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3 years ago

I need help plz help me out 10 points!!!!!!!

Physics

2 answers:

Sergeeva-Olga [200]3 years ago

8 0

Answer:

The answer is diffraction

Explanation:

I did the test! HOPE THIS HELPS!

mario62 [17]3 years ago

7 0

Answer:

The answer is diffraction

Explanation:

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4 years ago

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Help for physics

12345 [234]

Answer:

A. 5 m/s

Explanation:

From the graph, for the first 2 seconds, the graph is a straight line meaning that the slope is a constant.

Average speed of an object is the rate of change of position. Here, the position of the object changes from 0 m to 10 m for a time interval of 2 seconds.

The change in position ( $\Delta x$ ) and time interval ( $\Delta t$ ) are given as:

$\Delta x=10-0=10\ m\\\Delta t=2-0=2\ s$

Therefore, the average speed ( $s_{avg}$ ) is given as the ratio of the total change in position and the time interval for the change.

$s_{avg}=\frac{\Delta x}{\Delta t}=\frac{10-0}{2-0}=\frac{10}{2}=5\ m/s$

Hence, the average speed is 5 m/s.

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3 years ago

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3 years ago

What do we call the average motion of moving atoms and molecules in the atmosphere?

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3 years ago

A horizontal force of magnitude 31.3 N pushes a block of mass 4.30 kg across a floor where the coefficient of kinetic friction i

Ostrovityanka [42]

Answer:

$W_{F} = +119.6 J$ : Work done by horizontal force

$W_{Ff} = -96.42 J$ : Work done by friction force

$W_{n} = 23.18 J$ : Net work

Explanation:

Work theory

Work is the product of a force applied to a body and the displacement of the body in the direction of this force.

Thus. The work is equal to the product of the force by the distance (d) and by the cosine of the angle ( α) that exists between the direction of the force (F) and the direction that travels the point or the object that moves:

W= F*d*cosα Formula (1)

The work is positive (W+) if the force goes in the same direction of movement.

The work is negative (W-)if the force goes in the opposite direction to the movement

Net work (Wn)

Net work is defined as the sum of the work done from an initial place to an end point of the trajectory, for each of the individual forces to which the system is subjected.

Problem development

We determine the forces that perform work:

in formula (1 )we observe that the forces that perform work are the forces parallel to displacement. that is α = 0.

For this case they are the horizontal force (F) and the frictional force (Ff)

F=31.3 N : Horizontal force

Ff=μk*N , μk:coefficient of kinetic friction , N : Normal force

N=W

W: block weight W= m*g, m: block mass , g: acceleration due to gravity

W= 4.30 kg* 9.8 m/s² = 42.14 N

Ff= 0.599*42.14 N= 25.24 N : Frictional force

Calculating of the Net work (Wn)

$W_{n} = W_{F} +W_{Ff}$