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miss Akunina [59]
3 years ago
8

Janice has just measured the density of an object. Which value is possible? (Density: D = )

Physics
2 answers:
babunello [35]3 years ago
6 0
It is 6 g/cm3 because density cannot be negative, and it is not speed in which the unit would be m/s.
Naily [24]3 years ago
4 0

Answer:

density = 6 g/cm^3

Explanation:

as we know that density is the ratio of mass and volume of the object

density = \frac{mass}{volume}

here we know that mass as well as volume both are always positive so the ratio of both can not be negative.

so we know that mass is measured in gram and volume is measured in cubic cm

so the density is measured in

density = \frac{gram}{cm^3}

so the correct answer must be positive and in terms of above units

so we have

density = 6 g/cm^3

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6th grade science I mark as brainliest
Mila [183]

Answer:

8. organelle

Explanation:

9. Epithelial tissue

am i correct?

8 0
3 years ago
An open container holds ice of mass 0.555 kg at a temperature of -16.6 ∘C . The mass of the container can be ignored. Heat is su
s2008m [1.1K]

Answer: A. 23.59 minutes.

              B. 249.65 minutes

Explanation: This question involves the concept of Latent Heat and specific heat capacities of water in solid phase.

<em>Latent heat </em><em>of fusion </em>is the total amount of heat rejected from the unit mass of water at 0 degree Celsius to convert completely into ice of 0 degree Celsius (and the heat required for vice-versa process).

<em>Specific heat capacity</em> of a substance is the amount of heat required by the unit mass of a substance to raise its temperature by 1 kelvin.

Here, <u>given that</u>:

  • mass of ice, m= 0.555 kg
  • temperature of ice, T= -16.6°C
  • rate of heat transfer, q=820 J.min^{-1}
  • specific heat of ice, c_{i}= 2100 J.kg^{-1}.K^{-1}
  • latent heat of fusion of ice, L_{i}=334\times10^{3}J.kg^{-1}

<u>Asked:</u>

1. Time require for the ice to start melting.

2. Time required to raise the temperature above freezing point.

Sol.: 1.

<u>We have the formula:</u>

Q=mc\Delta T

Using above equation we find the total heat required to bring the ice from -16.6°C to 0°C.

Q= 0.555\times2100\times16.6

Q= 19347.3 J

Now, we require 19347.3 joules of heat to bring the ice to 0°C  and then on further addition of heat it starts melting.

∴The time required before the ice starts to melt is the time required to bring the ice to 0°C.

t=\frac{Q}{q}

=\frac{19347.3}{820}

= 23.59 minutes.

Sol.: 2.

Next we need to find the time it takes before the temperature rises above freezing from the time when heating begins.

<em>Now comes the concept of Latent  heat into the play, the temperature does not starts rising for the ice as soon as it reaches at 0°C it takes significant amount of time to raise the temperature because the heat energy is being used to convert the phase of the water molecules from solid to liquid.</em>

From the above solution we have concluded that 23.59 minutes is required for the given ice to come to 0°C, now we need some extra amount of energy to convert this ice to liquid water of 0°C.

<u>We have the equation:</u> latent heat, Q_{L}= mL_{i}

Q_{L}= 0.555\times334\times10^{3}= 185370 J

<u>Now  the time required for supply of 185370 J:</u>

t=\frac{Q_{L}}{q}

t=\frac{185370}{820}

t= 226.06 minutes

∴ The time it takes before the temperature rises above freezing from the time when heating begins= 226.06 + 23.59

= 249.65 minutes

8 0
3 years ago
Why is figure 5 an unhelpful visualization tool for this data set? <br><br> Please help!
Paraphin [41]

Explanation:

Because the temperature and the radiation are not correlated, they're not represented as functions of each other, they're represented as independent variables thus using graph 5 you cannot figure out how one affect another

8 0
2 years ago
A steel ball and a wooden ball of the same diameter are released together from the top of a tower. In the absence of air resista
ella [17]

Answer:

False

Explanation:

The steel ball and the wooden ball do not have the same force acting on them because their masses are different. But, they have the same acceleration which is the acceleration due to gravity g = 9.8 m/s².

Using the equation of motion under freefall, s = ut +1/2gt². Since u = 0,

s = 1/2gt² ⇒ t = √(2s/g)

Since. s = height is the same for both objects, they land at the same time neglecting air resistance.

8 0
3 years ago
The two-second rule applies to any speed __________ A. in all weather conditions. B. in ideal weather and road conditions. C. in
kumpel [21]

Answer:

The answer is A

Explanation:

A large risk of tailgating is the collision avoidance time being much less than the driver reaction time. Driving instructors advocate that drivers always use the "two-second rule" regardless of speed or the type of road. During adverse weather, downhill slopes, or hazardous conditions such as black ice, it is important to maintain an even greater distance.

5 0
3 years ago
Read 2 more answers
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