Is boiling water a non living or living organism

What is one measurement needed to calculate the speed of an object? Direction Mass Time Velocity

Vlad [161]

The answer is Time.

Formula: Distance/Time

3 0

3 years ago

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Air has a density of 1.3 kg/m³. Calculate the mass of 36 m³ of air in kilograms. Give your answer to 1 decimal place.

vredina [299]

Answer:

46.8 kg

Explanation:

Mass = (density)(volume)

= (1.3)(36)

Mass = 46.8 kg

3 0

3 years ago

Find the product of 30−−√ and 610−−√. Express it in standard form (i.

Helen [10]

The product of √30 and √610 is 10√183.

√30 = √(2×3×5)
and √610 = √(2×5×61

Since 61 can't be factorised further.
Therefore, the value of √30×√610 is
= √(2×3×5×2×5×61)
= 2×5×√(3×61)
=10√183

4 0

3 years ago

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To determine the specific heat of an object, a student heats it to 100 degrees C in boiling water. She then places the 110 g obj

kolezko [41]

Answer:

Cs=912.6J/kg/K

Explanation:

Sum of Heat gained=sum heat lośs

Definition of terms

Ms=mass of substance 0.11kg

Cs=specific heat capacity of the substance?

Ts= temperature of the substance

Tm= temperature of mixture 28.8°C

Tal= temperature of aluminium19.6°C

Tw=temperature of water 19.6°C

Mal=mass of aluminium 0.172kg

Mw=mass of water 0.148kg

Cw= specific heat capa of water 4200j/kgk

Cal=specific heat capacity of aluminium 903j/kgk

Having gotten all the parameters needed

Ms*Cs*(Ts-Tm)=Mw*Cw*(Tm-Tw) +Mal*Cal*(Tm-Tal)

0.11*Cs*(100-28.8)=0.148*4200*(28.8-19.6) +0.172*903*(28.8-19.6)

cs==((4467267)/(4895))

Cs=912.61Jkg–1K–1

8 0

3 years ago

Two small, identical metal balls with charges 7.0 µC and 14.0 µC are held in place 1.9 m apart. In an experiment, they are conne

Nostrana [21]

The new charge on 7.0 µC ball is 10.5 µC and the new charge on 14.0 µC ball is 10.5 µC.

The change in the electrostatic force after the experiment is 0.031 N.

The given parameters:

Charge on first metal ball, q1 = 7.0 µC
Charge on second metal ball, q2 = 14 µC
Distance between the charges, r = 1.9 m

<h3>Charge on each ball after experiment</h3>

After the experiment the charges will be at equilibrium, and the charge on each metal ball will be equal.

$Q_t = q_1 + q_2\\\\Q_t = 7 \mu C + 14 \mu C\\\\Q_t = 21 \ \mu C$

$Q_1 = Q_2 = \frac{Q_t}{2} = \frac{21 \mu C}{2} = 10.5 \ \mu C$

The new charge on 7.0 µC ball = 10.5 µC

The new charge on 14.0 µC ball = 10.5 µC

<h3>Change in electrostatic force</h3>

$F_1 = \frac{kQ_1Q_2}{r^2} \\\\F_1 = \frac{9\times 10^9}{1.9^2} (7 \times 10^{-6} \times 14 \times 10^{-6}) \ \\\\ F_1 = 0.244 \ N\\\\for \ new \ charges;\\\\F_2 = \frac{kQ_1Q_2}{r^2} \\\\F_2 = \frac{9\times 10^9}{1.9^2} (10.5 \times 10^{-6} \times 10.5 \times 10^{-6}) \ \\\\ F_2 = 0.275 \ N\\\\\Delta F = F_2 - F_2\\\\\Delta F = 0.275 \ N \ - \ 0.244 \ N\\\\\Delta F = 0.031 \ N$

Lear more about electrostatic force here: brainly.com/question/17692887

8 0

2 years ago