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DaniilM [7]
2 years ago
8

Why do we need to take care of the surface water/groundwater?

Physics
2 answers:
ANEK [815]2 years ago
3 0

Answer:

sghskm

rohit zimheisbwhx

kati45 [8]2 years ago
3 0
Groundwater supply the water needs of people when there is little water on the surface. This is important; we need better management.
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A 8.00g sample of substance (substance, molar mass = 152.0 g/mol) was combusted in a bomb calorimeter with a heat capacity of 6.
aleksandrvk [35]

Answer:

ΔH°comb=-5899.5 kJ/mol

Explanation:

First, consider the energy balance:

m_{c} *Cp*(T_{2}-T_{1})=-n_{s} *H_{c} Where m_{c} is the calorimeter mass and n_{s} is the number of moles of the samples; H_{c} is the combustion enthalpy. The energy balance says that the energy that the reaction release is employed in rise the temperature of the calorimeter, which is designed to be adiabatic, so it is suppose that the total energy is employed rising the calorimeter temperature.

The product m_{c} *Cp is the heat capacity, so the balance equation is:

6.21\frac{kJ}{K}*(75-25)=-8.00g*\frac{mol}{152.0g}*H_{c}

So, the enthalpy of combustion can be calculated:

H_{c}=-5899.5\frac{kJ}{mol}

I will be happy to solve any doubt you have.

4 0
3 years ago
As an admirer of Thomas Young, you perform a double-slit experiment in his honor. You set your slits 1.09 mm apart and position
kow [346]

Answer:

pretty dark

Explanation:

i dont know

4 0
3 years ago
WILL GIVE THAT CROWN THINGY!
frutty [35]
Conservation of energy explains that energy can only be transferred between different forms of energy
3 0
3 years ago
Read 2 more answers
a street light is mounted at the top of a 15 foot pole. A man 6 ft tall walks away from the pole wit a speed of 7 ft/s along a s
Mariulka [41]

Answer:

16.3 ft/s

Explanation:

Let d=distance

and

x = length of shadow.

Therfore,

x=(d + x)

 = 6/15

So,

    15x = 6x + 6d

     9x = 6d.

x = (2/3)d.

As we know that:

dx=dt

   = (2/3) (d/dt) 

Also,

Given:

d(d)=dt

     = 7 ft/s

Thus,

d(d + x)=dt

           = (7/3)d (d/dt)

Substitute, d= 7  

d(d + x) = 49/3 ft/s.

Hence,

d(d + x) = 16.3 ft/s.

4 0
3 years ago
A spring gun is made by compressing a spring in a tube and then latching the spring at the compressed position. A 4.97-g pellet
dimaraw [331]

Answer:

v  = 2.8898 \frac{m}{s}

Explanation:

This is a problem easily solve using energy conservation. As there are no non-conservative forces, we know that the energy is conserved.

When the spring is compressed downward, the spring has elastic potential energy. When the spring is relaxed, there is no elastic potential energy, but the pellet will have gained gravitational potential energy and kinetic energy. Lets see what are the terms for each of this.

<h3>Elastic potential energy</h3>

We know that a spring following Hooke's Law has a elastic potential energy:

E_{ep} = \frac{1}{2} k (\Delta x)^2

where \Delta x is the displacement from the relaxed length and k is the spring's constant.

To obtain the spring's constant, we know that Hooke's law states that the force made by the spring is :

\vec{F} = - k \Delta \vec{x}

as we need 9.12 N to compress 4.60 cm, this means:

k = \frac{9.12 \ N}{4.6 \ 10^{-2} \ m}

k = 198.26 \ \frac{ N}{m}

So, the elastic energy of the compressed spring is:

E_{ep} = \frac{1}{2} 198.26 \ \frac{ N}{m} (4.6 \ 10^{-2} \ m)^2

E_{ep} = 0.209759 \ Joules

And when the spring is relaxed, the elastic potential energy will be zero.

<h3>Gravitational potential energy</h3>

To see how much gravitational potential energy will the pellet win, we can use

\Delta E_{gp} = m g \Delta h

where m is the mass of the pellet, g is the acceleration due to gravity and \Delta h is the difference in height.

Taking all this together, the gravitational potential energy when the spring is relaxed will be:

\Delta E_{gp} = 4.97 \ 10^{-3} kg \ 9.8 \frac{m}{s^2} 4.6 \ 10^{-2} m

\Delta E_{gp} = 0.00224 \ Joules

<h3>Kinetic Energy</h3>

We know that the kinetic energy for a mass m moving at speed v is:

E_k = \frac{1}{2} m v^2

so, for the pellet will be

E_k = \frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2

<h3>All together</h3>

By conservation of energy, we know:

E_{ep} = \Delta E_{gp} + E_k

0.209759 \ Joules = 0.00224 \ Joules + \frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2

So

\frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2  = 0.209759 \ Joules - 0.00224 \ Joules

\frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2  = 0.207519 \ Joules

v  = \sqrt{ \frac{ 0.207519 \ Joules}{ \frac{1}{2} \ 4.97 \ 10^{-3} kg } }

v  = 2.8898 \frac{m}{s}

7 0
3 years ago
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