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crimeas [40]
3 years ago
6

Meeting URL: https://meet.google.com/goq-wpub-frp

Physics
2 answers:
kolbaska11 [484]3 years ago
4 0

Answer:

<h2>That's great</h2>

Explanation:

<h2>let me join</h2>
WITCHER [35]3 years ago
4 0
It’s not letting me





? What do I do
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A swimmer is capable of swimming 0.42 m/s in still water. part a if she aims her body directly across a 66-m-wide river whose cu
satela [25.4K]
<span>If the swimmer is swimming perpendicular to the current, it will take her 66m / 0.42 m/s = 157.14 seconds to cross the river. At the same time, the current will be taking her downstream at a rate of 0.32 m/s. So, when she reaches the opposite bank, her total downstream distance traveled will have been 0.32*157.14 = 50.28 meters.</span>
3 0
3 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
A 3.0kg weight W is initially at rest on incline AB, which is raised 40° above the horizontal. The effective coefficient is
lakkis [162]

(a) The acceleration of the system is determined as 1.58 m/s².

(b) The relative weight of P is pounds is determined as 0.14 lb.

<h3>Acceleration of the system</h3>

The acceleration of the system is calculated as follows;

W - T = m₂a --- (1)

T = m₁a ----(2)

μmgsinθ - m₁a = m₂a

(0.3 x 3 x 9.8 x sin40) - (0.4 + 0.2)a = 3a

5.67 - 0.6a = 3a

5.67 = 3.6a

a = 5.67/3.6

a = 1.58 m/s²

<h3> Relative Weight of P</h3>

W = ma

W = 0.4 x 1.58

W = 0.632 N = 0.14 lb

Learn more about weight here: brainly.com/question/2337612

#SPJ1

3 0
2 years ago
On what criteria is the Fujita scale for Tornadoes based?
svp [43]
B. Tornado destruction
It is based on the amount of damage
7 0
3 years ago
Eating eggs for breakfast increases grades. What is the constant?
Feliz [49]

Answer and Explanation:

The answer would be: <u>Breakfast</u>

<u></u>

Breakfast would be the constant of this scenario because this doesn't change. The time of day (for eating eggs) will always be in the morning, so that wouldn't change.

Eating eggs would be the independent variable because you can change this value. This could be a different food or an amount of the food that is being eaten.

The grades would be the dependent variable because the score <em>depends</em> on the eating of eggs.

<em><u>#teamtrees #PAW (Plant And Water)</u></em>

<em><u></u></em>

<em><u>I hope this helps!</u></em>

6 0
3 years ago
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