Question

3 which method could be used to classify objects by the property of density?
a mix blue and red food coloring with water
b dissolve sugar and salt in water
c place solid objects in water
d none of the above

4 according to the picture, what is the mass of the object? 5.4(a)
image of a balance scale
write your answer in the space.
box grams

materials conduct energy. the student places a separate containers of hot water. then the student takes to make sure both are the same temperature. what is (a)

Explanation:

Question 3

To classify objects by density, we observe if they sink or float in water (since density relative to water determines this).

• Option A: Mixing food coloring with water is about color mixing, not density classification.
• Option B: Dissolving sugar/salt in water is about solubility, not classifying objects by their density.
• Option C: Placing solid objects in water: if an object is denser than water, it sinks; if less dense, it floats. This uses density to classify (sink/float behavior).
• Option D is incorrect as C is valid.

Step1: Identify the balance parts

A triple - beam balance has a pan for the object and three beams with weights. The total mass is the sum of the masses on each beam. From the image, we assume the following (typical triple - beam balance setup: one beam with 100g, one with 50g, and the sliding weight on the third beam). Wait, looking at the scale, the top beam (usually the 100s or 500s, but in a typical school balance, let's re - examine. Wait, the lower scale (the sliding weight) and the fixed weights. Wait, maybe the fixed weights are 100g, 50g, and the sliding weight is at 3g? Wait, no, let's think again. Wait, the triple - beam balance: the first beam (usually the back one) has larger weights, the middle one medium, and the front one is the sliding weight (with marks for grams). Wait, from the image, let's assume that the fixed weights are 100g, 50g, and the sliding weight is at 3g? Wait, no, maybe the fixed weights are 100g and 50g, and the sliding weight on the front beam is at 3g. Wait, total mass = mass from fixed weights + mass from sliding weight. Wait, maybe the fixed weights are 100g, 50g, and the sliding weight is at 3g? Wait, no, let's check the scale. The front beam (the one with the sliding weight) has marks. Let's say the fixed weights are 100g, 50g, and the sliding weight is at 3g. Then total mass = 100 + 50+ 3=153? Wait, no, maybe the fixed weights are 100g and 50g, and the sliding weight is at 3g. Wait, or maybe the top beam (the one with the 100, 200, etc.) has a 100g weight, the middle beam (50, 100) has a 50g weight, and the front beam (the sliding one) has the weight at 3g. So total mass = 100 + 50+ 3 = 153 grams? Wait, no, maybe I misread. Wait, the standard triple - beam balance: the back beam (usually 0 - 600g in 100g increments), middle beam (0 - 100g in 50g increments), and front beam (0 - 10g in 0.1g increments, but in some basic ones, 0 - 10g in 1g increments). Wait, from the image, the front beam (the sliding weight) is at 3g, the middle beam has a 50g weight, and the back beam has a 100g weight. So total mass = 100+50 + 3=153 grams.

Step1: Sum the masses

The fixed weights (let's assume) are 100g and 50g, and the sliding weight is 3g. So total mass $M=100 + 50+3=153$ grams.

Answer:

C. Place solid objects in water

Question 4