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zhenek [66]
3 years ago
9

An example of an action and reaction is sitting on a chair. Your body pushes down on the chair because of the force of what?

Physics
1 answer:
vaieri [72.5K]3 years ago
8 0

Your body is pushing down on a chair because it is being attracted by gravity, the reason that your body is not moving down because there is a normal force acting on your body; together the net force of weight (m*g) and normal force is equal to zero

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The mass of the uniform cantilever is 1100 kg. Determine the force on the beam at A. Determine the force on the beam at B. Use C
Mashcka [7]

Answer:

Force A=-−2,697.75 N

Force B=13, 488.75 N

Explanation:

Taking moments at point A, the sum of clockwise and anticlockwise moments equal to zero.

25 mg-20Fb=0

25*1100g=20Fb

Fb=25*1100g/20=1375g

Taking g as 9.81 then Fb=1375*9.81=13,488.75 N

The sum of upward and downward forces are same hence Fa=1100g-1375g=-275g

-275*9.81=−2,697.75. Therefore, force A pulls downwards

Note that the centre of gravity is taken to be half the whole length hence half of 50 is 25 m because center of gravity is always at the middle

8 0
3 years ago
Why are most of the world’s deserts located between latitudes 10°n to 30°n and 10°s to 30°s.
RoseWind [281]

The bulk of the world's deserts are located at 30 degrees north latitude and 30 degrees south latitude, when the warm equatorial air begins to descend. The heavy, warm, descending air vaporises large amounts of water from the ground's surface. As a result, the environment is rather dry.

<h3>Why are the majority of the desert regions on Earth located between 20 and 30 degrees latitude?</h3>

The zones of falling air are those between 20 and 30 latitudes on the western borders of continents (high pressure and dry weather). As a result, the moisture continues to decrease as the air is compressed and warmed as it falls.

Where the scorching equatorial air starts to descend, the majority of the world's deserts are found between 30 degrees north and 30 degrees south latitude. Large volumes of water are vaporised off the surface of the ground by the thick, warming, falling air. As a result, the climate is extremely dry.

Learn more about latitude refer

brainly.com/question/1939015

#SPJ4

3 0
1 year ago
What is the best inference for the speed of the car after 3 seconds
SVEN [57.7K]

Answer:

B. 17m/s

Explanation:

This question contains a graph that illustrates the relationship between the speed of a car over time. The graph shows that one can make an inference of the amount of time it takes for the car to cover a particular speed and vice versa.

In this case, after 3 seconds, the speed of the car will be 17 m/s. This inference was got by tracing the position of 3s in the x-axis to the value on the y-axis. Doing this, the best inference for the speed of the car after 3 seconds is 17m/s.

8 0
3 years ago
I have multiple questions!
UNO [17]

I'm not sure about the rest but for question 2:

A theory is an attempt to come up with a big picture of all we know so far. It also drives future research as people do experiments to see if what the theory predicts actually happens. When experiments don’t support the theory, you have to change the theory and try again. That’s how science works. We come up with a “best guess” (theory), and then do research to test it’s accuracy. As we discover contradictions, we adjust the theory to take those into account, and then start testing the validity of the new theory.

3 0
3 years ago
Two vehicles approach a right angle intersection and then suddenly collide. After the collision, they become entangled. If their
geniusboy [140]

Answer:

a. 11 m/s at 76° with respect to the original direction of the lighter car.

Explanation:

In this exercise, since both cars make a right angle, let's assume that the lighter car only has a horizontal velocity component (vx) and that the heavier one only has a vertical velocity component (vy). The final velocities for both components for the system can be determined as:

m_{1} v_{x1}+m_{2}v_{x2}=(m_{1}+m_{2})v_{fx}\\m_{1} v_{y1}+m_{2}v_{y2} =(m_{1}+m_{2})v_{fy}

Assume that the lighter car has a 1kg mass and that the heavier car has a 4 kg mass.

1*13 + 4*0 = (1+4) v_{fx}\\v_{fx}=\frac{13}{5} =2.6\\1*0 + 4*13 = (1+4) v_{fy}\\v_{fy}=\frac{13*4}{5} =10.4\\

The magnitude of the final velocity of the wreck can be found as:

v_{f}^{2}= v_{fx}^{2}+ v_{fy}^{2}\\v_{f}=\sqrt[]{2.6^{2} + 10.4^{2}} \\v_{f}= 10.72

The final velocity has an intensity of roughly 11 m/s

As for the angle, it can be determined in respect to the lighter car (x axis) as follows:

\theta = cos^{-1}(\frac{v_{fx} }{v_{f}} )\\\theta = cos^{-1}(\frac{2.6}{10.7} )\\\theta = 76^{o}

Therefore, the wreck has a velocity with an intensity of 11 m/s at 76° with respect to the original direction of the lighter car.

5 0
3 years ago
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