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azamat
3 years ago
13

Can why hunting and megodinyong the stone Age​

Physics
2 answers:
Sladkaya [172]3 years ago
8 0
Woahhh can you rephrase that..
defon3 years ago
6 0
What are u trying to say
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Consider a river flowing toward a lake at an average speed of 3 m/s at a rate of 550 m3/s at a location 58 m above the lake surf
Vladimir [108]

Answer:

1. 0.574 kJ/kg

2. 315.7 MW

Explanation:

1. The mechanical energy per unit mass of the river is given by:

E_{m} = E_{k} + E_{p}

E_{m} = \frac{1}{2}v^{2} + gh

Where:

Ek is the kinetic energy

Ep is the potential energy

v is the speed of the river = 3 m/s

g is the gravity = 9.81 m/s²

h is the height = 58 m

E_{m} = \frac{1}{2}(3 m/s)^{2} + 9.81 m/s^{2}*58 m = 0.574 kJ/Kg

Hence, the total mechanical energy of the river is 0.574 kJ/kg.

2. The power generation potential on the river is:

P = m(t)E_{m} = \rho*V(t)*E_{m} = 1000 kg/m^{3}*550 m^{3}/s*0.574 kJ/kg = 315.7 MW

Therefore, the power generation potential of the entire river is 315.7 MW.

I hope it helps you!

4 0
3 years ago
Two point charges are arranged in a line with point P, which is on the right. The
MAVERICK [17]

Answer:

chicken

Explanation:

you like chicken and want to eat it

6 0
2 years ago
An artillery shell is fired with an initial velocity of 300 m/s at 52.0° above the horizontal. To clear an avalanche, it explode
sasho [114]

The x- and y-coordinates are 9142.57 m and -304.425 m

<u>Explanation:</u>

As the motion of the shell is in a plane (two dimensional space) and the acceleration is that due to gravity which is vertically downward, we resolve initial velocity of the shell v_{0} in horizontal and vertical directions. If the initial velocity of the shell is making angle with the horizontal, the horizontal component of initial velocity will be

                v_{x}=v_{0} \times \cos \theta

As the acceleration of the shell is vertical having no horizontal component, the shell may be considered to move horizontally with constant velocity of v_{x} and hence the horizontal distance covered (or the x coordinate of the shell with point of projection as origin) is given by

           v_{x}=v_{o} \times \cos \theta=300 \times \cos \left(52^{\circ}\right)=184.69 \mathrm{m} / \mathrm{s}

           v_{y}=v_{o} \times \sin \theta==300 \times \sin \left(52^{\circ}\right)=236.4 \mathrm{m} / \mathrm{s}

For motion with constant acceleration, we know

            s=s_{0}+v_{0} t+\left(\frac{(1)}{2}\right) a t^{2}

Along the horizontal, x-axis, we might write this as

            x=x_{0}+v_{x 0} t+\left(\frac{1}{2}\right) a_{x} t^{2}

Measuring distances relative to the firing point means

               x_{0}=0

we know that,

              a_{x}=0

or,

             v_{x}=v_{x 0}=\text { constant }

By applying the values, we get,

           x=0+(184.69 \times 49.5)+\left(\left(\frac{1}{2}\right) \times 0 \times(49.5)^{2}\right)=9142.57 \mathrm{m}

The acceleration of gravity is vertically downward and is g=-9.8 \mathrm{m} / \mathrm{s}^{2} , hence the vertical distance covered (or y coordinate of the shell) is given by the second equation of motion

           y=y_{0}+v_{y 0} t+\left(\frac{1}{2}\right) a_{y} t^{2}

we know, y_{0}=0 and a_{y}=-9.8 \mathrm{m} / \mathrm{s}^{2}, so,

          y=0+(236.4 \times 49.5)+\left(\left(\frac{1}{2}\right) \times(-9.8) \times(49.5)^{2}\right)

                 y = 11701.8 - 4.9(2450.25)= 11701.8 - 12006.225 = - 304.425 m

7 0
3 years ago
What is the relation between Physical states of solution and solvent
posledela

Explanation:

The solute does not have to be in the same physical state as the solvent, but the physical state of the solvent usually determines the state of the solution. As long as the solute and solvent combine to give a homogeneous solution, the solute is said to be soluble in the solvent.

4 0
3 years ago
A parallel-plate capacitor with circular plates of radius R is being charged by a battery, which provides a constant current. At
NikAS [45]

To solve this problem it is necessary to apply the concepts related to the magnetic field.

According to the information, the magnetic field INSIDE the plates is,

B=\frac{1}{2} \mu \epsilon_0 r

Where,

\mu =Permeability constant

\epsilon_0 =Electromotive force

r = Radius

From this deduction we can verify that the distance is proportional to the field

B \propto r

Then the distance relationship would be given by

\frac{r}{R} = \frac{B}{B_{max}}

r =\frac{B}{B_{max}} R

r = \frac{0.5B_{max}}{B_{max}}R

r = 0.5R

On the outside, however, it is defined by

B = \frac{\mu_0 i_d}{2\pi r}

Here the magnetic field is inversely proportional to the distance, that is

B \not\propto r

Then,

\frac{r}{R} = \frac{B_{max}{B}}

r = \frac{B_{max}{B}}R

r = \frac{B_{max}{0.5B_{max}}}R

r = 2R

7 0
3 years ago
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