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Anna11 [10]
3 years ago
13

nvestigations were carried out in a science lab to explore the topic of chemical and physical changes. Investigation A Step 1. A

dd 5 tsp. salt to 100 ml warm water and stir until most or all of the salt is no longer visible. Step 2. Heat the salt solution on a burner until only a white solid remains. Investigation B Step 1. Mix 10 tsp. white sugar into 100 ml water and stir until most or all of the sugar is no longer visible. Step 2. Heat the sugar solution on a burner until the solution thickens and turns brown. In which step(s) did a chemical change most likely occur?
Physics
1 answer:
MAVERICK [17]3 years ago
7 0

B Step 1. Mix 10 tsp. white sugar into 100 ml water and stir until most or all of the sugar is no longer visible.

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Which of the following is not a vector quantity?
maksim [4K]

Answer:

A. Speed

Explanation:

A vector quantity is a quantity which has both magnitude and direction. Here in the given options, speed is a scalar quantity but not the vector quantity.

7 0
3 years ago
A gyroscope flywheel of radius 3.25 cm is accelerated from rest at 11.6 rad/s2 until its angular speed is 1820 rev/min. (a) What
kati45 [8]

Explanation:

The given data is as follows.

           radius (r) = 3.25 cm,    \alpha = 11.6 rad/s^{2}

Now, we will calculate the tangential acceleration as follows.

          a_{tangential} = \alpha \times r

Putting the given values into the above formula as follows.

         a_{tangential} = \alpha \times r

                      = 11.6 rad/s^{2} \times 3.25 cm

                      = 37.7 rad cm/s^{2}

Thus, we can conclude that the tangential acceleration of a point on the rim of the flywheel during this spin-up process is 37.7 rad cm/s^{2}.

6 0
3 years ago
A 40-kg uniform semicircular sign 1.6 m in diameter is supported by two wires as shown. What
garri49 [273]

Answer:

T1 = 131.4 [N]

T2 = 261 [N]

Explanation:

To solve this problem we must make a sketch of how will be the semicircle, for this reason we conducted an internet search, to find the scheme of the problem. This scheme is attached in the first image.

Then we make a free body diagram, with this free body diagram, we raise the forces that act on the body. Since it is a problem involving static equilibrium, the sum of forces in any direction and moments must be equal to zero.

By performing a sum of forces on the Y axis equal to zero we can find an equation that relates the forces of tension T1 & T2.

The second equation can be determined by summing moments equal to zero, around the point of application of the T1 force. In this way we find the T2 force.

The value of T2, is replaced in the first equation and we can find the value for T1.

Therefore

T1 = 131.4 [N]

T2 = 261 [N]

The free body diagram and the developed equations can be seen in the second attached image.

4 0
3 years ago
A. Draw four rays parallel to the optical axis of your mirror. Two above the optic axis and two below it.
Katen [24]

Answer:

   f = q

Explanation:

In the attachment we can see a diagram of the parallel rays.

The dotted line represents the normal to the mirror surface

These rays when reflected using the constructor equation

        \frac{1}{f} = \frac{1}{p} + \frac{1}{q}

where p and q are the distance to the object and the image respectively.

Since the rays are parallel P = inf

          1 / f = 1 / inf + 1 / q

          f = q

this means that all the rays focus on one focal point.

6 0
3 years ago
Two people person A and person B lift the same amount of weight but person A lifts it in a shorter amount of time who has more p
const2013 [10]
Person A has more muscular strength then person B.  B has muscular endurance.
3 0
3 years ago
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