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EleoNora [17]
3 years ago
6

Nikolas, the fire extinguisher, and the skateboard have a combined mass of 50 kg. What force would the fire extinguisher have to

produce to propel nikolas if he wanted to accelerate at a rate of 1.2 m/s squared
Chemistry
2 answers:
maks197457 [2]3 years ago
6 0

Answer:

Force=60 N

Hope this helps ya'll!                                  

murzikaleks [220]3 years ago
3 0

Answer:

Force used by fire extinguisher = 60 N

Explanation:

Given:

Mass of skateboard with fire extinguisher = 50 kg

Acceleration of fire extinguisher = 1.2 m/s²

Find:

Force used by fire extinguisher = ?

Computation:

⇒ Force = Mass × Acceleration

⇒ Force used by fire extinguisher = Mass of skateboard with fire extinguisher × Acceleration of fire extinguisher

⇒ Force used by fire extinguisher = 50 kg × 1.2 m/s²

⇒ Force used by fire extinguisher = 60 N

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calculate the enthalpy change for converting 1.00 mol of liquid water at 100 C to water vapor at 145 degree celsius for water Δ
Svetllana [295]

Answer:

q1 = mCpΔT

    = 18.016g × 1.84J/g.K × (418.15-373.15)

     = 1491.72 J

q2 = n×ΔH vap

     =  1mol ×44.0kJ/mol

     = 44KJ

∴ qtotal = q1+ q2

             = 1.498kJ + 44.0kJ

             = 45.498KJ

Explanation: The heat flow can be separated into steps.all that is being observed at a constant pressure,the heat flow is equal to the enthalpy.

8 0
4 years ago
A piece of wood has a labeled length value of 63.2 cm. You measure its length three times and record the following data: 63.1 cm
Ulleksa [173]

Percent error (%)= \frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100

Accepted value is true value.

Measured values is calculated value.

In the question given Accepted value (true value) = 63.2 cm

Given Measured(calculated values) = 63.1 cm , 63.0 cm , 63.7 cm

1) Percent error (%) for first measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.1 cm

Percent error (%)= \frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100

Percent error = \frac{\left | 63.2 - 63.1 \right |}{63.2}\times 100

Percent error = \frac{0.1}{63.2}\times 100

Percent error = 0.00158\times 100

Percent error = 0.158 %

2) Percent error (%) for second measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.0 cm

Percent error (%)= \frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100

Percent error = \frac{\left | 63.2 - 63.0 \right |}{63.2}\times 100

Percent error = \frac{0.2}{63.2}\times 100

Percent error = 0.00316\times 100

Percent error = 0.316 %

3) Percent error (%) for third measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.7 cm

Percent error (%)= \frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100

Percent error = \frac{\left | 63.2 - 63.7 \right |}{63.2}\times 100

Percent error = \frac{\left | -0.5 \right |}{63.2}\times 100

Percent error = \frac{(0.5)}{63.2}\times 100

Percent error = 0.00791\times 100

Percent error = 0.791 %

Percent error for each measurement is :

63.1 cm = 0.158%

63.0 cm = 0.316%

63.7 cm = 0.791%




7 0
4 years ago
In which layer(s) would a solid cube with 6 cm sides and mass of 270 g float? Explain.
DIA [1.3K]
The buoyancy of an object is dictated by its density. So let us calculate for density, where:density = mass / volume
Calculate the volume first of a solid cube:volume = (6 cm)^3 = 216 cm^3 = 216 mL
Therefore density is:density = 270 g / 216 mLdensity = 1.25 g / mL
Therefore this object will float in the layer in which the density is more than 1.25 g / mL.
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alex41 [277]
I believe the answer is a because 2 times 12 is 24. Hope this helped
8 0
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Harman [31]

Explanation:

There are several ways to define acids and bases, but pH and pOH refer to hydrogen ion concentration and hydroxide ion concentration, respectively. The "p" in pH and pOH stands for "negative logarithm of" and is used to make it easier to work with extremely large or small values. pH and pOH are only meaningful when applied to aqueous (water-based) solutions. When water dissociates it yields a hydrogen ion and a hydroxide.

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