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UkoKoshka [18]
3 years ago
12

athlete running on a circular track of radius 5 m completes one round in 12 s. what its average speed and average displacement?

Physics
2 answers:
elixir [45]3 years ago
8 0

Average speed = total distance/time

That's 2.62 m/s .

You don't want average displacement. You want average velocity.

That's (displacement)/(time).

That's zero, because displacement = 0.

dangina [55]3 years ago
6 0
Here is your answer

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D, Metamorphism
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3 years ago
in a one dimensional collision, a 4kg object and 6 kg onject have initial velocity. calculate the magnitude of impulse
Greeley [361]

in a one dimensional collision, a 4kg object with 5ms^1 and 6 kg object with 2ms^1 have initial velocity, the magnitude of impulse is 12 , 18

given,

mass 1 = 4kg

mass 2 = 6kg

velocity 1 = 5ms^1

velocity 2 = 2ms^1

impulse 1 = 4*(5-2)

= 12

Impulse 2 = 6*(5-2)

= 18

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8 0
1 year ago
Calculate the number of moles of water molecules in 12 dm' of water<br>vapour at STP.<br><br>​
Vinvika [58]

Answer:

22.4 \:  {dm}^{3}  \: are \: occupied \: by \: 1 \: mole \\ 12 \:  {dm}^{3}  \: will \: be \: occupied \: by \: (  \frac{12}{22.4} ) \: moles \\  = 0.536 \: moles

4 0
2 years ago
To push a 26.0 kg crate up a frictionless incline, angled at 25.0° to the horizontal, a worker exerts a force of 209 N parallel
alukav5142 [94]

Answer:

(a) W = +397.1 J

(b) W = -204.6 J

(c) W = 0

(d) W= + 192.5 J

Explanation:

Work (W) is defined as the product of force (F) by the distance (d)the body travels due to this force. :

W= F*d Formula ( 1)

The forces that perform work on an object must be parallel to its displacement.

The forces perpendicular to the displacement of an object do not perform work on it.

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

The work is negative (W-) if the force has the opposite direction of the movement of the object.

Problem development

(a) Work performed by the worker's applied force on the box .

W= 209 N * 1.9 m = +397.1 J

(b) Work performed by the gravitational force on the crate

We calculate the weight component parallel to the displacement of the box:

We define the x-axis in the direction of the inclined plane ,25.0° to the horizontal.

We define the y-axis and in the direction of the plane perpendicular to the inclined plane.

W= m*g=26*9.8= 254.8N : total box weight

Wx= W*sen25.0°= 254.8*sen25.0°= 107.68 N

W = -Wx *d =107.68 N *1.9 m= -204.6 J

(c) Work performed by normal force (N) exerted by the incline on the crate

The force N is perpendicular to the displacement, then:

W=0

(d) Total work done on the crate

W = 397.1 J -204.6 J

W = 192.5 J

4 0
3 years ago
Why does the candle light has no shadow when light fall on it?​
andrezito [222]
Shadows are the absence of light, they are created when an object blocks light. In other words, shadows are the product of light particles, known as photons. These particles “bounce off” of the object without reaching the other side. Therefore light by itself will not form a shadow.
6 0
2 years ago
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