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Alborosie
3 years ago
6

someone help me please

Physics
1 answer:
fomenos3 years ago
5 0

Answer:

2g

Explanation:

Mass is force/area

16 N is the force because N is newtons and that measures force

8m/s^2 is the acceleration

They gave it to you right in front of you

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A block of metal which is 20 cm on a side has a mass of 20 kg. What is it’s density? Please answer in MkS (SI) units.
Ratling [72]

The density of an object is given by:

D = M/V

D = density, M = mass, V = volume

The volume of the cube V is given by:

V = s³

where s = side length

Make a substitution:

D = M/s³

Given values:

M = 20kg, s = 20×10⁻²m

Plug in and solve for D:

D = 20/(20×10⁻²)³

D = 2500kg/m³

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4 years ago
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3 years ago
A car traveling in a constant speed of 55km/h on a circular track what is the acceleration explain​
a_sh-v [17]

Answer:

See the explanation below

Explanation:

We must solve this problem by defining that when we have a constant velocity, the acceleration is equal to zero. That is, when there is no speed change, there is no acceleration. We can understand it very easily by means of the following equation of kinematics.

v_{f}=v_{o}+a*t

where:

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Vo = initial velocity = 55 [km/h]

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3 years ago
What makes soil a nonrenewable resource when new soil is constantly being made
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3 years ago
In an experiment, it took 300s to increase the temperature of 0.1kg of the liquid from 25°C to 50°C. During this period, the ene
klemol [59]

Answer:

<h2>4000 \textbf{J}\text{ }\textbf{kg}^{\textbf{-1}}\text{ }\textbf{K}^{\textbf{-1}} </h2>

Explanation:

        The temperature of 0.1 kg of liquid rises from 25°C to 50°C in 300 sec. Energy of 13,600 J was supplied during this time. Appartus was losing energy at the rate of 12 J/sec.

       Let us assume the Specific heat capacity as s.

      As there is no state change from liquid to gas, only Specific heat capacity is involved. Also, work done is approximately zero because volume does not change much. So,

       Energy gained = Energy required to rise the temperature

       Energy gained by liquid = 13600\text{ }J-(300\text{ }sec)\times(12\text{ }\frac{J}{sec})=10000\text{ }J

       10000\text{ }J=m\times s\times\Delta T=(0.1\text{ }kg)\times(s)\times(323K-298K)=2.5s\text{ }kgK\\s=4000\text{ }\frac{J}{kg.K}

∴ Specific heat capacity of liquid = 4000 \frac{J}{kg.K}

4 0
3 years ago
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