Which describes increasing the efficiency of energy resource

What do we mean by the observable universe?

mote1985 [20]

The observable universe<span> is a spherical region of the </span>Universe, <span>comprising all matter that can be observed from Earth at the present time, because light and other signals from these objects have had time to reach Earth since the beginning of the cosmological expansion.

</span>

4 0

3 years ago

A hungry rat is placed in a maze. It walks the following path to find a piece of cheese. 4.0m N, 7.5 m E, 6.8 m S, 3.7 m E, 3.6

storchak [24]

Answer:

Explanation:

We shall take the help of vector form of displacement . Taking east as i and north as j

4.0m N = 4 j

7.5 m E = 7.5 i

6.8 m S = - 6.8 j

3.7 m E, = 3.7 i

3.6 m S = - 3.6 j

5.3 m W = - 5.3 i

3.7 m N, = 3.7 j

5.6 m W = - 5.6 i

4.4 m S = - 4.4 j

4.9 m W = - 4.9 i

Total displacement = 4j +7.5 i -6.8j+3.7i-3.6j-5.3i+3.7j-5.6i-4.4j-4.9i

= -4.6 i -7.1 j

magnitude of displacement = $\sqrt{(4.6^2+7.1^2)}$

= 8.46 m

Direction

Tanθ = 7.1/ 4.6

θ = 57⁰ south of west .

distance walked = 4+7.5 +6.8+3.7+3.6+5.3+3.7+5.6+4.4+4.9

= 49.5 m

6 0

3 years ago

Read 2 more answers

WILL MARK BRAINLIST

ryzh [129]

Answer:

Definition of envelop

transitive verb

1: to enclose or enfold completely with or as if with a covering

2: to mount an attack on (an enemy's flank)

3 0

2 years ago

1. Which of the following would be a testable hypothesis?

Naddik [55]

B dropping a ball
C tentative and testable

7 0

3 years ago

Water (density = 1x10^3 kg/m^3) flows at 15.5 m/s through a pipe with radius 0.040 m. The pipe goes up to the second floor of th

RUDIKE [14]

Answer:

The speed of the water flow in the pipe on the second floor is approximately 13.1 meters per second.

Explanation:

By assuming that fluid is incompressible and there are no heat and work interaction through the line of current corresponding to the pipe, we can calculate the speed of the water floor in the pipe on the second floor by Bernoulli's Principle, whose model is:

$P_{1} + \frac{\rho\cdot v_{1}^{2}}{2}+\rho\cdot g\cdot z_{1} = P_{2} + \frac{\rho\cdot v_{2}^{2}}{2}+\rho\cdot g\cdot z_{2}$ (1)

Where:

$P_{1}$ , $P_{2}$ - Pressures of the water on the first and second floors, measured in pascals.

$\rho$ - Density of water, measured in kilograms per cubic meter.

$v_{1}$ , $v_{2}$ - Speed of the water on the first and second floors, measured in meters per second.

$z_{1}$ , $z_{2}$ - Heights of the water on the first and second floors, measured in meters.

Now we clear the final speed of the water flow:

$\frac{\rho\cdot v_{2}^{2}}{2} = P_{1}-P_{2}+\rho \cdot \left[\frac{v_{1}^{2}}{2}+g\cdot (z_{1}-z_{2}) \right]$

$\rho\cdot v_{2}^{2} = 2\cdot (P_{1}-P_{2})+\rho\cdot [v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2})]$

$v_{2}^{2}= \frac{2\cdot (P_{1}-P_{2})}{\rho}+v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2})$

$v_{2} = \sqrt{\frac{2\cdot (P_{1}-P_{2})}{\rho}+v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2}) }$ (2)

If we know that $P_{1}-P_{2} = 0\,Pa$ , $\rho=1000\,\frac{kg}{m^{3}}$ , $v_{1} = 15.5\,\frac{m}{s}$ , $g = 9.807\,\frac{m}{s^{2}}$ and $z_{1}-z_{2} = -3.5\,m$ , then the speed of the water flow in the pipe on the second floor is:

$v_{2}=\sqrt{\left(15.5\,\frac{m}{s} \right)^{2}+2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (-3.5\,m)}$

$v_{2} \approx 13.100\,\frac{m}{s}$

The speed of the water flow in the pipe on the second floor is approximately 13.1 meters per second.

4 0

3 years ago