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AlekseyPX
3 years ago
7

Bro i need help omg.

Physics
2 answers:
sergey [27]3 years ago
8 0

Answer:

25.

Explanation:

olga nikolaevna [1]3 years ago
4 0

5

because you're taking off the front number which is one

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Water is pumped steadily out of a flooded basement at a speed of 5.4 m/s through a uniform hose of radius 0.83 cm. The hose pass
Gala2k [10]

To solve this problem it is necessary to apply the concepts related to the flow as a function of the volume in a certain time, as well as the potential and kinetic energy that act on the pump and the fluid.

The work done would be defined as

\Delta W = \Delta PE + \Delta KE

Where,

PE = Potential Energy

KE = Kinetic Energy

\Delta W = (\Delta m)gh+\frac{1}{2}(\Delta m)v^2

Where,

m = Mass

g = Gravitational energy

h = Height

v = Velocity

Considering power as the change of energy as a function of time we will then have to

P = \frac{\Delta W}{\Delta t}

P = \frac{\Delta m}{\Delta t}(gh+\frac{1}{2}v^2)

The rate of mass flow is,

\frac{\Delta m}{\Delta t} = \rho_w Av

Where,

\rho_w = Density of water

A = Area of the hose \rightarrow A=\pi r^2

The given radius is 0.83cm or 0.83 * 10^{-2}m, so the Area would be

A = \pi (0.83*10^{-2})^2

A = 0.0002164m^2

We have then that,

\frac{\Delta m}{\Delta t} = \rho_w Av

\frac{\Delta m}{\Delta t} = (1000)(0.0002164)(5.4)

\frac{\Delta m}{\Delta t} = 1.16856kg/s

Final the power of the pump would be,

P = \frac{\Delta m}{\Delta t}(gh+\frac{1}{2}v^2)

P = (1.16856)((9.8)(3.5)+\frac{1}{2}5.4^2)

P = 57.1192W

Therefore the power of the pump is 57.11W

6 0
3 years ago
An unstrained horizontal spring has a length of 0.31 m and a spring constant of 220 N/m. Two small charged objects are attached
Kruka [31]
The solution you should use is Hooke's law: F=-kx

It should have the same signs because they repel due to the stretch of the spring. 

a. Since there is a constant energy within the spring, then Hooke's law will determine the possible algebraic signs. The solution should be 
<span>F = kx 
270 N/m x 0.38 m = 102.6 N 
</span>
b. Then use Coulomb's law; F=kq1q2/r^2 to find the charges produced in the force. 



8 0
3 years ago
LOTS OF BRAINLIST WILL BE GIVING TO THOSE WHO HELP
Alex_Xolod [135]
You know you can skip those and just submit them, they don’t even check them
4 0
3 years ago
Read 2 more answers
The 'little brain' attached to the rear of the brainstem is called the: (2 points)
11111nata11111 [884]

Answer:

The 'little brain' attached to the rear of the brainstem is called the cerebellum. See the explanation below, please.

Explanation:

The cerebellum is in the back of the brain, it is small in size. It is responsible for functions such as integrations of sensory and motor pathways, balance, coordination of movements, posture.

6 0
3 years ago
Read 2 more answers
A man attempts to pick up his suitcase of weight ws by pulling straight up on the handle. (See Figure 1) However, he is unable t
cupoosta [38]

Answer:

Normal force, N = W_s - F

Explanation:

Let w_s is the weight of suitcase. A man attempts to pick up his suitcase by pulling straight up on the handle. The weight of the suitcase in downward direction. The normal force is acting in upward direction. Let F is the force with which it is pulled straight up.

So, the normal force is given by :

N = W_s - F

N = mg - F

mg is the weight of the suitcase.

Hence, this is the required solution.

3 0
3 years ago
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