal force:

\begin{aligned}\mu_(k) &= \frac{(\text{magnitude of friction})}{(\text{magnitude of normal force})} \n &= \frac{F\, \cos(\theta)}{(\text{magnitude of weight}) + F\, \sin(\theta)} \n &= \frac{(378\; {\rm N})\, \cos(25.6^(\circ))}{(328\; {\rm N}) + (378\; {\rm N})\, \sin(25.6^(\circ))} \n &\approx 0.694\end{aligned}.
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