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Ivan
3 years ago
5

In the lab, you performed a trial where the fan was turned

Physics
2 answers:
beks73 [17]3 years ago
7 0

Answer:

Simple Response The first part of the graph is a straight diagonal line and the second part of the graph is a straight horizontal line. The slope of the first part of the graph is positive, indicating that the cart increased in speed, and was therefore accelerating for the first four seconds. After the fourth second, the slope of the line is zero, indicating that the cart moved at a constant speed and did not accelerate any more because the fan was turned off.

Explanation:

Ksju [112]3 years ago
6 0

Answer:

Sample Response: The first part of the graph is a straight diagonal line and the second part of the graph is a straight horizontal line. The slope of the first part of the graph is positive, indicating that the cart increased in speed, and was therefore accelerating for the first four seconds. After the fourth second, the slope of the line is zero, indicating that the cart moved at a constant speed and did not accelerate any more because the fan was turned off.

Explanation:

on edge :)

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What are two advantages of using renewable resources?
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One advantage is that whatever resource it is, it will never run out and you wont have to worry about not having it. A second is that there is going to be enough for everyone to use however much they want without there having to be a limit on how much you use.
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3 years ago
Constant Acceleration Kinematics: Car A is traveling at 22.0 m/s and car B at 29.0 m/s. Car A is 300 m behind car B when the dri
Ainat [17]

Answer:

The taken is  t_A  = 19.0 \ s

Explanation:

Frm the question we are told that

  The speed of car A is  v_A  =  22 \ m/s

   The speed of car B is  v_B  = 29.0 \ m/s

     The distance of car B  from A is  d = 300 \ m

     The acceleration of car A is  a_A  = 2.40 \ m/s^2

For A to overtake B

    The distance traveled by car B  =  The distance traveled by car A - 300m

Now the this distance traveled by car B before it is overtaken by A is  

          d = v_B * t_A

Where t_B is the time taken by car B

Now this can also be represented as using equation of motion as

      d = v_A t_A  + \frac{1}{2}a_A t_A^2 - 300

Now substituting values

       d = 22t_A  + \frac{1}{2} (2.40)^2 t_A^2 - 300

Equating the both d

       v_B * t_A = 22t_A  + \frac{1}{2} (2.40)^2 t_A^2 - 300

substituting values

   29 * t_A = 22t_A  + \frac{1}{2} (2.40)^2 t_A^2 - 300

   7 t_A = \frac{1}{2} (2.40)^2 t_A^2 - 300

  7 t_A =1.2 t_A^2 - 300

   1.2 t_A^2 - 7 t_A - 300  = 0

Solving this using quadratic formula we have that

     t_A  = 19.0 \ s

7 0
3 years ago
You and a friend work in buildings five equal-length blocks apart, and you plan to meet for lunch. Your friend strolls leisurely
Pavel [41]

Answer:

Your friend is 2.143 blocks from the restaurant.

You are 2.857 blocks from the restaurant.

Explanation:

Let t be the time both you and your friend take to walk to the restaurant.

The distance (m) from your building to the restaurant is your walking time t times your speed v1

s_1 = v_1t = 1.6t

Similarly the distance (m) from your friend building to the restaurant:

s_2 = v_2t = 1.2t

Let b be the length (in m) of a block, the total distance of 5 blocks is 5b

s_1 + s_2 = 5b

1.6t + 1.2t = 5b

2.8t = 5b

t = 5b/2.8 = 25b/14

s_2 = 1.2t = 1.2(25b/14) = 2.143b

So your friend are 2.143b meters from the restaurant, since each block is b meters long, 2.143b meters would equals to 2.143b/b = 2.143 blocks. And you are 5 - 2.143 = 2.857 blocks from the restaurant.

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CaHeK987 [17]

Answer:

speed is the gradient of the graph

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